Math

Relations An association between the elements of one set and the elements of another set.

The domains leads to the range.

We can rewrite relations as ordered pairs and as a table. Rewrite the relation above as ordered pairs. We use semi-colons and brackets to do this. We first write it like this. { That is the start of an ordered pair. Next, we write the first element in the relation in brackets. Domain goes first, range goes second. { (badminton, shuttlecock); Badminton has another range element, so we create another bracketed pair for badminton. { (badminton, shuttlecock); (badminton, racquet); We keep going until it's complete. Answer: { (badminton, shuttlecock); (badminton, racquet); (hockey, puck); (tennis, racquet); (tennis, ball); (soccer, ball) } Rewrite the above relation as a table. Creating this as a table is exactly like the relation image above except there are no circles.

Domain is the set of all valid inputs. Range is the set of all valid outputs.

Functions A special relation where each element in the domain is associated with exactly one element in the range. A function has a single response every single time. f(g) = function of g, a function that relates vehicles to number of wheels.

An independent variable is always the x-axis on a coordinate plane, domain, or in every case that involves time, "t". A dependent variable is always the y-axis on a coordinate plane and range. If g(x) = 1, what is x? We look at the function and find what equals one. That would be unicycle. x = unicycle If g(x) = 4, what is x? We look at the function and find what equals four. That would be car. x = car You can tell if it's a function or not by looking at the domain's arrow or line moving towards range. If there is more then one line connecting from domain to range, it is a relation, not a function. If there is only one line connecting from domain to range, it is a function. Is this a function?

You look at the arrows from domain to range. Is a domain element connecting more than once? No. So it's a function. Is this a function?

You look at the arrows from domain to range. Is a domain element connecting more than once? No. So it's a function. Is this a function?

You look at the arrows from domain to range. Is a domain element connecting more than once? Yes. So it's not a function.

Function Notation f(x) = 2x + 3 The x variable equals input. If f(x) = 15, what is x? Since f(x) = 2x + 3, then the problem can be 2x + 3 = 15 We take away the 3 on the left side and subtract it from 15. Now it's 2x = 12 Now we divide by 2. Now it's x = 6 Finally, f(6) = 15 If f(x) = 11, what is x? Since f(x) = 2x + 3, then the problem can be 2x + 3 = 11 We take away the 3 on the left side and subtract it from 15. Now it's 2x = 8 Now we divide by 2. Now it's x = 4 Finally, f(4) = 15

Graph of Functions/Vertical Line Test We can tell if a graphed relation is a function or not by using the vertical line test.

We draw lines on the vertical axis, and if the line goes through the line (or wave for this one) once, it's a function. If it were a circle on the graph, the vertical line would go through twice and that would not be a function.

Rate Of Change

What is the rate of change? The rate of change is when the change in dependent variable is divided by the change in independent variable. Since we know the independent variable is the x-axis, we start by finding that change in this graph. We select two random points.

The independent variable change is horizontal, so the change is +100. With the dependent variable change, it is vertical, so the change is +5. Next we create the formula: ROC = DV/INV ROC = 5/100 ROC = 0.05

For the graph above, what is the domain and range?

For domain, since it's the x-axis, we start at how low the domain can go and then how high. Since it's an ongoing line, we use the infinite sign.

Answer for domain: 0 ≤ x ≤ ∞

For range, since it's the y-axis, we start at how low the range can go and then how high. Since it's an ongoing line, we use the infinite sign.

Answer for range: 10 ≤ y ≤ ∞

We use the x and y for variables representing x-axis and y-axis. If there is a variable in the corners of the graph, use those variables instead.

If the rate of change is the same, the function is linear. If the rate of change is not the same, it is not a linear. Is this equation a linear? y = -3x + 25 We make a chart for x and y. We list down from 1 to 5 on the x's side and the value of y when x has value on the other. Find the rate of change afterwards. If it's the same, it's linear.

Graphing Linear Equations/Horizontal and Vertical Intercept

What is the horizontal and vertical intercept? The horizontal intercept is when the line touches the x-axis. In this case, the horizontal intercept is 4. The vertical intercept is when the line touches the y-axis. In this case, the vertical intercept is 6. You can make a point name for both by calling horizontal intercept (4, 0) and vertical intercept (0,6).

What is the horizontal and vertical intercept? The horizontal intercept is when the line touches the x-axis. In this case, the horizontal intercept is 0. The vertical intercept is when the line touches the y-axis. In this case, the vertical intercept is 0. You can make a point name for both by calling horizontal intercept (0, 0) and vertical intercept (0,0). Graph this linear function: f(x) = 4x + 3 We use the formula y = mx + b. m = 4 b = 3 Since b means the y-axis, we graph b first.

Next, we graph m by using the rise over run method. Since 4 is an integer, we have 1 for the run and 4 for the rise.

With run, since it's positive, we go right once and then with rise, since it's positive, we go up 4. We do this until there is no more coordinate plane left.

Another Way of Finding the Slope

With this formula, you can simply find the slope. Find the slope of this graph.

1. Label two points. 2. Use the formula above. 3. Label the top or bottom point as x1 y1 and the other one x2 y2. Here are the three steps but in equation.

Find the slope of this graph.

1. Label two points. 2. Use the formula above. 3. Label the top or bottom point as x1 y1 and the other one x2 y2. Here are the three steps but in equation.

Slope-Point Form Every line equation can also be written y – y1 = m(x – x1) m = slope y2 and x1 = any point Examples: y – 2 = 2/3(x – 1) This has a slope of 2/3. It passes through (1, 2).

y + 7 = 3/2(x + 4) This has a slope of 3/2. It passes through (-4, -7).

If you see y - # or x - #, the # is positive.

If you see y + # or x + #, the # is negative.

Graph y - 2 = 1/3(x + 4) We look at the x value, which is + 4. That would make it -4. We look at the y value, which is - 2. That would make it +2. We make our point (-4, 2) and put it on the graph.

Next we look at the slope, which is 1/3. We run 3 and go up 1. We can keep going if desired.

Graph y + 1 = -1/4(x + 3) We look at the x value, which is + 3. That would make it -3. We look at the y value, which is + 2. That would make it -2. We make our point (-3, -2) and put it on the graph.

Next we look at the slope, which is -1/4. We run 4 and go down 1. We can keep going if desired.

Write an equation in slope-point form for:

1. Find the slope using this formula.

2. After finding the slope, pick one point and rewrite the formula y - y2 = m(x - x1) Answer: y - 1 = 3/4(x - 3) or y + 2 = 3/4(x + 1) Convert y + 2 = 3/4(x + 1) into slope-intercept form. 1. Simplify 3/4(x + 1). It is just like multiplying binomials. Now it's y + 2 = 3/4x + 3/4. 2. Subtract two from both sides. Now it's y = 3/4x + 3/4 - 2. 3. Convert 2 into a fraction. Now it's y = 3/4x + 3/4 - 8/4. 4. Subtract the fraction. Finally: y = 3/4x - 5/4

General form: 8x - 3y - 52 = 0 Slope-intercept form: y = 8/3x - 17 Slope-point form: y + 17 = 8/3(x  ± 0) Standard form: 8x - 3y = 52

Index

Finding a missing angle: using tan-1 using sin-1 using cos-1

Finding a missing side: using tan using sin using cos

Solving a triangle

Solving two triangles

Rules SOH CAH TOA S = sin O = opposite side H = hypotenuse O divided by H C = cos A = adjacent H = hypotenuse A divided by H T = tan O = opposite side A = adjacent O divided by A Rule For When A Value Is Unknown

Finding the Tangent Ratio

Find the tangent ratio for angle D. To find the tangent ratio for angle D, we begin by determining which legs are opposite and adjacent to angle D.

Then, since this is the tangent ratio, we use the following equation..

