All practice questions

Question 2641Question →

In the standard $(x, y)$ coordinate plane, a parallelogram has vertices at $P(-2, -1)$ , $Q(4, -1)$ , $R(6, 4)$ , and $S(0, 4)$ . What is the area of parallelogram $PQRS$ , in square units?

15

18

24

30

40

Show answer & explanation

Answer: 30

Answer

30

The correct answer is 30. The area of a parallelogram is determined by multiplying its base by its perpendicular height. The base segment

PQ

is horizontal and extends from

x = -2

to

x = 4

, giving a length of

4 - (-2) = 6

units. The height is the vertical distance between the line containing the base

PQ

(

y = -1

) and the line containing the opposite side

SR

(

y = 4

). This distance is

4 - (-1) = 5

units. Multiplying the base of 6 units by the height of 5 units yields an area of 30 square units.

Step-by-Step Solution

1

Identify the base of the parallelogram by calculating the length of the horizontal side

PQ

.

PQ = 4 - (-2) = 6

units

The segment

PQ

lies on the horizontal line

y = -1

, so its length is the difference between the x-coordinates of its endpoints.

2

Identify the height of the parallelogram by calculating the vertical distance between the parallel horizontal sides

PQ

(on

y = -1

) and

SR

(on

y = 4

).

height = 4 - (-1) = 5

units

The height of a parallelogram is the perpendicular distance between its parallel bases.

3

Calculate the area of the parallelogram using the formula

Area = \text{base} \times \text{height}

.

Area = 6 \times 5 = 30

square units

Multiplying the base length by the vertical height gives the total area of the parallelogram.

Key Concept

Finding the area of a parallelogram on the coordinate plane using base and height calculations from coordinates.

Question 2642Question →

For the quadratic equation $0.4x^2 + bx - 4.8 = 0$ , where $b$ is a constant, the sum of the two solutions is equal to the product of the two solutions. What is the value of $b$ ?

Show answer & explanation

Answer: 4.8

Answer

The value of the constant

b

is

4.8

.

According to Vieta's formulas, the sum of the solutions to the quadratic equation

ax^2 + bx + c = 0

is given by

-\frac{b}{a}

and their product is given by

\frac{c}{a}

. For the given equation

0.4x^2 + bx - 4.8 = 0

, the product of the solutions is

\frac{-4.8}{0.4} = -12

. Setting the sum equal to the product yields the equation

-\frac{b}{0.4} = -12

. Multiplying both sides by

-0.4

isolates

b

, giving

b = 4.8

.

Step-by-Step Solution

1

Identify the coefficients of the quadratic equation.

a = 0.4

,

b = b

, and

c = -4.8

.

To apply formulas relating the coefficients to the solutions.

2

Express the sum and product of the solutions using Vieta's formulas.

Sum of solutions is

-\frac{b}{0.4}

and product of solutions is

\frac{-4.8}{0.4} = -12

.

To establish the mathematical relationship given in the problem.

3

Equate the sum and product of the solutions and solve for the constant

b

.

-\frac{b}{0.4} = -12 \implies b = -12 \times (-0.4) = 4.8

.

The problem states that the sum of the two solutions is equal to their product.

Key Concept

Sum and Product of Roots (Vieta's Formulas)

Question 2643Question →

An angle $\theta$ satisfies $\frac{\pi}{2} < \theta < \pi$ and $(\sin\theta - \cos\theta)^2 = \frac{17}{9}$ . What is the value of $\tan\theta + \cot\theta$ ?

1

-\frac{9}{8}

-\frac{9}{4}

\frac{9}{4}

-\frac{4}{9}

Show answer & explanation

Answer: $-\frac{9}{4}$

Answer

The value of the expression is

-\frac{9}{4}

Expanding the square of the difference

(\sin\theta - \cos\theta)^2

yields

\sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta

. Applying the Pythagorean identity

\sin^2\theta + \cos^2\theta = 1

simplifies this expression to

1 - 2\sin\theta\cos\theta

. Setting this equal to

\frac{17}{9}

and solving for the product of sine and cosine gives

\sin\theta\cos\theta = -\frac{4}{9}

. The target expression

\tan\theta + \cot\theta

can be rewritten using quotient and reciprocal identities as

\frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}

, which simplifies by finding a common denominator to

\frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}

. Substituting the value of

\sin\theta\cos\theta

into this expression results in

-\frac{9}{4}

.

Step-by-Step Solution

1

Expand the squared expression

(\sin\theta - \cos\theta)^2

and apply the Pythagorean identity.

(\sin\theta - \cos\theta)^2 = \sin^2\theta - 2\sin\theta\cos\theta + \cos^2\theta = 1 - 2\sin\theta\cos\theta

To express the squared binomial in terms of the product

\sin\theta\cos\theta

using the identity

\sin^2\theta + \cos^2\theta = 1

.

2

Equate the expanded form to the given value and solve for

\sin\theta\cos\theta

.

1 - 2\sin\theta\cos\theta = \frac{17}{9} \implies 2\sin\theta\cos\theta = -\frac{8}{9} \implies \sin\theta\cos\theta = -\frac{4}{9}

To find the value of the product

\sin\theta\cos\theta

from the given equation.

3

Rewrite the target expression

\tan\theta + \cot\theta

in terms of sine and cosine.

\tan\theta + \cot\theta = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}

To simplify the target sum of ratios using quotient and reciprocal identities so that it depends only on the product

\sin\theta\cos\theta

.

4

Substitute the value of

\sin\theta\cos\theta

into the simplified expression.

\tan\theta + \cot\theta = \frac{1}{-4/9} = -\frac{9}{4}

To compute the final numerical value of the expression.

Key Concept

Simplifying trigonometric expressions using fundamental Pythagorean, quotient, and reciprocal identities, and solving for unknown products of trigonometric functions.

