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Sample SAT Math Practice Test 1

Hard SAT Math Questions

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An equation of a circle in the xy-plane with radius 2 and center at (5, −4) is x² + y² + ax + by = c. What is the value of c ?

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Hint: The equation of a circle in standard form is (x − h)² + (y − k)² = r².

Explanation

Correct Answer: -37

The equation of a circle in standard form is (x − h)² + (y − k)² = r², where (h, k) is the center and r is the length of the radius.

The given circle has radius 2 and center at (5, −4), so plug in r = 2 and (h, k) = (5, −4) to get an equation of the circle in standard form. Then rewrite the equation in the form x² + y² + ax + by = c to identify the value of c.

(x − h)² + (y − k)² = r²	Standard form
(x − 5)² + (y − (−4))² = 2²	Plug in (h, k) = (5, −4) and r = 2
(x − 5)² + (y + 4)² = 4	Simplify

Now expand the perfect squares (x − 5)² and (y + 4)² and then rearrange the equation to the form x² + y² + ax + by = c.

(x − 5)² + (y + 4)² = 4
x² − 10x + 25 + y² + 8y + 16 = 4	Expand (x − 5)² and (y + 4)²
x² + y² − 10x + 8y + 41 = 4	Rearrange and combine constant terms on left
x² + y² − 10x + 8y = −37	Subtract 41 from both sides

Compare the resulting equation to the given equation to identify c.

Things to remember:

The equation of a circle in standard form is (x − h)² + (y − k)² = r², where (h, k) is the center and r is the length of the radius.
To expand (a + b)², use the identity (a + b)² = a² + 2ab + b² or rewrite the expression (a + b)² as (a + b)(a + b) and multiply.

In a warehouse, there are 7 shelves of cardboard boxes and 8 shelves of plastic boxes. Each shelf of cardboard boxes has 12 boxes that weigh 25 pounds or more and 8 boxes that weigh less than 25 pounds. Each shelf of plastic boxes has 11 boxes that weigh 25 pounds or more and 9 boxes that weigh less than 25 pounds. A box from one of these shelves will be selected at random. What is the probability of selecting a plastic box, given that the box is 25 pounds or heavier?

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Hint : To find the probability of a desired event, use the following formula:

P = number of desired outcomes / total number of possible outcomes

Explanation

Correct Answer: 22/43

The question asks for the probability of selecting a plastic box, given that the box is 25 pounds or heavier.

To find the probability of a desired event, use the following formula:

P = number of desired outcomes / total number of possible outcomes

The total number of possible outcomes is the total number of boxes that are 25 pounds or heavier.
The number of desired outcomes is the number of those boxes that are plastic.

The total boxes that weigh 25 pounds or more is the sum of the numbers of plastic boxes and cardboard boxes that weigh 25 pounds or more.

It is given that there are 8 shelves with 11 plastic boxes on each shelf that weigh 25 pounds or more and that there are 7 shelves with 12 cardboard boxes on each shelf that weigh 25 pounds or more.

Multiply each number of shelves by the number of boxes on each shelf to see that there are 8 ∙ 11 = 88 plastic boxes and there are 7 ∙ 12 = 84 cardboard boxes that weigh 25 pounds or more.

P = 88 / 88 + 84

Now simplify the probability.

$P = \frac{88}{88 + 84}$
$P = \frac{88}{172}$	Add in denominator
$P = \frac{22}{43}$	Simplify 88 / 172 = 4 . 22 / 4 . 43

The probability of selecting a plastic box, given that the box is 25 pounds or heavier, is 22 / 43.

Things to remember:
To find the probability of a desired event, use the following formula:

P = number of desired outcomes / total number of possible outcomes

The table shows three values of x and their corresponding values of y, where y = f(x) + 6 and f is a quadratic function. What is the y-coordinate of the y-intercept of the graph of y = f(x) in the xy-plane?

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Hint: For any function f(x), the graph of y = f(x − h) + k is the graph of y = f(x) transformed by a horizontal shift of h units and a vertical shift of k units.