Replace opposite and adjacent with the length values given in the triangle diagram.

Using Tan-1

Find G. First, we find the tangent ratio of angle G.

Lastly, we put in the calculator: Tan-1 0.8 To get the answer of.. 38.65°

Using Sin-1

Find I. We label the marked sides corresponding to angle I.

We notice with the chart that this is sin, because it is opposite and hypotenuse. (SOH) Next, we find the sine ratio of angle I.

Lastly, we put in the calculator: Sin-1 (12/19) To get the answer of.. 39.2° We could use the decimal, but that would give us an answer one digit short. It's better to stay with fraction form.

Using Cos-1

Find J. We label the marked sides corresponding to angle J.

We notice with the chart that this is cos, because it is adjacent and hypotenuse. (CAH) Next, we find the cosine ratio of angle J.

Lastly, we put in the calculator: Cos-1 (12/19) To get the answer of.. 50.8° We could use the decimal, but that would give us an answer one digit short. It's better to stay with fraction form.

Using Tan

Find x. We will use a given angle measure to find x. A's angle is 15°. We label the marked sides that correspond to angle A.

We notice it's opposite and adjacent. (O and A) That would mean, looking from the chart above, that we should use tan. Our equation begins..

Since the x value is on top, we multiply tan15 with 3.4. We put in our calculator: 3.4 tan 15 Therefore, x = 0.911 miles. If the fraction above (x/3.4) was 3.4/x, you divide 3.4 with tan15. Which looks like this in your calculator: 3.4 / tan 15

Using Sin

Find x. We will use a given angle measure to find x. C's angle is 31°. We label the marked sides that correspond to angle C.

We notice it's opposite and hypotenuse. (O and H) That would mean, looking from the chart above, that we should use sin. Our equation begins..

Since the x value is on top, we multiply sin31 with 12. We put in our calculator: 12 sin 31 Therefore, x = 6.2 If the fraction above (x/12) was 12/x, you divide 12 with sin31. Which looks like this in your calculator: 12 / sin 31

Using Cos

Find x. We will use a given angle measure to find x. X's angle is 43°. We label the marked sides that correspond to angle X.

We notice it's adjacent and hypotenuse. (A and H) That would mean, looking from the chart above, that we should use cos. Our equation begins..

Since the x value is on top, we multiply cos43 with 6. We put in our calculator: 6 cos 43 Therefore, x = 4.4 If the fraction above (x/6) was 6/x, you divide 6 with cos43. Which looks like this in your calculator: 6 / cos 43

Solve:

Solve means to find all angles and lengths of a triangle.

Let us start off with the angles. We know two angles so far, H's 39° angle and G's 90° angle. To solve for J, we subtract 39° from 90° and get 51°.

Now we will do one of the lengths. We labeled the one we will work on first with x.

We know that angle H is 39°. We need to find x.

We label the sides that are marked. x would be named adjacent of H (ADJ) and 9 would be named opposite of H (OPP).

Since we are using ADJ and OPP, we look at the chart and define angle 39° as tan39.

Now the equation is

tan H = OPP/ADJ tan H = 9/x 9/tan39 = x

To find x, we put in the calculator

9 / tan39

Answer for length HG: 11.1

Lastly, we will do HJ's length.

We know that angle J is 51°. We need to find x.

We label the sides that are marked. x would be named hypotenuse and 9 would be named adjacent to J (ADJ).

Since we are using ADJ and HYP, we look at the chart and define angle 51° as cos51.

Now the equation is

cos J = ADJ/HYP cos J = 9/x 9/cos51 = x

To find x, we put in the calculator

9 / cos51

Answer for length HJ: 14.3

We have completely solved the triangle.

Find x (or length YX).

Now we are dealing with two triangles. We cannot find the length of YX with one step, since we only know a degree and no lengths to the smaller triangle.

We start off by finding the length of YW. We use the bigger triangle for help.

We know that angle Z is 20°. We need to find the length of YW.

We label the sides that are marked. YW would be named opposite of Z (OPP) and 8.4 would be named hypotenuse (HYP).

Since we are using OPP and HYP, we look at the chart and define angle 20° as sin20.

Now the equation is

sin Z = OPP/HYP sin Z = x/8.4 (8.4)sin20 = YW

To find YW, we put in the calculator

8.4 sin 20

Answer for YW: 2.9

Now it can finally find the length of YX (x).

We know that angle Y is 22°. We need to find the length of YX.

We label the sides that are marked. YX would be named the hypotenuse (HYP) and 2.9 would be named the adjacent of Y (ADJ).

Since we are using ADJ and HYP, we look at the chart and define angle 22° as cos22.

Now the equation is

cos Y = ADJ/HYP cos Y = 2.9/x (2.9)/cos22 = YX

To find YX, we put in the calculator

2.9 / cos22

Answer for YX (x): 3.1

The "slant" graph of a function is called its slope.

We assign a number to a slope to indicate how steep the slope is and whether the slope is upward or downward.

If the slope is upward, the number is positive. If the slope is downward, the number is negative. If the graph is horizontal, the slope is neither positive nor negative; it is zero. If the graph is vertical, the slope cannot be determined.

State whether the slope of each line is positive, negative, zero, or cannot be determined.

a)

b)

c)

d)

To determine the sign of the slope, follow the graph of the function with your eyes from left to right as though you were reading.

a) From left to right, the graphed line rises, so the slope is positive. b) From left to right, the graphed line does not rise or fall, so the slope is zero. c) From left to right, the graphed line slopes downward, so the slope is negative. d) There is no left to right component of the graphed line, so we cannot determine if the line is rising or falling. The slope is not positive, not negative, and not zero. The slope of the vertical line cannot be determined.

To determine the numerical value of the slope of a line, it is helpful to draw a right triangle using the background grid of the coordinate plane and a portion of the graphed line.

First we look for points where the graphed line crosses intersections of the grid. We have circled some of these points on the graph below.

Next we select two points from the graphed line and, following the background grid, sketch the legs of a right triangle so that the legs intersect the chosen points.

It is helpful to draw the horizontal leg to the right. Then draw the vertical leg starting from that horizontal leg.

We use the words run and rise to describe the two legs of the right triangle.

The run is the length of the horizontal leg, and the rise is the length of the vertical leg.

We assign a positive sign to the rise if it goes up to meet the graphed line and a negative sign if it goes down to meet the graphed line.

In the graph, the run is 5 and the rise is +4. We use these numbers to write the slope of each graphed line.

So the slope of the graphed line is this ratio:

The slope of a line is the ratio of its rise to its run. ("rise over run") Remember to always make the right triangle on the right side of the line.

A positive slope with a line that has equal run and rise has a slope of 1. A negative slope with a line that has equal run and rise has a slope of -1.

Transform this equation so that it is in slope-intercept form. 3x + y = 6

We subtract 3x from both sides of the equal sign.

(3x + y) - 3x = y 6 - 3x = 6 - 3x

Now we have y = 6 - 3x.

We use commutative property to rearrange the terms on the right side of the equal sign so that the x-terms precedes the constant term.

Answer: y = -3x + 6

Write the following equation in slope-intercept form: 2x + y - 3 = 0

We start by adding 3 to both sides.

(2x + y - 3) + 3 = 2x + y 0 + 3 = 3

Now we have 2x + y = 3. Next we subtract 2x from both sides.

(2x + y) - 2x = y 3 - 2x = 3 - 2x

Now we have y = 3 - 2x. We use commutative property because the x-intercept should be in front.

Answer: y = -2x + 3

Write the following equation in slope-intercept form: x + y = 4 - x

We use commutative property for the right side.

4 - x turns into -x + 4

Now we have x + y = -x + 4. We subtract x from both sides.