Alternative Method

Alternatively, one can find the individual values of

\sin\theta

and

\cos\theta

. Since

(\sin\theta - \cos\theta)^2 = \frac{17}{9}

and

\sin\theta\cos\theta = -\frac{4}{9}

, we can use the identity

(\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta = \frac{1}{9}

. In Quadrant II,

\sin\theta > 0

and

\cos\theta < 0

, and since

-\cos\theta > 0

, we have

\sin\theta - \cos\theta = \frac{\sqrt{17}}{3}

. Solving the system of equations for

\sin\theta

and

\cos\theta

yields

\sin\theta = \frac{\sqrt{17} - 1}{6}

(since it must be positive) and

\cos\theta = \frac{-\sqrt{17} - 1}{6}

(since it must be negative). Substituting these into

\tan\theta + \cot\theta = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}

will yield the same result of

-\frac{9}{4}

, though this method involves significantly more algebraic work.

Estimated Time:3m 0s

Question 2644Question →

A terminal ray of an angle $\theta$ in standard position passes through the point $\left(-\frac{1}{2}, y\right)$ on the unit circle in Quadrant III. What is the value of $\theta$ in radians, where $0 \le \theta < 2\pi$ ?

\frac{2\pi}{3}

\frac{4\pi}{3}

\frac{5\pi}{4}

\frac{5\pi}{3}

\frac{7\pi}{6}

Show answer & explanation

Answer: $\frac{4\pi}{3}$

Answer

The correct answer is the option containing the radian measure

\frac{4\pi}{3}

.

The correct answer is the option containing the radian measure

\frac{4\pi}{3}

. Any point

(x, y)

on the unit circle satisfies

x^2 + y^2 = 1

. Substituting

x = -\frac{1}{2}

gives

\frac{1}{4} + y^2 = 1

, which simplifies to

y^2 = \frac{3}{4}

, so

y = \pm \frac{\sqrt{3}}{2}

. Since the terminal ray lies in Quadrant III, the y-coordinate must be negative, so

y = -\frac{\sqrt{3}}{2}

. The angle

\theta

in

[0, 2\pi)

with

\cos(\theta) = -\frac{1}{2}

and

\sin(\theta) = -\frac{\sqrt{3}}{2}

is

\theta = \frac{4\pi}{3}

radians.

Step-by-Step Solution

1

Use the equation of the unit circle,

x^2 + y^2 = 1

, to find the y-coordinate of the point.

Since

x = -\frac{1}{2}

, we have

\left(-\frac{1}{2}\right)^2 + y^2 = 1 \Rightarrow \frac{1}{4} + y^2 = 1 \Rightarrow y^2 = \frac{3}{4}

.

The coordinates of any point on the unit circle must satisfy the equation

x^2 + y^2 = 1

.

2

Determine the correct sign of the y-coordinate using the quadrant information.

y = -\frac{\sqrt{3}}{2}

Since the point lies in Quadrant III, both the x and y coordinates must be negative.

3

Identify the angle

\theta

in the interval

[0, 2\pi)

that corresponds to these coordinates on the unit circle.

\theta = \frac{4\pi}{3}

radians

On the unit circle,

\cos(\theta) = x = -\frac{1}{2}

and

\sin(\theta) = y = -\frac{\sqrt{3}}{2}

. In Quadrant III, the angle with a reference angle of

\frac{\pi}{3}

is

\pi + \frac{\pi}{3} = \frac{4\pi}{3}

.

Key Concept

Finding an angle in standard position on the unit circle given its x-coordinate and quadrant

Alternative Method

Alternatively, one can convert the options from radians to degrees and evaluate their cosine values. The correct angle must satisfy

\cos(\theta) = -1/2

. Since

\cos(2\pi/3) = -1/2

and

\cos(4\pi/3) = -1/2

, we only need to check these two options. Among these, only

\frac{4\pi}{3}

lies in Quadrant III, which confirms it as the correct answer.

Estimated Time:1m 30s

Question 2645Question →

In the standard $(x, y)$ coordinate plane, the vertices of a right triangle are $P(1, 2)$ , $Q(5, 5)$ , and $R(k, 9)$ . If the right angle of the triangle is at vertex $Q$ , what is the value of the constant $k$ ?

Show answer & explanation

Answer: 2

Answer

The value of

k

is

2

.

Because the right angle of the triangle is at vertex

Q

, segment

PQ

must be perpendicular to segment

QR

. The slope of

PQ

is

\frac{5 - 2}{5 - 1} = \frac{3}{4}

. The slope of a perpendicular line is the negative reciprocal, so the slope of

QR

must be

-\frac{4}{3}

. Expressing the slope of

QR

using the coordinates of

Q(5, 5)

and

R(k, 9)

gives

\frac{9 - 5}{k - 5} = \frac{4}{k - 5}

. Setting this equal to

-\frac{4}{3}

and solving for

k

yields

k - 5 = -3

, which means

k = 2

.

Step-by-Step Solution

1

Use the slope formula

m = \frac{y_2 - y_1}{x_2 - x_1}

to calculate the slope of the line segment

PQ

with endpoints

P(1, 2)

and

Q(5, 5)

.

m_{PQ} = \frac{5 - 2}{5 - 1} = \frac{3}{4}

This establishes the direction of the first leg of the right triangle.

2

Find the slope of segment

QR

. Because the right angle is at vertex

Q

, the segment

PQ

is perpendicular to segment

QR

.

m_{QR} = -\frac{4}{3}

Perpendicular lines have slopes that are negative reciprocals of each other (

m_1 \cdot m_2 = -1

).

3

Write the slope of segment

QR

in terms of

k

using coordinates

Q(5, 5)

and

R(k, 9)

.

m_{QR} = \frac{9 - 5}{k - 5} = \frac{4}{k - 5}

This sets up an equation to find the unknown coordinate value.

4

Equate the two expressions for the slope of

QR

and solve for

k

.

\frac{4}{k - 5} = -\frac{4}{3} \Rightarrow k - 5 = -3 \Rightarrow k = 2

Solving the rational equation yields the correct coordinate parameter.

Key Concept

Perpendicular lines in a coordinate plane have slopes that are negative reciprocals of each other.

Estimated Time:1m 30s

Question 2646Question →

What is the period of the function $f(x) = 4\cos\left(\frac{1}{3}x\right)$ ?