The equation of a quadratic function in vertex form is y = a(x − h)² + k, where (h, k) is the vertex of the parabola and a is the scale factor.

Explanation

Correct Answer: 1075

The equation y = f(x) + 6 is in terms of f(x), so it is the result of a transformation to the function f(x). The constant 6 is added to the output f(x), so it represents a vertical shift 6 units up.

Each row of the table represents an (x, y) point on the graph of y = f(x) + 6. The question asks for the y-intercept of f(x), so apply the transformation to the given table to get a table of values for y = f(x).

Every point on the graph of y = f(x) + 6 is 6 units above a point on the graph of y = f(x), so subtract 6 from each y-value to get a table that represents points on the graph of y = f(x).

It is given that f(x) is a quadratic function, so its graph is a parabola. A vertical axis of symmetry passes through the vertex and is located halfway between any two points on a parabola that have the same y-value.

Notice that two points from the table have the same y-value (4). Therefore, the point with an x-value halfway between 17 and 21 is the vertex (h, k). This point is in the table (19, −8).

Plug (h, k) = (19, −8) into the vertex form equation y = a(x − h)² + k to see that an equation for the graph of f(x) is y = a(x − 19)² − 8. Plug in a different point from the table to solve for a. Choose (x, y) = (17, 4).

y = a(x − 19)² − 8
4 = a(17 − 19)² − 8	Plug in x = 17 and y = 4
4 = a(−2)² − 8	Subtract inside parentheses
4 = 4a − 8	Apply the exponent
12 = 4a	Add 8 to both sides
3 = a	Divide by 4 on both sides

Plug a = 3 into y = a(x − 19)² − 8 to see that an equation for f(x) is y = 3(x − 19)² − 8. To find the y-coordinate of the y-intercept, plug in x = 0 and solve for y.

y = 3(x − 19)² − 8
y= 3(0 − 19)² − 8	Plug in x = 0
y = 3(−19)² − 8	Subtract inside parentheses
y = 3(361) − 8	Apply the exponent
y = 1,075	Simplify

Note: It is also possible to use the given table of values to see that the equation of the transformed function is y = 3(x − 19)² − 2 and has y-intercept 1,081. Then subtract 6 to get the y-coordinate of the y-intercept of f(x).

Things to remember:

For any function f(x), the graph of y = f(x − h) + k is the graph of y = f(x) transformed by a horizontal shift of h units and a vertical shift of k units.
A vertical axis of symmetry passes through the vertex (h, k) of a parabola and is located halfway between any two points on a parabola that have the same y-value.
The equation of a quadratic function in vertex form is y = a(x − h)² + k, where (h, k) is the vertex of the parabola and a is the scale factor.

$\frac{13}{20} y - \frac{11}{20} x = \frac{10}{13} - \frac{13}{20} y$

$\frac{11}{10} x + \frac{13}{10} = r y + \frac{33}{10}$

In the system of equations shown, r is a constant. If the system has no solution, what is the value of r ?

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Hint : A system of linear equations has no solution if the same expression of x and y is equal to two different numbers.

Explanation

Correct Answer: 26/10

It is given that the system has no solution, so its graph must consist of lines that do not intersect. Therefore, the lines must be parallel. Parallel lines have the same slope but different y-intercepts.

Rewrite the given system of linear equations in slope-intercept form y = mx + b, where m is the slope and b is the y-intercept. Isolate the y-term in the first equation.

Rewrite the second equation to align like terms, then compare the two equations to identify the slopes and y-intercepts.

Set the slopes equal and solve for r to find the value of r that makes the slopes the same.

11 / 26 = 11 / 10r	Set slopes equal
( 11 / 26 )r = ( 11 / 10r )r	Multiply both sides by r
26 / 11 ( 11 / 26 )r = ( 11 / 10 ) 26 / 11	Multiply both sides by 26 / 11 , the reciprocal of 11 / 26
r = 26 / 10	Simplify

For the y-intercepts to be different and the slopes to be the same, the value of r must be 26 / 10 .

Note: It is possible to plug r = 26 / 10 into the second equation to verify that the y-intercepts are different (graph).