(x + y) - x = y (-x + 4) - x = -2x + 4

It's -2x because (-x) - (+x) = -2x.

Answer: y = -2x + 4

What is the equation of line L in slope-intercept form?

We first find the rise and run, which is 2.

Then we find what number the line intersects on the y-axis, which is -4.

We use the equation "y = nx + b" and fill in the blank of n and b. The n represents the rise/run, and the b represents the intersecting y-axis number.

So it look like this now:

y = 2x - 4

Answer: y = 2x - 4

What is the equation of line M in slope-intercept form?

We first find the rise and run, which is -1/2.

Then we find what number the line intersects on the y-axis, which is 1.

We use the equation "y = nx + b" and fill in the blank of n and b. The n represents the rise/run, and the b represents the intersecting y-axis number.

So it look like this now:

y = -1/2x + 1

Answer: y = -1/2 + 1

A slope-intercept form should always look like this:

y = nx + b

In this model, the n stands for the slope and the b stands for the y-intercept.

Graph y = -3x + 6 using the slope and y-intercept.

The slope of the graph is coefficient of x, which is -3.

The y-intercept is +6, which is located at +6 on the y-axis.

We start at +6 on the y-axis and move 1 unit to the right and down 3 units because the slope is -3.

This gives us another point on the line.

Using only slope and y-intercept, graph y = x - 2.

The slope is coefficient to x, which is +1.

The y-intercept is -2.

We begin at -2 on the y-axis and sketch a line that has a slope of +1.

Find the area of the shaded sector of this circle.

The central angle of the shaded sector measures 60° (the full circle, 360°, minus the given angle, 300°). Since 60° is 1/6 of a circle (60/360 = 1/6), the area of the sector is 1/6 of the area of the circle.

First we find the area of the circle then divide it by 6.

We multiply radius with radius and get 36. We then multiply 36 with π (3.14) and get 113.04. Lastly, we divide 113.04 by 6 and get 18.84.

Answer: 18.84

An arc is part of the circumference of a circle. In the following figure, we see two arcs between point A and point B.

The arc from A clockwise to B is called a minor arc because it is less than a semicircle.

The arc from A counterclockwise to B is called a major arc because it is greater than a semicircle.

Major arcs are named with three letters. The major arc between point A and point B may be named ACB.

We can measure an arc in°. The number of° in an arc equals the measure of the central angle of the arc. If the minor arc AB in the figure above measures 120°, then the measure of the major arc ACB is 240°, because the sum of the measures of the major arc and minor arc must be 360°.

In this figure, central angle AOC measures 70°. What is the measure of major arc ADC?

We know the measures of AOC, which is 70°. To finish the last measure that finishes the whole circle, we only need to subtract 70 from 360°.

A semicircle is half of a circle. Thus the length of a semicircle is half the circumference of a circle with the same radius. The area enclosed by a semicircle and its diameter is half the area of the full circle.

We will practice finding the lengths of semicircles and the areas they enclose by calculating the perimeters and areas of figures that contain semicircles.

Find the perimeter of this figure.

The perimeter of the semicircle is half the perimeter of a whole circle. This semicircle has a radius of 5, so the diameter must be 10.

We multiply 10 by π (3.14) then divide by 2.

10 x 3.14 = 31.4 divided by 2 = 15.7

We can now label all the dimensions on the figure and add them to find the perimeter.

The circular side is 15.7, the straight side is the diameter (10). Now we add.

15.7 + 10 + 4 + 4 = 33.7

Answer: 33.7

Find the area of this figure.

We divide the figure into two parts. Because of the dashed line already dividing the figure, we don't have to do much but label some dimensions and take away the radius mark.

We first find the area of a1. Here are the steps. Remember the fact that the radius is 5.

We square the radius and get 25. We multiply 25 with 3.14 (π). Then we divide by 2.

5e2 = 25 x 3.14 divided by 2 = 39.25

Now we find the area of a2 by multiplying 10 with 4 and get 40.

Lastly, we add those two areas.

39.25 + 40 = 79.25

A transversal is a line that intersects one or more other lines in a plane. In this tutorial we will pay particular attention to the angles formed when a transversal intersects a pair of parallel lines. Notice the eight angles that are formed.

In this figure there are four acute angles numbered 2, 3, 6, and 7, and there are four obtuse angles numbered 1, 4, 5, and 8. All of the acute angles have the same measures, and all of the obtuse angles have the same measure.

Using the figure above, 2 is 105°. Find 1's measures and 5's.

Each acute angle is a supplement of an obtuse angle, so the measure for 1 is

180 = 105 = 75

Answer for 1: 75°

1 and 5 are corresponding angles, so they both have the same measure.

Answer for 5: 105°

When a transversal intersects a pair of lines, special pairs of angles are formed. In the figure above, 1 and 5 are corresponding angles because the position of 5 corresponds to the position of 1. Other pairs of corresponding angles are 3 and 7; 2 and 6; 4 and 8.

Angle 1 and angle 8 also form a special pair of angles. They are on alternate sides of the transversal and are outside of (not between) the parallel lines. So 1 and 8 are called alternate exterior angles. Angle 2 and angle 7 are another pair of alternate exterior angles.

Angle 4 and 5 are alternate interior angles because they are on alternate sides of the transversal and in the interior of (between) the parallel lines. Angle 3 and 6 are are another pair of alternate interior angles.

Name four pairs of corresponding angles.

Answer: <1 and <5, <3 and <7, <2 and <6, <4 and <8

Name two pairs of alternate exterior angles.

Answer: <2 and <7, <1 and <8

Name two pairs of alternate interior angles.

The longest sides of a right triangle is called the hypotenuse. The other two sides are called legs. Every right triangle has a property that makes right triangles very important in mathematics.

The area of the hypotenuse is the sum of the areas of the legs.

The triangle on the left is a right triangle. On the right we have drawn a square on each side of the triangle.

We have divided the squares into units and can see that their areas are 9, 16, and 25. Notice that the area of the largest square equals the sum of the areas of the other two squares.

25 = 16 + 9

This property of right triangles was known to the Egyptians as early as 2000 BC, but it is named for a Greek mathematician who lived about 550 BC. The Greek's name was Pythagoras, and the property is called Pythagorean theorem. The Greeks are so proud of Pythagoras that they have issued a postage stamp that illustrates the theorem.

Find c.

To find c, we square each leg measure and add them together.

6 x 6 = 36 8 x 8 = 64 36 + 64 = 100

If the answer to c were 100, the measures 6 and 8 would instead be 36 and 64. We are looking for the square root measure, so instead of 100, the answer is the square root.

Answer: 10

Find a.

To find a, we first find out the areas of the leg and hypotenuse.

13 x 13 = 169 12 x 12 = 144

Since 13 represents the hypotenuse, we do not add these products together. We need to find the measure of the other leg, and subtracting 144 from 169 will help solve that.

169 - 144 = 25

If the answer was 25, the measures of the sides would instead be 169 and 144, not 13 and 12. So we are looking for the square root of the answer, which is 5.

Answer: 5

Find the perimeter of this right triangle.

In order to find the perimeter, we need to know all three measures of this triangle. So far we only know two. We use Pythagorean theorem to help find the missing measure.

4 x 4 = 16 3 x 3 = 9 16 + 9 = 25√25 = 5

The missing measure is 5, so we add 5 with 3 and 4.

5 + 3 + 4 = 12

Answer 12

The numbers 5, 12, and 13 are a Pythagorean triplet because 5² + 12² = 13². What are the next three multiples of this Pythagorean triplet?

10, 24, 26 15, 36, 39 20, 48, 52

Notice the first row is multiples of 5 (bold), the second row multiples of 12 (italic) and the third row multiples of 13 (strike).