\frac{2\pi}{3}

2\pi

4

6\pi

\frac{4\pi}{3}

Show answer & explanation

Answer: $6\pi$

Answer

The period of the function is

6\pi

.

The standard period of the cosine function

y = \cos(x)

is

2\pi

. For a transformed trigonometric function of the form

y = A\cos(Bx)

, the period is given by the formula

\frac{2\pi}{|B|}

. In the function

f(x) = 4\cos\left(\frac{1}{3}x\right)

, the coefficient of

x

is

B = \frac{1}{3}

. Dividing the standard period

2\pi

by

\frac{1}{3}

yields a period of

2\pi \times 3 = 6\pi

.

Step-by-Step Solution

1

Identify the standard period of the parent cosine function and the coefficient

B

of the variable

x

in the given function.

The parent function is

y = \cos(x)

, which has a standard period of

2\pi

. In the function

f(x) = 4\cos\left(\frac{1}{3}x\right)

, the coefficient of

x

is

B = \frac{1}{3}

.

This sets up the parameters needed for the period formula

\text{Period} = \frac{2\pi}{|B|}

.

2

Substitute the value of

B

into the period formula and simplify.

The period is

\frac{2\pi}{\frac{1}{3}} = 2\pi \times 3 = 6\pi

.

Dividing by a fraction is equivalent to multiplying by its reciprocal, which gives the final horizontal distance for one complete cycle.

Key Concept

Graphs of Trigonometric Functions

Question 2647Question →

A right triangle has a hypotenuse of length $13\text{ inches}$ . One of the legs is $7\text{ inches}$ longer than the other leg. What is the length, in inches, of the shorter leg?

3.0

5.0

6.0

8.6

11.0

Show answer & explanation

Answer: 5.0

Answer

The length of the shorter leg is 5.0 inches.

The correct answer is the option representing 5.0. By setting the shorter leg as

x

, the longer leg is

x + 7

. Applying the Pythagorean theorem gives

x^2 + (x + 7)^2 = 13^2

, which simplifies to

2x^2 + 14x - 120 = 0

. Factoring the divided equation

x^2 + 7x - 60 = 0

gives

(x - 5)(x + 12) = 0

. Since length must be positive, the shorter leg is 5.0 inches.

Step-by-Step Solution

1

Define variables for the side lengths of the right triangle based on the problem statement.

Let the length of the shorter leg be

x

inches. The length of the longer leg is

x + 7

inches, and the hypotenuse is

13

inches.

This translates the verbal descriptions into mathematical expressions using a single variable.

2

Apply the Pythagorean theorem to set up an equation relating the side lengths.

x^2 + (x + 7)^2 = 13^2

For any right triangle, the sum of the squares of the legs equals the square of the hypotenuse.

3

Expand the squared binomial and simplify the equation into standard quadratic form.

x^2 + (x^2 + 14x + 49) = 169 \Rightarrow 2x^2 + 14x - 120 = 0

Expanding allows us to group like terms and solve for the variable

x

.

4

Divide the quadratic equation by 2 and factor the resulting expression.

x^2 + 7x - 60 = 0 \Rightarrow (x + 12)(x - 5) = 0

Factoring is the most efficient way to find the roots of this quadratic equation.

5

Solve for

x

and select the mathematically and physically valid solution.

x = 5

or

x = -12

. Since a side length must be positive,

x = 5

inches.

Lengths in geometry must be positive, so we discard the negative root.

Key Concept

Applying the Pythagorean theorem to solve for unknown side lengths of a right triangle given algebraic relationships between the sides.

Estimated Time:1m 30s

Question 2648Question →

An angle $\theta$ in standard position is rotated counterclockwise by $225^\circ$ . The terminal ray of the resulting angle lies in the fourth quadrant along the line $y = -x\sqrt{3}$ . If the radian measure of the smallest positive angle $\theta$ is written in simplest form as $\frac{a\pi}{b}$ , where $a$ and $b$ are positive integers, what is the value of $a + b$ ?

Show answer & explanation

Answer: 17

Answer

The value of

a + b

is

17

.

The resulting angle

\alpha

lies along the line

y = -x\sqrt{3}

in the fourth quadrant, meaning its measure is

300^\circ

plus any multiple of

360^\circ

. Subtracting the counterclockwise rotation of

225^\circ

gives the original angle

\theta = 75^\circ

(for the smallest positive angle). Converting

75^\circ

to radians by multiplying by

\frac{\pi}{180^\circ}

yields

\frac{5\pi}{12}

. Since the fraction is in simplest form,

a = 5

and

b = 12

, and their sum is

17

.

Step-by-Step Solution

1

Find the angle of the terminal ray after rotation from its equation and quadrant

The terminal ray after rotation is at an angle of

300^\circ

(or

\frac{5\pi}{3}

radians)

The line

y = -x\sqrt{3}

has a slope of

-\sqrt{3}

, so the angle

\alpha

in Quadrant IV satisfies

\tan\alpha = -\sqrt{3}

, which means

\alpha = 300^\circ

.

2

Set up and solve the equation for the original angle

\theta

before the counterclockwise rotation of

225^\circ

\theta = 75^\circ + 360^\circ k

Since the angle was rotated counterclockwise by

225^\circ

to reach the final position of

300^\circ

, we have

\theta + 225^\circ = 300^\circ + 360^\circ k

.

3

Determine the smallest positive value of

\theta

\theta = 75^\circ

Setting

k = 0

gives the smallest positive angle of

75^\circ

.

4

Convert the angle

\theta

from degrees to radians

\theta = \frac{5\pi}{12}

radians

To convert degrees to radians, multiply by

\frac{\pi}{180^\circ}

, yielding

\frac{75\pi}{180} = \frac{5\pi}{12}

.

5

Calculate the sum of the numerator and denominator of the simplified radian fraction

a + b = 17

The fraction

\frac{5}{12}

is in simplest form, so

a = 5

and

b = 12

. The sum is

5 + 12 = 17

.