Things to remember:

A system of linear equations has no solution if its graph consists of two parallel lines because parallel lines never intersect.
Parallel lines have the same slope but different y-intercepts.

Alternate Method :

A system has no solution if the same expression of x and y is equal to two different numbers.

Rewrite the given system of linear equations in standard form so that the constant terms are isolated.

Now each equation is written in standard form. Notice that $- \frac{11}{20} x$ is a multiple of $- \frac{11}{10} x$ . Multiply the first equation by 2 so that the x-terms are the same ( $(- \frac{11}{10} x)$ ) and $- \frac{11}{10} x$ .

The x-terms are the same ( $(- \frac{11}{10} x)$ and $- \frac{11}{10} x$ ) and the constant terms are different ( $(\frac{20}{13} \neq - 2)$ ), so identify the value of r that makes the y-terms the same.

Therefore, the value of r must be $\frac{26}{10}$ for the system to have no solution.

Things to remember:
A system has no solution if the same expression of x and y is equal to two different numbers.

In triangle LMN, the measure of angle L is 30°, the length of LM is 10 units and the length of LN is 8 units. What is the area, in square units, of triangle LMN?

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Hint : The height of a triangle is the perpendicular distance from the base of the triangle to the opposite vertex.

Explanation

Correct Answer: 20

The formula for the area of a triangle is $A = \frac{1}{2} bh$ , where b is the length of the base and h is the height.

Draw triangle LMN and label it with the given information. Let LM with a given length of 10 be the base of triangle and notice the height of the triangle is unknown.

The height of a triangle is the perpendicular distance from the base of the triangle to the opposite vertex.

Draw a line segment from vertex N down and perpendicular to base LM . Label the new point of intersection P. The length of NP is the height of triangle LMN.

Triangle LNP is a right triangle with a given angle of 30°, so it is a 30°-60°-90° triangle.

In a 30°-60°-90° triangle, the hypotenuse is twice the length of the shorter leg. Compare the side lengths of the general triangle to triangle LNP to see that hypotenuse LN must be twice the height NP.

The length of the hypotenuse is 8, so set 2x equal to 8 and divide by 2 to find the length of the shorter leg x.

2x = 8	Length of hypotenuse of special right triangle
x = 4	Divide by 2 on both sides

The length of the shorter leg is 4, so the height of triangle LMN is 4. Plug the length of the base (10) and the height (4) of into the formula for the area of a triangle and simplify.

The area of triangle LMN, in square units, is 20.

Things to remember:

The formula for the area of a triangle is $A = \frac{1}{2} bh$ , where b is the length of the base and h is the height.
A 30°-60°-90° triangle has the following ratio of side lengths:

The height of a triangle is the perpendicular distance from the base of the triangle to the opposite vertex.

Sample SAT Math Practice Test 2

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Sample SAT Math Practice Test 4

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SAT Math Grid-In Questions

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y = −x² + 4x + p

y = −6

In the system of equations shown, p is a constant. For which value of p does the system have exactly one distinct real solution?

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Hint: A solution to a system of equations is a point of intersection (x, y) in the xy-plane. In the given system, y = −6 represents a horizontal line and y = −x² + 4x + p represents a parabola.

Explanation

Correct Answer: -10

A solution to a system of equations is a point of intersection (x, y) in the xy-plane. In the given system, y = −6 represents a horizontal line and y = −x² + 4x + p represents a parabola.

A parabola intersects a horizontal line at 0, 1, or 2 points. If a parabola intersects a horizontal line at a single point, then that point must be the vertex of the parabola (proof).

To find the value of p, first find the coordinates of the vertex of the parabola. Use the vertex formula for an equation in standard form y = ax² + bx + c to find the x-coordinate of the vertex.

Compare the equation of the parabola y = −x² + 4x + p to standard form y = ax² + bx + c to see that a = −1 and b = 4. Plug these values into the vertex formula and simplify.