A roof is being built over a 24-ft-wide room. The slope of the roof is 4 in 12. Calculate the length of the rafters needed for the roof. (Include 2 ft for the rafter tail.)

We consider the rafter to be the hypotenuse of a right triangle.

The width of the room is 24 ft, but a rafter spans only half the width of the room. So the base of the right triangle is 12 ft.

The slope of the roof, 4 in 12, means that for every 12 horizontal units, the roof rises (or falls) 4 vertical units. Thus, since the base of the triangle is 12 ft, its height is 4 ft.

We use the Pythagorean theorem to calculate the hypotenuse by using its formula.

a² + b² = c²                formula (4 ft)² + (12 ft)² = c²     equation 16 ft² + 144 ft² = c²     squared 4 and 12 160 ft² = c²                 added 16 and 144 √160 ft = c                 square root 160 12.65 ≈ c                    squared root of 160 is about 12.65

We can use a calculator to find the square root of 160.

Remember in the question it told us to include 2 ft for the rafter tail, so we add 2 ft to 12.65.

12.65 + 2 ft = 14.65 ft

We convert 0.65 ft into inches by multiplying 0.65 with 12.

0.65 x 12 = 7.8 inches

We round 7.8 inches to the nearest inch, which is 8.

Answer: 14 ft 8 in.

Serena went to a level field to fly a kite. She let out all 200 ft of the string and tied it to a stake. Then she walked out on the field until she was directly under the kite, 150 feet from the stake. About how high was the kite?

We begin by sketching the problem.

The length of the kite (200 ft) is the hypotenuse of a right triangle.

The distance between Serena and the stake is one leg of the triangle.

We use the Pythagorean theorem to find the remaining leg, which is the height of the kite.

a² + b² = c²                        formula a² + (150 ft)² = (200 ft)²        equation a² + 22,500 ft² = 40,000 ft²   squared 150 and 200 a² = 17,500 ft²                     subtracted 22,500 from 40,000 a = √17,500 ft                     square root of 17,500 a = 132 ft                           square root of 17,500 is 132

We can use a calculator to find the square root of 17,500.

To indirectly measure the height of a light pole on her street, Jenna measured the length of the shadow cast by the pole (24 ft) and the length of the shadow cast by a vertical meter stick (40 cm). About how tall was the light pole?

We create a proportion to find out.

100 cm/? = 40 cm/24 ft

We multiply 100 with 24 and get 2400. Then we divide 2400 by 40.

What is the measure of each interior angle of this octagon?

An interior angle is the angles inside a polygon.

Since all angles inside a polygon with 4 or more angles equal 360°, we just divide 360 with octagon's total angles (8).

Answer: 45°

Using the picture of the octagon above, what is the measure of each exterior angle?

An exterior angle is the angles made outside of a polygon.

We know each interior angle in the octagon is 45°, and the angle opposite to each interior angle is supplementary to each other, so the exterior angle of a 45 degree interior angle totals 180°.

If there is one term (example: -2x), the type of expression is called monomial.

If there is two terms (example: a² - 4b²), the type of expression is called binomial.

If there are three terms (example: 3x² - x - 4), the type of expression is called trinomial.

Constant/coefficient term is the number, variable is the letter.

One step to solve equations is to collect like terms. In this equation we can collect variable terms and constant terms as the first step of the solution.

3x - x - 5 + 8 = 17

Combine 3x - x and make 2x. Combine -5 + 8 and make 3.

Simplify and solve this equation: 3x + 5 - x = 17 + x - x

We collect like terms on each side of the equal sign. On the left side, combining 3x and -x gives us 2x. On the right side, combing +x and -x gives us zero.

simplified: 2x + 5 = 17

Now we solve the simplified equation.

equation: 2x + 5 = 17 subtracted 5 from both sides: 2x + 5 - 5 = 2x // 17 - 5 = 12 simplified: 2x = 12 divide 12 by 2.

Answer: 6

Simplify and solve this equation: 5x - 17 = 2x - 5

We see an x-term on both sides of the equal sign. We may remove the x-term from either side. We choose to remove the x-term from the right side. We do this by subtracting 2x from both sides of the equation.

left side: 5x - 17 - 2x = 3x - 17 right side: 2x - 5 - 2x = -5

Now we have:

3x - 17 = -5

Now we solve the equation.

add 17 on the left side: 3x - 17 + 17 = 3x add 17 on the right side: -5 + 17 = 12 now it's: 3x = 12 divide 12 by 3.

Answer: 4

Solve: 3x + 2(x - 4) = 32

We first apply distributive property to clear parentheses.

equation: 3x + 2(x - 4) = 32 distributive property: 3x + 2x - 8 = 32 added 3x and 2x: 5x - 8 = 32 added 8 on both sides: 5x = 40 divide 40 by 5.

Answer: 8

Dividing Terms

We divide terms by removing pairs of factors that equal 1.

Answer: 4a²b

Answer: 4xz²

Answer:

Literal Equations

A literal equation is an equation that contains letters instead of numbers.

Solve for x: x + a = b

We solve for x by isolating x on one side of the equal sign. We do this by taking away a from each side.

x + a - a = b - a x = b - a

Answer: b - a

Another Way

Solving for x is like saying what can we do to change this equation to make the answer x instead of b? Since this is addition, we can use the reverse, which is subtraction.

So it's basically like using the rules for solving number problems but instead there are letters.

Solve for x: ax = b

Solving for x is like saying what can we do to change this equation to make the answer x instead of b? Since this is multiplication, we can use the reverse, which is division.

Remember that b is the bigger number, because a and x multiplied together make b, so b would be the dividend and a the divisor.

Answer: b/a

3x² + 1 = 28

We will subtract +1 from both sides of the equal sign. Now it looks like this.

3x² = 27

Next, we divide both sides by 3. Now it looks like this.

x² = 9

What two numbers multiplied together make 9? That would be 3.

Answer: 3, -3

Notice that there are two solutions, 3 and -3. Both solutions satisfy the equation.

3x² - 1 = 47

We will add +1 from both sides of the equal sign. Now it looks like this.

3x² = 48

Next, we divide both sides by 3. Now it looks like this.

x² = 16

What two numbers multiplied together make 16? That would be 4.

Answer: 4, -4

2x² = 10

We divide both sides by 2. Now it looks like this.

x² = 5

What is the square root of 5? √5, -√5.

Since √5 is an irrational number, we leave it in its radical form. The negative of √5 is -√5 and not √-5.

Factoring Terms

Factor the monomial 12a²b³c.

We factor 12 as (2)(2)(3), and we factor a²b³c as aabbbc.

Answer: (2)(2)(3)aabbbc

Factor 3x²y + 6xy².

First, we lay the factors out.

3 x x y + 2 3 x y y

To find the answer, we find the GCF, greatest common factor. You need to just find a pair on each side that look alike. We have circled the GCF below.

3xy is the GCF for both sides. We divide 3xy out of both sides, leaving us with..

x + 2y

Lastly, we put the GCF in front of it and bracket x + 2y.

Answer: 3xy(x + 2y)

Factor 6a²b + 4ab² + 2ab.

We first lay the factors out.

2 3 a a b + 2 2 a b b + 2 a b

To find the answer, we find the GCF, greatest common factor. You need to just find a pair on each side that look alike. We have circled the GCF below.

2ab is the GCF for each side. We divide 2ab out of each side, leaving us with..

3a + 2b + 1

Notice that the third term is 1, not zero. This is because we divided 2ab by 2ab; we did not subtract.

Remember to put the remaining terms in brackets and have the GCF in front.