Key Concept

Converting degree measures to radian measures and finding coterminal angles on the unit circle

Question 2649Question →

A convex octagon has interior angles whose measures, in degrees, are eight consecutive even integers. What is the measure, in degrees, of the largest interior angle of this octagon?

Show answer & explanation

Answer: 142

Answer

The correct answer is 142.

The sum of the interior angles of a convex octagon is

(8-2) \times 180^\circ = 1080^\circ

. If we represent the eight consecutive even integer angle measures as

x, x+2, x+4, x+6, x+8, x+10, x+12,

and

x+14

, their sum is

8x + 56

. Setting this equal to

1080^\circ

and solving for

x

yields

x = 128

. The largest angle is

x + 14

, which equals

128 + 14 = 142^\circ

.

Step-by-Step Solution

1

Calculate the sum of the interior angles of a convex octagon.

The sum of the interior angles is

1080^\circ

.

The formula for the sum of the interior angles of an

n

-sided polygon is

(n-2) \times 180^\circ

. For an octagon (

n=8

), the sum is

(8-2) \times 180^\circ = 6 \times 180^\circ = 1080^\circ

.

2

Set up an equation representing the sum of the eight consecutive even integer angle measures.

The equation is

8x + 56 = 1080

.

Letting the smallest angle measure be

x

, the eight consecutive even integer angle measures are

x, x+2, x+4, x+6, x+8, x+10, x+12,

and

x+14

. Their sum is

8x + 56

, which must equal the total sum of the interior angles (

1080^\circ

).

3

Solve the equation for the smallest angle measure,

x

.

x = 128

Subtracting 56 from both sides of the equation yields

8x = 1024

. Dividing both sides by 8 gives

x = 128

.

4

Calculate the measure of the largest interior angle.

The measure of the largest angle is

142^\circ

.

The largest angle is represented by the expression

x + 14

. Substituting

128

for

x

gives

128 + 14 = 142

.

Key Concept

Calculating the sum of the interior angles of a convex polygon and using algebraic methods to find unknown angle measures.

Question 2650Question →

In the standard $(x, y)$ coordinate plane, a line passes through the points $(a, 2)$ and $(10, a - 1)$ . If the slope of the line is $-\frac{1}{3}$ , what is the value of $a$ ?

Show answer & explanation

Answer: -0.5

Answer

The value of

a

is

-0.5

(or

-\frac{1}{2}

)

Applying the slope formula

m = \frac{y_2 - y_1}{x_2 - x_1}

to the points

(a, 2)

and

(10, a - 1)

with slope

-\frac{1}{3}

yields the equation

-\frac{1}{3} = \frac{a - 3}{10 - a}

. Solving this linear equation correctly yields

a = -0.5

.

Step-by-Step Solution

1

Apply the slope formula using the given coordinates.

-\frac{1}{3} = \frac{(a - 1) - 2}{10 - a}

The slope of a line passing through

(x_1, y_1)

and

(x_2, y_2)

is defined as

m = \frac{y_2 - y_1}{x_2 - x_1}

.

2

Simplify the numerator.

-\frac{1}{3} = \frac{a - 3}{10 - a}

Subtracting

2

from

a - 1

simplifies the numerator to

a - 3

.

3

Cross-multiply to solve the rational equation.

-1(10 - a) = 3(a - 3)

Multiplying both sides by the denominators eliminates the fractions.

4

Solve the linear equation for

a

.

a = -0.5

Distributing on both sides gives

-10 + a = 3a - 9

. Rearranging terms yields

2a = -1

, which simplifies to

a = -0.5

.

Key Concept

Slope of a Line

Question 2651Question →

In the standard $(x, y)$ coordinate plane, a quadrilateral $ABCD$ is an isosceles trapezoid with parallel bases $AB$ and $CD$ . The coordinates of three of the vertices are $A(-15, -20)$ , $B(15, 20)$ , and $C(-7, 24)$ . If $ABCD$ is NOT a parallelogram, what are the coordinates of the fourth vertex, $D$ ?

(-37, -16)

(-25, 0)

(-7, -24)

(7, -24)

(23, 64)

Show answer & explanation

Answer: $(-25, 0)$

Answer

(-25, 0)

The correct answer is

(-25, 0)

. First, we find the slope of the parallel bases

AB

and

CD

to be

\frac{4}{3}

, which gives the equation of the line containing

CD

as

4x - 3y + 100 = 0

. Since the trapezoid is isosceles, the leg lengths are equal, meaning

AD^2 = BC^2 = 500

. Substituting the line equation into the distance equation yields two potential coordinates for

D

:

(-25, 0)

and

(-37, -16)

. Since the problem specifies that

ABCD

is not a parallelogram, we eliminate

(-37, -16)

(which makes

AD \parallel BC

) to conclude that

D

must be

(-25, 0)

.

Step-by-Step Solution

1

Calculate the slope of the base

AB

.

The slope of

AB

is

\frac{20 - (-20)}{15 - (-15)} = \frac{40}{30} = \frac{4}{3}

.

Since

AB \parallel CD

, the line containing base

CD

must also have a slope of

\frac{4}{3}

.

2

Set up the equation for the line containing

CD

.

Using the point-slope form with

C(-7, 24)

and slope

\frac{4}{3}

, the equation of the line is

y - 24 = \frac{4}{3}(x + 7) \implies 4x - 3y + 100 = 0

.

The fourth vertex

D(x, y)

must lie on this line, so its coordinates satisfy

x = \frac{3y - 100}{4}

.

3

Set up the distance equation for the equal leg lengths.

The square of the leg length

BC^2 = (-7 - 15)^2 + (24 - 20)^2 = (-22)^2 + 4^2 = 484 + 16 = 500

. Since

AD = BC

, the distance equation is

(x + 15)^2 + (y + 20)^2 = 500

.

In an isosceles trapezoid, the non-parallel sides (legs)

AD

and

BC

must have equal lengths.

4

Solve the system of equations for the coordinates of

D

.