$x = - \frac{b}{2 a}$	x-coordinate of vertex
$x = - \frac{4}{2 (- 1)}$	Plug in a = −1 and b = 4
$x = 2$	Simplify

The x-coordinate of the vertex is 2. The parabola intersects the line y = −6 when the y-coordinate of the vertex is equal to −6.

Therefore, the given system of equations has exactly one distinct real solution when the vertex of the parabola is (2, −6). Plug (x, y) = (2, −6) into the equation y = −x² + 4x + p and solve for p.

y = −x² + 4x + p	Given equation
−6 = −(2)² + 4(2) + p	Plug in (x, y) = (2, −6)
−6 = −4 + 8 + p	Apply exponent and multiply
−6 = 4 + p	Combine constant terms
−10 = p	Subtract 4 from both sides

Things to remember:

A solution to a system of equations is a point where the graphs of the equations intersect in the xy-plane.
If a horizontal line intersects a parabola at only one point, it must intersect at the vertex of the parabola.
The vertex formula defines the x-coordinate of the vertex of a parabola with a quadratic equation in standard form y = ax² + bx + c:

Alternate Method :

It is also possible to use the substitution method to create a single quadratic equation in terms of x, then use the discriminant to determine what value of p results in one distinct real solution.

It is given that y = −6, so substitute −6 for y in the first equation y = −x² + 4x + p and rewrite it in standard form y = ax² + bx + c.

y = −x² + 4x + p	Given first equation
−6 = −x² + 4x + p	Plug in y = −6
0 = −x² + 4x + p + 6	Add 6 on both sides

A quadratic equation in standard form 0 = ax² + bx + c has exactly one solution when the discriminant b² − 4ac is equal to 0. The discriminant is the expression under the square root in the quadratic formula.

Compare the equation 0 = −x² + 4x + p + 6 to standard form to identify the values of a, b, and c.
Note: It is given that p is a constant, so p + 6 is also a constant.

Plug these values into the discriminant b² − 4ac and simplify.

b² − 4ac	Discriminant
4² − 4(−1)(p + 6)	Plug in a = −1, b = 4, and c = p + 6
16 + 4(p + 6)	Apply the exponent and simplify
16 + 4p + 24	Distribute 4 to (p + 6)
40 + 4p	Combine like terms

The discriminant is 40 + 4p, so the system has exactly one solution when 40 + 4p is equal to 0. Set the discriminant equal to 0 and solve for p.

40 + 4p = 0	Set discriminant equal to 0
4p = −40	Subtract 40 from both sides
p = −10	Divide by 4 on both sides

Things to remember:

To find the number of solutions to a linear and quadratic system, use the substitution method to create a single equation in terms of only x.
A quadratic equation in standard form has exactly one solution when the discriminant b² − 4ac is equal to 0.

The circle shown has center O, diameters AD , BE , and CF , and the length of BO is 9. If the length of minor arc $\overset{⏜}{AB}$ is $kπ$ , what is the value of $k$ ?

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Hint : The length of an arc divided by the circumference of the circle must equal the measure of its intercepting central angle divided by the measure of the entire circle.

Explanation

Correct Answer: 2

Ratios of the part to the whole are equivalent between the length of a circular arc and the measure of its intercepting central angle.

The length of an arc (part) divided by the circumference of the entire circle (whole) must equal the measure of its intercepting central angle (part) divided by the measure of the entire circle (whole), forming the following proportion:

The formula for the circumference of a circle is $C = 2 πr$ , where r is the length of the radius. The angle measure on the given figure is in degrees, so use 360° for the measure of a whole circle (instead of $2 π$ radians).

The central angle that intercepts $\overset{⏜}{AB}$ is ∠AOB , so write the proportion as follows:

To find the measure of $∠ AOB$ , first identify that $∠ BOC$ and $∠ EOF$ are a pair of vertical angles. When two lines intersect, the vertical angles formed by those lines are congruent.

Therefore, $∠ BOC$ also has a measure of 50°. Now notice that $∠ BOC$ forms a straight line $(\bar{AD})$ when combined with $∠ AOB$ and $∠ COD$ .

Angles that combine to form a straight line have measures that sum to 180°. Set the sum of $m ∠ AOB$ , 50, and 90 equal to 180, and then solve for $m ∠ AOB$ .