Answer: 2ab(3a + 2b +1)

Multiplying Polynomials (Calculus)

There is a single rule for multiplying polynomials. "To multiply polynomials, multiply every term of the first polynomial by every term of the second polynomial." 4d(2 - 3d) 4d · 2 = 8d 4d · -3d = -12d² Put it together. 8d - 12d² x(x - 2) x · x = x² x · -2 = -2x Put it together. x² - 2x 3(x - 2) 3 · x = 3x 3 · -2 = -6 Put it together. 3x - 6 (x + 3)(x - 2) Consider the first parentheses #1 and the second parentheses #2. We start off by multiplying the first term in #1 (x) by the terms in #2 (x, -2). x · x = x² x · -2 = -2x We do the same but with the second term in #1 (+3). 3 · x = 3x 3 · -2 = -6 Now we put it together from the order of when you multiplied them. x² - 2x + 3x - 6 We can simplify this problem if we notice like terms. 1. There are no like terms for x², so we put x² as x². So far: x² 2. There is a like term for -2x, which is +3x, so that would be 1x or x. So far: x² + x 3. There are no like terms for -6, so we put -6 as -6. Finally: x² + x - 6 (x + 4)(x + 1) Consider the first parentheses #1 and the second parentheses #2. We start off by multiplying the first term in #1 (x) by the terms in #2 (x, +1). x · x = x² x · 1 = 1x or x We do the same but with the second term in #1 (+4). 4 · x = 4x 4 · 1 = 4 Now we put it together from the order of when you multiplied them. x² + x + 4x + 4 We can simplify this problem if we notice like terms. 1. There are no like terms for x², so we put x² as x². So far: x² 2. There is a like term for x, which is 4x, so that would be 5x. So far: x² + 5x 3. There are no like terms for 4, so we put 4 as 4. Finally: x² + 5x + 4 (a - 3)(a + 2) Consider the first parentheses #1 and the second parentheses #2. We start off by multiplying the first term in #1 (a) by the terms in #2 (a, +2). a · a = a² a · 2 = 2a We do the same but with the second term in #1 (-3). -3 · a = -3a -3 · 2 = -6 Now we put it together from the order of when you multiplied them. a² + 2a - 3a - 6 We can simplify this problem if we notice like terms. 1. There are no like terms for a², so we put a² as a². So far: a² 2. There is a like term for 2a, which is -3a, so that would be -1a or -a. So far: a² - a 3. There are no like terms for -6, so we put -6 as -6. Finally: a² - a - 6 (x - y)² (x - y) squared would look like (x - y)(x - y), so that is really the problem if you lay it out. Consider the first parentheses #1 and the second parentheses #2. We start off by multiplying the first term in #1 (x) by the terms in #2 (x, -y). x · x = x² x · -y = -xy We do the same but with the second term in #1 (-y). -y · x = -yx (or -xy, it's the same) -y · -y = y² Now we put it together from the order of when you multiplied them. x² - xy - xy + y² We can simplify this problem if we notice like terms. 1. There are no like terms for x², so we put x² as x². So far: x² 2. There is a like term for -xy, which is -xy, so that would be xy². So far: x² + xy² 3. There are no like terms for y², so we put y² as y². Finally: x² + xy² + y² (2h + 5)(h² + 3h - 4) Consider the first parentheses #1 and the second parentheses #2. We start off by multiplying the first term in #1 (2h) by the terms in #2 (h², +3h, -4). 2h · h² = 2h³ 2h · 3h = 6h² 2h · -4 = -8h We do the same but with the second term in #1 (+5). 5 · h² = 5h² 5 · 3h = 15h 5 · -4 = -20 Now we put it together from the order of when you multiplied them. 2h³ + 6h² - 8h + 5h² + 15h - 20 We can simplify this problem if we notice like terms. 1. There are no like terms for 2h³, so we put 2h³ as 2h³. So far: 2h³ 2. There is a like term for 6h², which is 5h², so that would be 11h². So far: 2h² + 11h² (not 11h

because we are adding, not multiplying) 3. There is a like term for -8h, which is 15h so that would be 7h. So far: 2h² + 11h² + 7h 4. There are no like terms for -20, so we put -20 as -20. Finally: 2h² + 11h² + 7h - 20 (-3f² + 3f - 2)(4f² - f - 6) Consider the first parentheses #1 and the second parentheses #2. We start off by multiplying the first term in #1 (-3f²) by the terms in #2 (4f², -f, -6). -3f² · 4f² = -12f

-3f² · -f = 3f³ -3f² · -6 = 18f² We do the same but with the second term in #1 (+3f). 3f · 4f² = 12f³ 3f · -f = -3f² 3f · -6 = -18f We do the same but with the third term in #1 (-2). -2 · 4f² = -8f² -2 · -f = 2f -2 · -6 = 12 Now we put it together from the order of when you multiplied them. -12f

+ 3f³ + 18f² + 12f³ - 3f² - 18f - 8f² + 2f + 12 We can simplify this problem if we notice like terms. 1. There are no like terms for -12f

, so we put -12f

as -12f

. So far: -12f

2. There is a like term for 3f³, which is 12f³, so that would be 15f³. So far: -12f

+ 15f³ 3. There are two like terms for 18f², which is -3f² and -8f² so that would be 7f². (18f² - 3f² = 15f² - 8f² = 7f²) So far: -12f

+ 15f³ + 7f² 4. There is one like term for -18f, which is 2f so that would be -16f. So far: -12f

+ 15f³ + 7f² - 16f 5. There are no like terms for 12, so we put 12 as 12. Finally: -12f

+ 15f³ + 7f² - 16f + 12 (3x - 2y)(x + y) Consider the first parentheses #1 and the second parentheses #2. We start off by multiplying the first term in #1 (+3x) by the terms in #2 (+x, +y). 3x · x = 3x² 3x · y = 3xy (or yx, it's the same) We do the same but with the second term in #1 (-2y). -2y · x = -2xy -2y · y = -2y² Now we put it together from the order of when you multiplied them. 3x² + 3xy - 2xy - 2y² We can simplify this problem if we notice like terms. 1. There are no like terms for 3x², so we put 3x² as 3x². So far: 3x² 2. There is one like term for 3xy, which is -2xy, so that would be 1xy or xy. So far: 3x² + xy 3. There are no like terms for -2y², so we put -2y² as -2y². Finally: 3x² + xy - 2y² (3x + y - 1)(2x - 4) - (3x + 2y)² Let's lay out the squared parentheses. (3x + y - 1)(2x - 4) - (3x + 2y)(3x + 2y) Consider the parentheses as #1A, #2A and #1B, #2B. First, before we subtract, we will find the answer to #1A and #2A. We start off by multiplying the first term in #1A (+3x) by the terms in #2A (+2x, -4). 3x · 2x = 6x² 3x · -4 = -12x We do the same but with the second term in #1A (+y). y · 2x = 2xy y · -4 = -4y We do the same but with the third term in #1A (-1) -1 · 2x = -2x -1 · -4 = 4 Now we put it together from the order of when you multiplied them. 6x² - 12x + 2xy - 4y - 2x + 4 We can simplify this problem if we notice like terms. 1. There are no like terms for 6x², so we put 6x² as 6x². So far: 6x² 2. There is a like term for -12x, which is -2x, so that would be -14x. So far: 6x² - 14x 3. There are no like terms for 2xy, -4y or 4, so we just add those in. Finally: 6x² - 14x + 2xy - 4y + 4 We have finished parentheses #1A and #2A, now it's time for #1B and #2B. We start off by multiplying the first term in #1B (+3x) by the terms in #2B (+3x, +2y). 3x · 3x = 9x² 3x · 2y = 6xy We do the same but with the second term in #1B (2y). 2y · 3x = 6xy 2y · 2y = 4y² Now we put it together from the order of when you multiplied them. 9x² + 6xy + 6xy + 4y² We can simplify this problem if we notice like terms. 1. There are no like terms for 9x², so we put 9x² as 9x². So far: 9x² 2. There is a like term for 6xy, which is 6xy, so that would be 12xy. So far: 9x² + 12xy 3. There are no like terms for 4y², so we put 4y² as 4y². Finally: 9x² + 12xy + 4y² Now we have finished #1B and #2B. We put the final products back with the subtraction sign. From (3x + y - 1)(2x - 4) - (3x + 2y)(3x + 2y) To (6x² - 14x + 2xy - 4y + 4) - (9x² + 12xy + 4y²) Consider the parentheses as #1 and #2. Since this is subtraction, we turn every term in parentheses #2 that is positive to negative and the negatives to positives. So now it's (6x² - 14x + 2xy - 4y + 4) - (-9x² - 12xy - 4y²) Now we simplify by collecting like terms. 1. There is a like term for 6x², which is -9x², so that would be -3x². So far: -3x² 2. There are no like terms for -14x, so we put -14x as -14x. So far: -3x² - 14x 3. There is a like term for 2xy, which is -12xy, so that would be -10xy. So far: -3x² - 14x - 10xy 4. There are no like terms for -4y, 4 or -4y², so we just add those in. Finally: -3x² - 14x - 10xy - 4y + 4 - 4y²