Substitute

x = \frac{3y - 100}{4}

into the distance equation:

(\frac{3y - 40}{4})^2 + (y + 20)^2 = 500 \implies 25y^2 + 400y = 0

. This yields

y = 0

or

y = -16

. The corresponding coordinates are

D_1(-25, 0)

and

D_2(-37, -16)

.

Substituting the linear relationship into the quadratic distance equation gives the two mathematically possible locations for

D

.

5

Verify which solution satisfies the non-parallelogram constraint.

For

D_2(-37, -16)

, the slope of

AD

is

\frac{-16 - (-20)}{-37 - (-15)} = -\frac{2}{11}

, which is equal to the slope of

BC

. This makes

ABCD

a parallelogram. For

D_1(-25, 0)

, the slope of

AD

is

-2 \neq -\frac{2}{11}

, which forms a trapezoid.

The problem states that

ABCD

is not a parallelogram, so

D

must be

(-25, 0)

.

Key Concept

Identifying vertices of an isosceles trapezoid using coordinate geometry, slopes of parallel lines, distance formula, and distinguishing a trapezoid from a parallelogram.

Question 2652Question →

In right triangle $PQR$ , the right angle is at vertex $Q$ . If the side lengths are $PQ = 15$ and $QR = 8$ , what is the value of $\tan(P)$ ?

\frac{15}{8}

\frac{15}{17}

\frac{8}{15}

\frac{8}{17}

\frac{17}{8}

Show answer & explanation

Answer: $\frac{8}{15}$

Answer

The tangent of angle

P

is

\frac{8}{15}

.

The tangent of angle

P

is the ratio of the opposite side (

QR = 8

) to the adjacent side (

PQ = 15

). This gives the value

\frac{8}{15}

.

Step-by-Step Solution

1

Identify the reference angle and the sides of the right triangle relative to it.

The reference angle is

P

. The side opposite to angle

P

is

QR

with a length of

8

. The side adjacent to angle

P

is

PQ

with a length of

15

. The hypotenuse is

PR

.

To calculate a trigonometric ratio, we must first determine which sides are opposite, adjacent, and the hypotenuse relative to the target angle.

2

Recall the definition of the tangent ratio in a right triangle.

The tangent of an angle is defined as the ratio of the length of the opposite side to the length of the adjacent side:

\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}

.

The question asks for the tangent of angle

P

.

3

Substitute the identified side lengths into the tangent formula.

\tan(P) = \frac{QR}{PQ} = \frac{8}{15}

.

Plugging the lengths of the opposite side (

8

) and the adjacent side (

15

) into the tangent ratio yields the final value.

Key Concept

Right triangle trigonometry ratios (SOHCAHTOA), specifically the tangent ratio definition.

Estimated Time:45s

Question 2653Question →

A circular archery target has a radius of $12\text{ inches}$ . A sector of this target has a central angle of $150^\circ$ . What is the area, in square inches, of this sector?

10\pi

15\pi

60\pi

120\pi

240\pi

Show answer & explanation

Answer: $60\pi$

Answer

The correct area of the sector is

60\pi

square inches.

To find the area of a sector, first calculate the total area of the circle, which is

\pi r^2 = \pi (12)^2 = 144\pi

square inches. Then, multiply this total area by the fraction of the circle that the sector represents:

\frac{150^\circ}{360^\circ} = \frac{5}{12}

. Calculating the product gives

\frac{5}{12} \times 144\pi = 60\pi

square inches.

Step-by-Step Solution

1

Find the total area of the circular archery target using the area formula

A = \pi r^2

with a radius of

12\text{ inches}

.

The total area of the circle is

\pi (12)^2 = 144\pi

square inches.

The area of a sector is a fractional part of the circle's total area.

2

Calculate the fraction of the circle represented by a central angle of

150^\circ

.

The fraction is

\frac{150^\circ}{360^\circ} = \frac{5}{12}

.

A complete circle has a central angle of

360^\circ

.

3

Multiply the total area of the circle by the fraction of the circle to determine the sector area.

The sector area is

\frac{5}{12} \times 144\pi = 60\pi

square inches.

Applying the fraction to the total area yields the area of the sector.

Key Concept

Calculating the area of a circle sector using the formula

A = \frac{\theta}{360^\circ} \pi r^2

.

Estimated Time:1m 0s

Question 2654Question →

A circular sector with radius $R$ and central angle $\theta$ (measured in radians) has the same area and the same perimeter as a square with side length $s$ . What is the value of $\theta$ ?

2

4

8

\frac{1}{\pi}

2\pi

Show answer & explanation

Answer: 2

Answer

The value of theta must be 2

The correct value is 2. By equating the sector's area and perimeter to the square's area and perimeter, we set up a system of equations:

\frac{1}{2} R^2 \theta = s^2

and

2R + R\theta = 4s

. Solving for

s

from the area equation gives

s = R\sqrt{\theta/2}

. Substituting this into the perimeter equation and dividing by

R

yields

2 + \theta = 2\sqrt{2\theta}

. Squaring both sides and simplifying leads to the quadratic equation

\theta^2 - 4\theta + 4 = 0

, which has a single solution of

\theta = 2

radians.

Step-by-Step Solution

1

Write down the equations for the area and perimeter of both shapes.

For the circular sector: Area =

\frac{1}{2}R^2\theta

, Perimeter =

2R + R\theta

(where

\theta

is in radians). For the square: Area =

s^2

, Perimeter =

4s

.

These are the standard geometric formulas for a sector in radians and a square.

2

Set the corresponding areas and perimeters equal to each other to form a system of equations.

Equation 1:

\frac{1}{2} R^2 \theta = s^2

Equation 2:

2R + R\theta = 4s

The problem statement specifies that the two shapes have equal areas and equal perimeters.

3

Solve Equation 1 for the side length

s

of the square.

s = R \sqrt{\frac{\theta}{2}}

This allows us to substitute

s

in Equation 2 and solve for

\theta

in terms of

R

.

4

Substitute the expression for

s

into Equation 2 and simplify.

2R + R\theta = 4\left(R \sqrt{\frac{\theta}{2}}\right) \implies 2 + \theta = 4\sqrt{\frac{\theta}{2}}

Since the radius

R

is a positive length, we can divide both sides of the equation by

R

.