The measure of $∠ AOB$ is 40°. Notice that BO is a radius of the circle and has a given length of 9. Plug r = 9 and $m ∠ AOB = 40$ into the proportion, and then solve for the length of $\overset{⏜}{AB}$ .

The length of $\overset{⏜}{AB}$ is $2 π$ , so the value of k is 2.

Things to remember:

Ratios of the part to the whole are equivalent between the area of a circular sector and the measure of its intercepting central angle. The following proportion relates circular sectors and central angles of a circle:

The formula for the circumference of a circle is $C = 2 πr$ , where r is the length of the radius.

Russell's toy box contains 15 solid-colored blocks: 6 red blocks, 4 blue blocks, and 5 purple blocks. There is also a single number printed on each block. Each number from 1 through 6 is printed on a red block, each number from 1 through 4 is printed on a blue block, and each number from 1 through 5 is printed on a purple block. If Russell chooses a block at random from his toy box, what is the probability that the block will be purple or will have a 3 printed on it?

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Hint: To find the probability of a desired event, use the following formula:

Explanation

Correct Answer: 7/15

An item with one quality OR another quality has at least one of the qualities. To find the probability of a desired event, divide the number of desired outcomes by the total number of possible outcomes.

The number of desired outcomes is the number of blocks that are purple OR printed with a 3.
The number of possible outcomes is the total number of blocks in the toy box.

It is given that there are 15 total blocks in the toy box: 6 red blocks printed with 1 through 6, 4 blue blocks printed with 1 through 4, and 5 purple blocks printed with 1 through 5.

There are 5 purple blocks, and there are 3 blocks printed with a 3. Add 3 and 5, then subtract 1 for the 1 block that is both purple and printed with a 3 to see that there 7 blocks that are purple or printed with a 3.

Note: There is 1 block that is both purple and printed with a 3, so it is necessary to subtract 1 from the sum of the 5 purple blocks and 3 blocks printed with a 3 to avoid double-counting.

Of the 15 total blocks, there are 7 blocks that are purple or printed with a 3.

$P = \frac{purple or with a 3}{total blocks}$	Probability of purple or with a 3
$P = \frac{7}{15}$	Plug in values

The probability that Russell chooses a block that is purple or printed with a 3 is $\frac{7}{15}$ .

Things to remember:

To find the probability of a desired event, use the following formula:

An item that has one quality OR another quality has at least one of the qualities (possibly both).

Fifteen tropical farms with equal land areas grow a combination of coconuts and pineapples. The number of coconuts and pineapples grown by each farm is shown in the scatterplot. How many farms grew at least 4 times as many coconuts as pineapples?

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Hint : The question asks for the number of farms that grew at least 4 times as many coconuts as pineapples, so the number of coconuts must be greater than or equal to 4 times the number of pineapples.

Explanation

Correct Answer: 6

On the given scatterplot, the number of coconuts is given on the y-axis (vertical) and the number of pineapples is given on the x-axis (horizontal).

The question asks for the number of farms that grew at least 4 times as many coconuts as pineapples, so the number of coconuts (y) must be greater than or equal to (≥) 4 times the number of pineapples (x).

Therefore, the inequality y ≥ 4x represents the given relationship. To determine which points on the scatterplot satisfy the inequality, consider the graph of the boundary line y = 4x (see how).

The inequality y ≥ 4x means that any point (x, y) on or above the line y = 4x represents a farm that grew at least 4 times as many coconuts as pineapples. Count the number of points on or above the line.

There were 6 farms that grew at least 4 times as many coconuts as pineapples.

Things to remember:
To find points on a scatterplot that satisfy a relationship between two variables x and y, consider the graph of the equation or inequality on the scatterplot and identify which points satisfy the relationship.

The population of a certain town decreased by 4% from 2005 to 2015. If the population in 2015 is k times the population in 2005, what is the value of k ?

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Hint : To decrease a value by p%, multiply it by ( $1 - \frac{p}{100}$ )

Explanation

Correct Answer: 0.96

The question states that the population in 2015 is k times the population in 2005.