Short Version (for another idea of how to do the last subtraction problem above)

Equation: (2a + 1)(4a - 3) - (a - 2)(a - 2) Multiplied the left parentheses: (8a² - 6a + 4a - 3) - (a - 2)(a - 2) Multiplied the right parentheses: (8a² - 6a + 4a - 3) - (a² - 2a - 2a + 4) Collected like terms on both sides: (8a² - 2a - 3) - (a² - 4a + 4) Converted negs to pos' and vice versa on the right side: (8a² - 2a - 3) - (-a² + 4a - 4) Collected like terms for all: (7a² + 2a - 7)

Factoring Terms (Calculus) There are a few important rules when factoring expressions. ~Always remember to find the GCF. Greatest, not smallest. ~Even the letters have common factors. Factor the following: 6n + 9 We first find the GCF for 6 and 9. That would be 3. Since there's only variables in 6, we can't have a common factor of n. Now it should look like this: 3(6n + 9) We divide by 3. Finally: 3(2n + 3) 6c + 4c² We first find the GCF for 6 and 4. That would be 2. There are variables on both sides (c), so we take the smallest value of them, which is c. Now it should look like this: 2c(6c + 4c2) We divide by 2c. Finally: 2c(3 + 2c) 7x² - 49x We first find the GCF for 7 and 49. That would be 7. There are variables on both sides (x), so we take the smallest value of them, which is x. Now it should look like this: 7x(7x² - 49x) We divide by 7x. Finally: 7x(x - 7) 16a³ + 20a² We first find the GCF for 16 and 20. That would be 4. There are variables on both sides (a), so we take the smallest value of them, which is a². Now it should look like this: 4a²(16a³ + 20a²) We divide by 4a². Finally: 4a²(4a + 5) 5 - 10z - 5z² We first find the GCF for 5, 5 and 10. That would be 5. Since there's only variables in 10 and -5, we can't have a common factor of z. Now it should look like this: 5(5 - 10z - 5z²) We divide 5, 10z, 5² by 5. Finally: 5(1 - 2z - z²) 6 + 12x - 18x² We first find the GCF for 6, 12 and 18. That would be 6. Since there's only variables in 12 and -18, we can't have a common factor of x. Now it should look like this: 6(6 + 12x - 18x²) We divide by 6. Finally: 6(1 + 2x - 3x²) 14c³ + 21c² - 70c We first find the GCF for 14, 21 and 70. That would be 7. There are variables on all sides (c), so we take the smallest value of them, which is c. Now it should look like this: 7c(14c³ + 21c² - 70c) We divide by 7c. Finally: 7c(2c² + 3c - 10) 50d³ - 60d² + 100d We first find the GCF for 50, 60 and 100. That would be 10. There are variables on all sides (d), so we take the smallest value of them, which is d. Now it should look like this: 10d(50d³ - 60d² + 100d) We divide by 10d. Finally: 10d(5d² - 6d + 10) 45a

+ 90a³ - 27a² We first find the GCF for 45, 90, 27. That would be 9. There are variables on all sides (a), so we take the smallest value of them, which is a². Now it should look like this: 9a²(45a

+ 90a³ - 27a²) We divide by 9a². Finally: 9a²(5a³ + 10a - 3) -64b

- 16b

- 32b

We first find the GCF for 64, 16 and 32. That would be 16. There are variables on all sides (b), so we take the smallest value of them, which is b

. Now it should look like this: -16b

(-64b

- 16b

- 32b

) Or it could look like this. The end result for both are correct. 16b

(-64b

- 16b

- 32b

) We divide by 16 or -16. We'll do 16. Finally: 16b

(-4b³ - b - 2) The other answer is -16b

(4b³ + b + 2) -12x³y + 20xy² - 16x²y² We first find the GCF for 12, 20 and 16. That would be 4. There are two variables on all sides (x and y), so we take the smallest value of them, which is x and y. Now it should look like this: 4xy(-12x³y + 20xy² - 16x²y²) We divide by 4xy. Finally: 4xy(-3x² + 5y - 4xy)

Factoring Simple Trinomials (Calculus) Here are some basic trinomials multiplied. (x + 3)(x - 2) = (x² - 2x + 3x - 6) = (x² + x - 6) (x - 1)(x + 5) = (x² + 5x + x + 5) = (x² + 6x + 5)   (x - 1)(x + 4) = (x² + 4x - x - 4) = (x² + 3x - 4) (x - 2)(x - 3) = (x² - 3x - 2x + 6) = (x² - 5x + 6) You notice one thing in common. They all have at least one x² or variable squared. If that is the case, you have a simple trinomial. Which you can solve very fast.

The terms underlined in red are called the B value. The terms underlined in blue are called the C value. The B value and C value go together to find the answer. x² + x - 6 We know it's a simple trinomial because there is only one variable squared. We look at the C value, which is -6. We list down what multiplication equations would equal -6. 1 · -6 -1 · 6 2 · -3 -2 · 3 Now we look at the B value, which is +x (or 1). Out of all of the equations that equal -6, which one would equal the sum of +x (or 1)? That would be -2 · 3. So we make the equation: (x - 2)(x + 3) x² - 5x + 6 We know it's a simple trinomial because there is only one variable squared. We look at the C value, which is +6. We list down what multiplication equations would equal +6. 1 · 6 -1 · -6 2 · 3 -2 · -3 Now we look at the B value, which is -5. Out of all of the equations that equal +6, which one would equal the sum of -5? That would be -2 · -3. So we make the equation: (x - 2)(x - 3) x² + 3x + 2 We know it's a simple trinomial because there is only one variable squared. We look at the C value, which is +2. We list down what multiplication equations would equal +2. 1 · 2 -1 · -2 Now we look at the B value, which is +3. Out of all of the equations that equal +2, which one would equal the sum of +3? That would be 1 · 2. So we make the equation: (x + 1)(x + 2) x² + 7x + 6 We know it's a simple trinomial because there is only one variable squared. We look at the C value, which is +6. We list down what multiplication equations would equal +6. 1 · 6 -1 · -6 2 · 3 -2 · -3 Now we look at the B value, which is +7. Out of all of the equations that equal +6, which one would equal the sum of +7? That would be 1 · 6. So we make the equation: (x + 1)(x + 6) 4t² + 16t - 128 We know it's a simple trinomial because there is only one variable squared. But we notice that there is a number a part of the squared variable, which is 4. So, before we factor the trinomial, we will find the GCF. We first find the GCF for 4, 16 and 128. That would be 4. Since there’s only variables in 4 and 16, we can’t have a common factor of t. Now it should look like this: 4(4t² + 16t - 128) We divide by 4. Finally: 4(t² + 4t - 32) 4 will be staying for the rest of the problem. Now that we simplified it, we ignore 4 out of parentheses and look at the C value, which is -32. We list down what multiplication equations would equal -32. 1 · -32 -1 · 32 2 · -16 -2 · 16 -4 · 8 4 · -8 Now we look at the B value, which is +4. Out of all of the equations that equal -32, which one would equal the sum of +4? That would be -4 · 8. (not 4 · -8 because 4 + -8 = -4) So we make the equation: 4(x - 4)(x + 8) -5h² - 20h + 60 We know it's a simple trinomial because there is only one variable squared. But we notice that there is a number a part of the squared variable, which is -5. So, before we factor the trinomial, we will find the GCF. We first find the GCF for 5, 20 and 60. That would be 5. We could use 5, but it's easier using -5, because then the variable squared turns positive. Since there’s only variables in 5 and 20, we can’t have a common factor of h. Now it should look like this: -5(5h² - 20h + 60) We divide by -5. Finally: -5(h² + 4h - 12) -5 will be staying for the rest of the problem. Now that we simplified it, we ignore -5 out of parentheses and look at the C value, which is -12. We list down what multiplication equations would equal -12. 1 · -12 -1 · 12 2 · -6 -2 · 6 -4 · 3 4 · -3 Now we look at the B value, which is +4. Out of all of the equations that equal -12, which one would equal the sum of +4? That would be -2 · 6. So we make the equation: -5(x - 2)(x + 6)