5

Solve the simplified equation for

\theta

by squaring both sides.

2 + \theta = 2\sqrt{2\theta} \implies (2+\theta)^2 = (2\sqrt{2\theta})^2 \implies 4 + 4\theta + \theta^2 = 8\theta \implies \theta^2 - 4\theta + 4 = 0 \implies (\theta - 2)^2 = 0 \implies \theta = 2

Squaring both sides and setting the quadratic equation to zero yields a perfect square trinomial with a single real solution.

Key Concept

Relating the area and perimeter of a circular sector using radian measures to those of a square.

Estimated Time:3m 0s

Question 2655Question →

The passage below is adapted from an essay on the history of sound recording.

In the late nineteenth century, the quest to capture and preserve sound transformed the landscape of human communication. Before this era, music, speech, and performance were ephemeral, surviving only in memory or through written transcriptions. The invention that shattered this limitation was the phonograph, conceived by Thomas Alva Edison in the autumn of 1877. Edison’s breakthrough occurred while he was working on improvements to the telegraph and telephone. He realized that the vibrations of the human voice could be used to indent a moving medium, which could then be replayed by tracing a needle along the same grooves.

Edison’s first prototype utilized a sheet of tinfoil wrapped around a hand-cranked grooved metal cylinder. To record sound, a speaker shouted into a mouthpiece, causing a diaphragm to vibrate. This vibration drove a sharp stylus into the tinfoil, leaving varying indentations. To play back the recording, the stylus was returned to the starting point, and as the cylinder rotated, the stylus followed the indentations, vibrating the diaphragm and reproducing the sound. The first recorded words were Edison’s own recital of 'Mary Had a Little Lamb.' While the device was a marvel, the tinfoil was fragile and degraded after only a few playbacks, making the original phonograph little more than a novelty.

Recognizing the limitations of tinfoil, other inventors sought to improve upon Edison’s design. In the early 1880s, Chichester Bell and Charles Sumner Tainter, working at the Volta Laboratory established by Alexander Graham Bell, introduced the graphophone. Unlike Edison’s machine, which used tinfoil on a metal cylinder, the graphophone recorded onto wax-coated cardboard cylinders. The wax was soft enough to be easily engraved by a cutting stylus, yet durable enough to withstand repeated playback without significant degradation of sound quality. This shift from indenting tinfoil to engraving wax marked a major technological advancement, turning sound recording into a commercially viable enterprise.

Shortly thereafter, Emile Berliner introduced another radical departure from Edison's design: the gramophone. Patented in 1887, the gramophone abandoned cylinders entirely in favor of flat zinc discs coated with beeswax. Rather than employing a vertical 'hill-and-dale' engraving method, where the stylus moved up and down to cut grooves of varying depths, Berliner used a lateral-cut method, where the stylus moved from side to side in a groove of constant depth. This lateral recording technique was easier to replicate on a mass scale. Berliner developed a process to electroplate the original zinc disc to create a metal stamper, which could then press duplicate discs into hard rubber (and later, shellac). This ability to mass-produce recordings from a single master disc laid the groundwork for the modern music industry.

Despite these competing designs, Edison did not abandon his invention. In 1887, he returned to sound recording and introduced the 'Perfected Phonograph,' which utilized solid wax cylinders instead of tinfoil. This design allowed users to shave off the outer layer of a recorded cylinder to reuse it for a new recording. The battle between cylinders and discs continued for decades, but by the 1920s, the flat disc format championed by Berliner had largely triumphed due to its ease of storage and superior mass-production capabilities.

According to the passage, which material did Chichester Bell and Charles Sumner Tainter use to record sound on the graphophone?

Tinfoil wrapped around a metal cylinder

Wax-coated cardboard cylinders

Flat zinc discs coated with beeswax

Reusable solid wax cylinders

Show answer & explanation

Answer: Wax-coated cardboard cylinders

Answer

Wax-coated cardboard cylinders

The correct answer is correct because paragraph three explicitly states that the graphophone recorded onto wax-coated cardboard cylinders, distinguishing it from Edison's tinfoil design.

Step-by-Step Solution

1

Locate the portion of the passage that discusses Chichester Bell, Charles Sumner Tainter, and the graphophone.

The third paragraph of the passage introduces these inventors and their device.

The question specifically asks about the materials used by Bell and Tainter on the graphophone, so the search must focus on where this device is described.

2

Read the sentences describing the physical construction and materials of the graphophone in paragraph three.

The text states: 'Unlike Edison’s machine, which used tinfoil on a metal cylinder, the graphophone recorded onto wax-coated cardboard cylinders.'

This direct comparison explicitly names the material used for the graphophone.

3

Match this explicit detail to the correct option.

The option stating 'wax-coated cardboard cylinders' matches the details given in the text.

Literal comprehension questions require finding a direct match to facts explicitly stated in the passage.

Key Concept

Identifying Explicit Details

Question 2656Question →

For an angle $\theta$ in the interval $\frac{3\pi}{2} < \theta < 2\pi$ , the expression $\sec\theta - \tan\theta$ is equal to $3$ . What is the value of $\csc\theta + \cot\theta$ ?

Show answer & explanation

Answer: -2

Answer

The value of

\csc\theta + \cot\theta

is

-2

.

By using the difference of squares on the identity

\sec^2\theta - \tan^2\theta = 1

, we get

(\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1

. Substituting

\sec\theta - \tan\theta = 3

gives

\sec\theta + \tan\theta = \frac{1}{3}

. Solving the system of equations gives

\sec\theta = \frac{5}{3}

and

\tan\theta = -\frac{4}{3}

. Since

\theta

lies in Quadrant IV,

\cos\theta = \frac{3}{5}

and

\sin\theta = -\frac{4}{5}

. We then find

\csc\theta = -\frac{5}{4}

and

\cot\theta = -\frac{3}{4}

, which sum to

-2

.

Step-by-Step Solution

1

Use the Pythagorean identity

\sec^2\theta - \tan^2\theta = 1

, which factors into

(\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1

.