It is also given that the population decreased by 4% from 2005 to 2015. To decrease a value by p%, multiply it by ( $1 - \frac{p}{100}$ ). Therefore, the 2015 population is equal to ( $1 - \frac{4}{100}$ ) multiplied by the 2005 population.

Therefore, k must equal $1 - \frac{4}{100}$ . Simplify $1 - \frac{4}{100}$ to write the value of k as a decimal.

$1 - \frac{4}{100}$	Value of k
$1 - 0.04$	Divide
$0.96$	Subtract

The value of k is 0.96.

Note: It is also possible to rewrite $1 - \frac{4}{100}$ as a fraction and simplify the result.

Things to remember:

To increase a value by p%, multiply it by ( $1 + \frac{p}{100}$ ).
To decrease a value by p%, multiply it by ( $1 - \frac{p}{100}$ ).

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Frequently Asked Questions (FAQs)

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Yes. Each SAT math problem features comprehensive explanations so you understand the reasoning behind correct and incorrect answers and avoid repeating the same mistakes on future questions.

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Yes. Our SAT math prep questions mirror the current SAT Math section format: multiple-choice and student-produced response questions that reflect the official SAT math exam structure. Students can use the full-length practice exam in our SAT Prep Course to match the 2 modules format or create custom ones focused on key topics like the sample SAT math practice tests we’ve included above.

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How should I time myself during SAT math practice?

On the real SAT math exam, you’ll have 70 minutes for the Math section (35 minutes per module). Practice SAT math problems under these conditions or with smaller timed sets to build pacing and accuracy.

What concepts are covered on the SAT Math section?

You’ll encounter SAT math questions testing algebra, problem-solving, data analysis, geometry, trigonometry, and math in real-world contexts. Both calculator and no-calculator SAT math problems are required.

How many questions are on the SAT Math section?

There are 44 SAT math questions total, split across 2 modules. You’ll answer multiple-choice and student-produced response SAT math practice questions that mirror sample SAT math questions from official tests.

What's a good score on the SAT Math section?

SAT math test scores range from 200 to 800. Competitive colleges often look for 650–780 in Math, but your target should align with the requirements of your chosen schools. Regular SAT math practice with challenging SAT math practice problems helps achieve higher scores.

(a − b)(a² − b²)	Given expression
(a − b)(a + b)(a − b)	Factor a² − b²

(a − b)(a + b)(a − b)
(3)(4)(3)	Substitute (a − b) = 3 and (a + b) = 4
36	Multiply

x² − 9 / x − 3 = −1	Given equation
(x² − 9)(x − 3) / x − 3 = −1(x − 3)	Multiply both sides by (x − 3)
(x² − 9)(x − 3) / x − 3 = −1(x − 3)	Cancel factors to clear fraction
x² − 9 = −1(x − 3)	Simplify

x² − 9 = −1(x − 3)
x² − 9 = −x + 3	Distribute −1 to (x − 3)
x² + x − 12 = 0	Add x and subtract 3 on both sides
(x + 4)(x −3) = 0	Factor

x + 4 = 0	Set each factor equal to 0	x − 3 = 0
x = −4	Solve for x	x = 3

$\frac{sum of points scored}{number of games}$	Mean number of points
$\frac{14 + 9 + 23 + 22}{4}$	Plug in values
$\frac{68}{4}$	Add in numerator
$17$	Divide

(16s)²
16²s²	Use the distributive property of exponents
256s²	Simplify

$\frac{k}{1}$ = ( $\frac{16}{1}$ )²
$\frac{k}{1} = \frac{256}{1}$	Apply the exponent
k = 256	Simplify

$\frac{7 π}{12} + \frac{3 π}{4}$	Measure of angle Q
$\frac{7 π}{12} + \frac{9 π}{12}$	Rewrite $\frac{3 π}{4}$ to get a common denominator: $\frac{3 π}{4} \cdot \frac{3}{3} = \frac{9 π}{12}$
$\frac{16 π}{12}$	Add