Factoring Complex Trinomials (Calculus) A complex trinomial is when the squared variable does not equal 1. Simple trinomial: x² + 4x + 5 Complex trinomial: 4x² + 4x + 5 There are two ways of factoring complex trinomials. Decomposition and guessing then checking. Let's do two problems for the hardest, longest one. Decomposition. Factor the following: 7x² + 2x - 5 First, multiply value A and value C. That would be 7 and -5. 7 · -5 = -35 Now we take -35 and make an equation with the B value, 2x. What number would be the product of -35 and also the sum of 2? We list down all of the ways to make the product of -35 and choose the one that will equal 2. 1 · -35 -1 · 35 5 · -7 -5 · 7 We choose -5 · 7 to use. Not 5 · -7, because the sum of that is negative 2, not positive. Next we make the equation. Always put the new terms in with a variable. 7x² - 5x + 7x - 5 It can be in any order, because they are both right in the end, just it may look backward if not in chronological order. We take a look at the first two terms, 7x² and -5x. What is their GCF? 1x. So now it's 1x() We take a look at the second pair of terms, 7x and -5. What is their GCF? 1. So now it's 1x() + 1() So, let's start with 1x. What does 1x need to multiply with to get 7x²? 7x. What does 1x need to multiply with to get -5x? -5. We have finished the first parentheses. 1x(7x - 5) Next, what does 1 need to multiply with to get 7x? 7x. What does 1 need to multiply with to get -5? -5. We have finished the last parentheses. Now it's 1x(7x - 5) + 1(7x - 5) We put the numbers outside parentheses together. (1x + 1)() Then the twin numbers in the parentheses in the second parentheses just once. Finally: (1x + 1)(7x - 5) or (x + 1)(7x - 5) Factor the following: 77x² + 143x + 66 We have to find the GCF first, since all of these constant values have one. Which is 11. Now it's 11(7x² + 13x + 6) The 11 stays for the rest of the problem. First, multiply value A and value C. That would be 7 and 6. 7 · 6 = 42 Now we take 42 and make an equation with the B value, 13x. What number would be the product of 42 and also the sum of 13? We list down all of the ways to make the product of 42 and choose the one that will equal 13. 1 · 42 -1 · -42 -6 · -7 6 · 7 (Not all of the multiplies) We choose 6 · 7 to use. Not -6 · -7, because the sum of that is negative 13, not positive. Next we make the equation. Always put the new terms in with a variable. 7x² + 6x + 7x + 6 It can be in any order, because they are both right in the end, just it may look backward if not in chronological order. We take a look at the first two terms, 7x² and 6x. What is their GCF? 1x. So now it's 1x() We take a look at the second pair of terms, 7x and 6. What is their GCF? 1. So now it's 1x() + 1() So, let's start with 1x. What does 1x need to multiply with to get 7x²? 7x. What does 1x need to multiply with to get 6x? 6. We have finished the first parentheses. 1x(7x + 6) Next, what does 1 need to multiply with to get 7x? 7x. What does 1 need to multiply with to get 6? 6. We have finished the last parentheses. Now it's 1x(7x + 6) + 1(7x + 6) We put the numbers outside parentheses together. (1x and 1) (1x + 1)() Then the twin numbers in the parentheses in the second parentheses just once. (7x and 6) (1x + 1)(7x + 6) or (x + 1)(7x + 6) Remember to add the 11. Finally: 11(x + 1)(7x + 6) Guessing Then Checking This is probably more faster, but the only problem is, you can basically only do this with prime numbers as value C. Factor the following: 3x² + 5x + 2 We look at 3x². What is the first, most common way to get that product? 3x · x. We put it in starting brackets. (3x)(x) Next we look at 2. What is the first, most common way to get that product? 1 · 2 or 2 · 1 Now it's the guessing part. Which one goes first? The 1 or 2? Could it be either (3x + 1)(x + 2) or (3x + 2)(x + 1)? We check by multiplying one of them and seeing if it's the answer above. If not, it's the second choice. Option A: (3x + 1)(x + 2) = 3x² + 6x + 1x + 6 Option B: (3x + 2)(x + 1) = 3x² + 5x + 2 Which is correct? Option B. Factor the following: 2x² + 15x + 7 We look at 2x². What is the first, most common way to get that product? 2x · x. We put it in starting brackets. (2x)(x) Next we look at 7. What is the first, most common way to get that product? 1 · 7 or 7 · 1 Now it's the guessing part. Which one goes first? The 1 or 7? Could it be either (2x + 1)(x + 7) or (2x + 7)(x + 1)? We check by multiplying one of them and seeing if it's the answer above. If not, it's the second choice. Option A: (2x + 1)(x + 7) = 2x² + 14x + 1x + 7 Option B: (2x + 7)(x + 1) = 2x² + 2x + 7x + 7 Which is correct? Option A.

Special Factoring (Calculus) Special factoring is like regular factoring but with two variables. One variable: (2x + 3)(3x - 2) = 6x² + 5x - 6 Two variables: (2x + 3y)(3x - 2y) = 6x² + 5xy - 6y² Factor: 2x² + 7xy + 3y² First, multiply value A and value C. That would be 2 and 3. 2 · 3 = 6 Now we take 6 and make an equation with the B value, 7xy. What number would be the product of 6 and also the sum of 7? We list down all of the ways to make the product of 6 and choose the one that will equal 7. -1 · -6 1 · 6 We choose 1 · 6 to use. Not -1 · -6, because the sum of that is negative 7, not positive. Next we make the equation. Always put the new terms in with a variable. 2x² + 1xy + 6xy + 3y² It can be in any order, because they are both right in the end, just it may look backward if not in chronological order. We take a look at the first two terms, 2x² and 1xy. What is their GCF? 1x. So now it’s 1x() We take a look at the second pair of terms, 6xy and 3y². What is their GCF? 3y. So now it’s 1x() + 3y() So, let’s start with 1x. What does 1x need to multiply with to get 2x²? 2x. What does 1x need to multiply with to get 1xy? y. We have finished the first parentheses. 1x(2x + y) Next, what does 3y need to multiply with to get 6xy? 2x. What does 3y need to multiply with to get 3y²? y. We have finished the last parentheses. Now it’s 1x(2x + y) + 3y(2x + y) We put the numbers outside parentheses together. (1x + 3y)() Then the twin numbers in the parentheses in the second parentheses just once. Finally: (1x + 3y)(2x + y) or (x + 3y)(2x + y)