Since

\sec\theta - \tan\theta = 3

, we have

3(\sec\theta + \tan\theta) = 1 \implies \sec\theta + \tan\theta = \frac{1}{3}

.

To establish a second linear equation in terms of

\sec\theta

and

\tan\theta

.

2

Add and subtract the two equations:

\sec\theta - \tan\theta = 3

and

\sec\theta + \tan\theta = \frac{1}{3}

.

Adding them gives

2\sec\theta = \frac{10}{3} \implies \sec\theta = \frac{5}{3}

. Subtracting the first from the second gives

2\tan\theta = -\frac{8}{3} \implies \tan\theta = -\frac{4}{3}

.

To isolate the values of

\sec\theta

and

\tan\theta

.

3

Find

\cos\theta

and

\sin\theta

using

\cos\theta = \frac{1}{\sec\theta}

and

\sin\theta = \tan\theta\cos\theta

.

\cos\theta = \frac{3}{5}

and

\sin\theta = -\frac{4}{5}

. Since

\frac{3\pi}{2} < \theta < 2\pi

(Quadrant IV), cosine is positive and sine is negative, which matches these values.

To find the primary trigonometric values needed for the reciprocal functions.

4

Calculate

\csc\theta

and

\cot\theta

using reciprocal identities.

\csc\theta = \frac{1}{\sin\theta} = -\frac{5}{4}

and

\cot\theta = \frac{1}{\tan\theta} = -\frac{3}{4}

.

To obtain the terms of the required sum.

5

Sum the values of

\csc\theta

and

\cot\theta

.

\csc\theta + \cot\theta = -\frac{5}{4} + \left(-\frac{3}{4}\right) = -\frac{8}{4} = -2

.

To obtain the final value requested by the question.

Key Concept

Pythagorean and Reciprocal Trigonometric Identities

Question 2657Question →

Three vertices of a parallelogram are $A(1, 3)$ , $B(-2, -1)$ , and $C(4, -1)$ in the standard $(x, y)$ coordinate plane. If the fourth vertex, $D$ , is located in the fourth quadrant, what is the $y$ -coordinate of $D$ ?

Show answer & explanation

Answer: -5

Answer

The

y

-coordinate of the fourth vertex of the parallelogram in the fourth quadrant is

-5

.

By applying the midpoint formula to the diagonals of the three possible parallelograms formed by the vertices

A(1, 3)

,

B(-2, -1)

, and

C(4, -1)

, we find the candidate points for the fourth vertex

D

to be

(7, 3)

,

(-5, 3)

, and

(1, -5)

. The point

(1, -5)

is the only candidate that lies in the fourth quadrant, where

x > 0

and

y < 0

. Therefore, the

y

-coordinate of the fourth vertex is

-5

.

Step-by-Step Solution

1

Set up equations based on the property that the diagonals of a parallelogram bisect each other (have the same midpoint).

Three possible configurations of diagonals lead to three potential sets of coordinates for the fourth vertex

D(x, y)

:

(7, 3)

,

(-5, 3)

, and

(1, -5)

.

Three points in a coordinate plane can form three distinct parallelograms depending on which pairs are connected as diagonals.

2

Identify the signs of the coordinates for each potential vertex to determine which quadrant it lies in.

The point

(7, 3)

is in Quadrant I (

x > 0, y > 0

). The point

(-5, 3)

is in Quadrant II (

x < 0, y > 0

). The point

(1, -5)

is in Quadrant IV (

x > 0, y < 0

).

The fourth quadrant is defined by positive

x

-values and negative

y

-values.

3

Select the correct coordinate value corresponding to the question's requirement.

The

y

-coordinate of the vertex in the fourth quadrant,

(1, -5)

, is

-5

.

The question asks specifically for the

y

-coordinate of the fourth vertex

D

.

Key Concept

Finding a missing vertex of a parallelogram on the coordinate plane using midpoint properties

Question 2658Question →

For an angle $\theta$ such that $\pi < \theta < \frac{3\pi}{2}$ , the equation $\frac{1 - \sin\theta}{\cos\theta} + \frac{\cos\theta}{1 - \sin\theta} = -\frac{10}{3}$ is satisfied. What is the value of $\sin\theta$ ?

-\frac{4}{5}

-\frac{3}{5}

\frac{4}{5}

-\frac{3}{4}

\frac{3}{5}

Show answer & explanation

Answer: $-\frac{4}{5}$

Answer

The value of

\sin\theta

is

-\frac{4}{5}

.

The correct answer is determined by first rewriting the given equation by finding a common denominator, which simplifies the numerator to

2(1-\sin\theta)

using the Pythagorean identity

\sin^2\theta + \cos^2\theta = 1

. The term

(1-\sin\theta)

cancels out, resulting in

\frac{2}{\cos\theta} = -\frac{10}{3}

, which implies

\cos\theta = -\frac{3}{5}

. In Quadrant III, sine is negative, so using the identity

\sin\theta = -\sqrt{1 - \cos^2\theta}

yields the correct value.

Step-by-Step Solution

1

Find a common denominator to add the fractions on the left-hand side of the equation.

The expression becomes

\frac{(1 - \sin\theta)^2 + \cos^2\theta}{\cos\theta(1 - \sin\theta)}

.

To combine the algebraic terms into a single rational expression.

2

Expand the numerator and apply the Pythagorean identity

\sin^2\theta + \cos^2\theta = 1

.

The numerator simplifies to

1 - 2\sin\theta + \sin^2\theta + \cos^2\theta = 1 - 2\sin\theta + 1 = 2 - 2\sin\theta = 2(1 - \sin\theta)

.

To reduce the numerator's complexity using fundamental trigonometric identities.

3

Cancel the common factor

(1 - \sin\theta)

from the numerator and denominator, and equate the simplified term to

-\frac{10}{3}

.

The equation simplifies to

\frac{2}{\cos\theta} = -\frac{10}{3}

, which gives

\cos\theta = -\frac{3}{5}

.

To solve for the cosine of the angle

\theta

.

4

Use the Pythagorean identity

\sin^2\theta + \cos^2\theta = 1

to calculate

\sin\theta

, keeping the quadrant sign rules in mind.

Since

\pi < \theta < \frac{3\pi}{2}

(Quadrant III),

\sin\theta

is negative. Therefore,

\sin\theta = -\sqrt{1 - \cos^2\theta} = -\sqrt{1 - \left(-\frac{3}{5}\right)^2} = -\frac{4}{5}

.

To find the correct value and sign of the sine ratio for the given quadrant.

Key Concept

Fundamental Trigonometric Identities

Question 2659Question →

In the standard $(x, y)$ coordinate plane, line $L_1$ passes through the points $(1, 3)$ and $(4, 8)$ . Line $L_2$ is perpendicular to line $L_1$ . If line $L_2$ passes through the points $(5, k)$ and $(10, 1)$ , what is the value of $k$ ?

Show answer & explanation

Answer: 4

Answer

The value of

k

is

4

.

The correct answer is

4

. First, determine the slope of line

L_1

using the points

(1, 3)

and

(4, 8)

, which is

\frac{8 - 3}{4 - 1} = \frac{5}{3}

. Because line

L_2

is perpendicular to

L_1

, its slope must be the negative reciprocal of

\frac{5}{3}

, which is

-\frac{3}{5}

. Next, set up the slope equation for

L_2

with the points

(5, k)

and

(10, 1)

, yielding

\frac{1 - k}{10 - 5} = -\frac{3}{5}

. Simplifying the equation gives

\frac{1 - k}{5} = -\frac{3}{5}

, which reduces to

1 - k = -3

. Solving for

k

gives

k = 4

.

Step-by-Step Solution

1

Calculate the slope of line

L_1

using the coordinates of the two given points,

(1, 3)

and

(4, 8)

.

The slope of line

L_1

is

\frac{5}{3}

.

Using the slope formula

m = \frac{y_2 - y_1}{x_2 - x_1}

, we find

m_1 = \frac{8 - 3}{4 - 1} = \frac{5}{3}

.

2

Determine the slope of line

L_2

based on the perpendicular relationship between

L_1

and

L_2

.

The slope of line

L_2

is

-\frac{3}{5}

.

Perpendicular lines have slopes that are negative reciprocals of each other. The negative reciprocal of

\frac{5}{3}

is

-\frac{3}{5}

.

3

Use the coordinates

(5, k)

and

(10, 1)

on line

L_2

to write an expression for its slope, set it equal to

-\frac{3}{5}

, and solve for

k

.

k = 4

The slope expression is

\frac{1 - k}{10 - 5} = \frac{1 - k}{5}

. Setting this equal to the perpendicular slope gives

\frac{1 - k}{5} = -\frac{3}{5}

. Multiplying both sides by

5

results in

1 - k = -3

. Adding

k

to both sides and adding

3

to both sides yields

k = 4

.

Key Concept

The slopes of perpendicular lines in a coordinate plane are negative reciprocals of each other, meaning their product is

-1

(

m_1 \cdot m_2 = -1

).

Estimated Time:1m 15s

Question 2660Question →

Match each of the degree measures of angles in standard position on the left with its mathematically equivalent radian measure on the right. Which radian measure corresponds to each degree measure?

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are:

-135^\circ

matches with

-\frac{3\pi}{4}

radians;

480^\circ

matches with

\frac{8\pi}{3}

radians;

-300^\circ

matches with

-\frac{5\pi}{3}

radians; and

585^\circ

matches with

\frac{13\pi}{4}

radians.

Each degree measure is multiplied by

\frac{\pi}{180^\circ}

and simplified to find its equivalent radian measure. This process yields the unique matching pairs:

-135^\circ

to

-\frac{3\pi}{4}

radians,

480^\circ

to

\frac{8\pi}{3}

radians,

-300^\circ

to

-\frac{5\pi}{3}

radians, and

585^\circ

to

\frac{13\pi}{4}

radians.

Step-by-Step Solution

1

Recall the formula to convert degrees to radians.

Radian measure = Degree measure

\times \frac{\pi}{180^\circ}

Since a full circle is

360^\circ

or

2\pi

radians, the conversion ratio simplifies to

\pi

radians per

180^\circ

.

2

Convert the first degree measure,

-135^\circ

, to radians.

-135^\circ \times \frac{\pi}{180^\circ} = -\frac{135\pi}{180} = -\frac{3\pi}{4}

radians

Dividing the numerator and denominator by their greatest common divisor,

45

, simplifies the fraction to

-\frac{3}{4}

.

3

Convert the second degree measure,

480^\circ

, to radians.

480^\circ \times \frac{\pi}{180^\circ} = \frac{480\pi}{180} = \frac{8\pi}{3}

radians

Dividing the numerator and denominator by their greatest common divisor,

60

, simplifies the fraction to

\frac{8}{3}

.

4

Convert the third degree measure,

-300^\circ

, to radians.

-300^\circ \times \frac{\pi}{180^\circ} = -\frac{300\pi}{180} = -\frac{5\pi}{3}

radians

Dividing the numerator and denominator by their greatest common divisor,

60

, simplifies the fraction to

-\frac{5}{3}

.

5

Convert the fourth degree measure,

585^\circ

, to radians.

585^\circ \times \frac{\pi}{180^\circ} = \frac{585\pi}{180} = \frac{13\pi}{4}

radians

Dividing the numerator and denominator by their greatest common divisor,

45

, simplifies the fraction to

\frac{13}{4}

.

Key Concept

Converting degree measures to equivalent radian measures using the conversion factor

\frac{\pi}{180^\circ}

.

Alternative Method

Alternatively, you can recall key benchmark angles on the unit circle (such as

45^\circ = \frac{\pi}{4}

radians and

60^\circ = \frac{\pi}{3}

radians) and express each angle as an integer multiple of these benchmarks. For example,

-135^\circ

is

-3 \times 45^\circ

, which corresponds to

-3 \times \frac{\pi}{4} = -\frac{3\pi}{4}

radians.

Estimated Time:1m 30s

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