Difference of Squares (Calculus) (2x + 5)(2x - 5) Notice how the bolded terms are square numbers. Their product is 4x². Same with the italic terms. Their product is -25. 4 and 25 are both square numbers. Factor: x² - 25 What is the square root of x²? x and x. Now it's (x)(x) What is the square root of -25? +5 and -5. Finally it's (x + 5)(x - 5) Factor: x² - 81 What is the square root of x²? x and x. Now it's (x)(x) What is the square root of -81? +9 and -9. Finally it's (x + 9)(x - 9) Factor 4x² - 25y² What is the square root of 4x²? 2x and 2x. Now it's (2x)(2x) What is the square root of -25y²? +5y and -5y. Finally it's (2x + 5y)(2x - 5y) Factor: x

- y

What is the square root of x

? x

and x

. Now it's (x

)(x

) What is the square root of -y

? +y³ and -y³. Finally it's (x

+ y³)(x

- y³) Factor: 81x

- 16y

What is the square root of 81x

? 9x² and 9x². Now it's (9x²)(9x²) What is the square root of -16y

What is the area of a circle if the radius is 10 cm? Use 3.14 for π.

We first multiply the radius by the radius.

10 x 10 = 100

Then we multiply the area of that corner in the circle with π.

100 x 3.14 = 314 cm²

Answer: 314 cm²

What is the area of a circle if the radius is 7 inches? Use 22/7 for π.

We first multiply the radius with the radius.

7 x 7 = 49

Then we multiply 49 with 22.

49 x 22 = 1078

Then we divide 1078 by 7.

1078 divided by 7 = 154

Answer: 154 in.²

What is the area of a circle if the diameter is 12 ft? Leave π as π.

Since the diameter is 12 ft, we know the radius is 6 ft. We multiply the radius with the radius.

6 x 6 = 36

It says to leave π as π, so we do nothing to change 36 but add π at the end.

Recall that two figures are congruent if they are the same shape and size. Two triangles can be congruent, but they don't have to be in the same position.

We can use three types of position change to move a shape.

We can flip a shape over, we can slide a shape to the right or left or up or down, and we can turn a shape clockwise or counterclockwise at any angle.

These terms "flip, slide, and turn" are called transformations.

We have used <, >, =, ≠, and ≈ for comparing numbers.

But there is another sign we should learn, which is ≤ and ≥.

≥ is read "greater than or equal to" and

≤ is read "lesser than or equal to".

Graph on a number line: x ≤ 4

We start by making a number line and labeling the main integer 4 with a dot.

Instead of making a segment from 4 to -2, we make a ray, because we are labeling all the numbers lesser or equal to 4, which is endless beginning at four and going down, so the ray will tell us the numbers keep going even if they aren't shown on the number line.

Graph on a number line: x > 4

We are told to graph numbers greater than four, but it doesn't say to include four itself. And we don't start at 5 on the number line either. We still need to count for the in between values before four such as 4 1/2 or 4.001. So, we still put a dot on 4 but we make it an empty dot.

Remember the line should be a ray, because there is no end for how many numbers are greater than four.

2x - 5 ≥ 1

Since term 2x is positive, we add the second term (5) to 2x - 5 and 1.

2x - 5 + 5 = 2x 1 + 5 = 6

Now we have 2x ≥ 6.

We can simplify by dividing them both by 2.

2x divided by 2 = x 6 divided by 2 = 3

Simplify.

A complex fraction is a fraction that contains one or more fractions in the numerator or denominator.

To simplify this complex fraction, we take the denominator (2/3) and turn it into its reciprocal (3/2), then multiply it with the numerator (3/5).

3/5 x 3/2 = 9/10

Answer: 9/10

You notice complex fractions are like another way of dividing fractions. We usually use a division sign, but complex fractions are dividing fractions or mixed numbers with a division bar instead of a division sign.

Simplify.

To simplify this complex fraction, we take the denominator (100) and turn it into its reciprocal (1/100), then make the numerator (25 2/3) into an improper fraction (77/3) and multiply them.

77/3 x 1/100 = 77/300

Answer: 77/300

Simplify.

To simplify this complex fraction, we take the denominator (7 1/3) and turn it into an improper fraction (22/3) and make it into it's reciprocal (3/22), then make the numerator (15) into a fraction (15/1) and multiply them.

15/1 x 3/22 = 45/22 divided by 22 = 2 1/22

Answer: 2 1/22

Make 83 1/3 percent into a fraction and simplify.

A percent is a fraction that has the denominator of 100. So, we turn 83 1/3 percent into

To simplify this complex fraction, we take the denominator (100) and turn it into its reciprocal (1/100) then make the numerator (83 1/3) into an improper fraction (250/3) and multiply them.

250/3 x 1/100 = 250/300

Reduced: 5/6

What is the area of this trapezoid?

To find the area, we divide the trapezoid in half, making two triangles. Remember the height is 6, so that will help find the area for both triangles.

We start by finding the area of the bigger triangle.

Area of bigger triangle: 6 x 10 / 2 = 30

Area of smaller triangle: 6 x 7 / 2 = 21

What is the area of this figure?

When we find the area of figures with right angles by dividing it into smaller rectangles and calculate the area, we are doing the same thing with complex figures, except instead of only rectangles, we also divide into triangles.

The smartest and easiest way to divide this figure is separating the obvious looking rectangle and triangle. Then labeling the dimension of the dashed line.

Then we add the area of the rectangle and the area of the triangle together to find the total area of this figure.

Area of rectangle: 7 x 10 = 70

Area of triangle: 6 x 9 / 2 = 27

Total area: 97

What is the area of this figure?

We start by dividing this figure into polygons. And the left over spaces should look like polygons too. If not, keep dividing until you see polygons for every region.

You can divide this figure into many ways, but try and find the easiest and most obvious.

Area of bigger rectangle: 10 x 6 = 60

Area of smaller rectangle: 3 x 4 = 12

Area of triangle: 7 x 4 / 2 = 14

Total area: 86

A square sheet of paper with an area 81 inches has a corner cut off, forming a pentagon as shown above.

What is the perimeter?

A sheet with an area of 81 inches could only mean that the measurements on each side were 9 inches. We create the cut off corner into a triangle and label the sides.

Then we count the sides.

9 + 9 + 6 + 5 + 3 + 4 = 34 inches

Answer: 34 inches

What is the area of the pentagon above?

We knew before it was 81 inches in area, but since a corner was cut off, we have to subtract from the original area of 81 inches. We first find the area of the triangle.

3 x 4 / 2 = 6

So we subtract 6 from 81 and get 75 inches.

Three fifths of the fish in the pond are bluegills. If there are 45 bluegills in the pond, how many fish are in the pond?

Divide 45 by the numerator of 3/5, which is 3.

45 divided by 3 = 15

Multiply 15 with the denominator of 3/5, which is 5.

15 x 5 = 75

Answer: 75

When Juan finished page 51, he was 3/8 of the way through his book. His book had how many pages?

Divide 51 by the numerator of 3/8, which is 3.

51 divided by 3 = 17

Multiply 17 with the denominator of 3/8, which is 8.

17 x 8 = 136

Answer: 136

As Sakura went from room to room she found that 3/5 of the lights were on and that 30 lights were off. How many lights were on?

3/5 were on, making 2/5 of the lights off.

We already know 30 lights are off, so we divide 30 by the numerator of its fraction 2/5, which is 2.

30 divided by 2 = 15

Then we find out how many lights were on by multiplying 15 by the numerator of 3/5, the fraction stating how many lights are on.

15 x 3 = 45