All practice questions

Question 2621Question →

A right triangle has legs of length $x - 0.5$ inches and $2x$ inches, and a hypotenuse of length $2x + 0.5$ inches. What is the value of $x$ ?

1

1.5

2

3

4

Show answer & explanation

Answer: 3

Answer

3

Applying the Pythagorean theorem to the right triangle yields

(x - 0.5)^2 + (2x)^2 = (2x + 0.5)^2

. Expanding the terms gives

x^2 - x + 0.25 + 4x^2 = 4x^2 + 2x + 0.25

. Subtracting

4x^2

and

0.25

from both sides simplifies the equation to

x^2 - x = 2x

. Subtracting

2x

from both sides gives the standard quadratic equation

x^2 - 3x = 0

. Factoring this expression gives

x(x - 3) = 0

, which yields solutions

x = 0

and

x = 3

. Since the side length

x - 0.5

must be positive,

x

must be greater than

0.5

. Therefore, the only valid solution is

3

.

Step-by-Step Solution

1

Set up the equation using the Pythagorean theorem,

a^2 + b^2 = c^2

, with the given side lengths.

(x - 0.5)^2 + (2x)^2 = (2x + 0.5)^2

The Pythagorean theorem relates the legs and hypotenuse of any right triangle.

2

Expand each squared term algebraically.

(x^2 - x + 0.25) + 4x^2 = 4x^2 + 2x + 0.25

Expanding the binomials allows us to combine like terms and simplify the equation.

3

Subtract

4x^2

and

0.25

from both sides of the equation.

x^2 - x = 2x

Simplifying the equation makes it easier to solve.

4

Move all terms to the left side to write the quadratic equation in standard form.

x^2 - 3x = 0

A quadratic equation must be set to zero to be solved by factoring.

5

Factor the quadratic expression.

x(x - 3) = 0

Factoring allows us to find the roots of the equation.

6

Solve for

x

and choose the value that makes all side lengths positive.

x = 3

(since

x = 0

is not a valid length because a side length

x - 0.5

must be greater than

0

)

Only a positive value of

x

greater than

0.5

yields physically possible side lengths for the triangle.

Key Concept

Setting up and solving quadratic equations using algebraic expansion and the Pythagorean theorem

Estimated Time:1m 30s

Question 2622Question →

In the standard $(x, y)$ coordinate plane, a line segment has endpoints at $P(2, 1)$ and $Q(8, 5)$ . Line $L$ is parallel to segment $PQ$ and passes through the point $(3, -2)$ . If the point $(9, y)$ also lies on line $L$ , what is the value of $y$ ?

Show answer & explanation

Answer: 2

Answer

The value of

y

is

2

.

The slope of segment

PQ

is

\frac{5 - 1}{8 - 2} = \frac{2}{3}

. Since line

L

is parallel to segment

PQ

, its slope is also

\frac{2}{3}

. The slope of line

L

through

(3, -2)

and

(9, y)

is given by

\frac{y - (-2)}{9 - 3} = \frac{y + 2}{6}

. Equating the two slopes yields

\frac{y + 2}{6} = \frac{2}{3}

, which simplifies to

y + 2 = 4

, so

y = 2

.

Step-by-Step Solution

1

Calculate the slope of segment

PQ

.

Slope of

PQ = \frac{2}{3}

Since parallel lines have equal slopes, finding the slope of the reference segment

PQ

is the first step in determining the slope of line

L

.

2

Set up the slope equation for line

L

using its parallel relationship to segment

PQ

.

\frac{y - (-2)}{9 - 3} = \frac{2}{3}

Because line

L

is parallel to segment

PQ

, its slope must also be

\frac{2}{3}

. The slope of line

L

is calculated using the points

(3, -2)

and

(9, y)

.

3

Solve the equation for

y

.

y = 2

Simplifying the numerator gives

y + 2

and the denominator gives

6

. Multiplying both sides of

\frac{y + 2}{6} = \frac{2}{3}

by

6

yields

y + 2 = 4

, which solves to

y = 2

.

Key Concept

Parallel lines have equal slopes in the coordinate plane.

Question 2623Question →

Given that $\cos\theta = \frac{4}{5}$ and the terminal side of angle $\theta$ lies in Quadrant IV, what is the value of $\tan\theta$ ?

-\frac{4}{3}

-\frac{3}{4}

-\frac{3}{5}

\frac{3}{4}

\frac{4}{3}

Show answer & explanation

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

Answer

-\frac{3}{4}

The correct answer is

-\frac{3}{4}

. Since the angle

\theta

has its terminal side in Quadrant IV, its cosine is positive and its sine is negative. Using the Pythagorean identity

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

, we find

\sin^2\theta + \left(\frac{4}{5}\right)^2 = 1

, which simplifies to

\sin^2\theta = \frac{9}{25}

. Because sine is negative in Quadrant IV,

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

. Finally, applying the quotient identity

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

, we get

\tan\theta = \frac{-3/5}{4/5} = -\frac{3}{4}

.

Step-by-Step Solution

1

Determine the sign of

\sin\theta

in Quadrant IV.

\sin\theta < 0

In Quadrant IV, the x-coordinates (representing cosine) are positive, and the y-coordinates (representing sine) are negative.

2

Use the Pythagorean identity

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

to calculate the value of

\sin\theta

.

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

Substituting

\cos\theta = \frac{4}{5}

gives

\sin^2\theta + \left(\frac{4}{5}\right)^2 = 1 \implies \sin^2\theta = 1 - \frac{16}{25} = \frac{9}{25}

. Taking the negative square root because sine is negative in Quadrant IV yields

-\frac{3}{5}

.

3

Use the quotient identity

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

to calculate

\tan\theta

.

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

Dividing the value of

\sin\theta

by

\cos\theta

yields

\frac{-3/5}{4/5} = -\frac{3}{4}

.

Key Concept

Fundamental Trigonometric Identities

Question 2624Question →

A dog is tied to a post in the center of a flat, grassy yard with a leash that is $12\text{ feet}$ long. If the dog walks along an arc formed by a central angle of $\frac{\pi}{3}$ radians, what is the length, in feet, of the path the dog travels?

2\pi

4\pi

8\pi

12\pi

24\pi

Show answer & explanation

Answer: $4\pi$

Answer

The correct option is the one stating

4\pi

.

The length of the path traveled by the dog is the arc length of a circle with a radius of

12

feet and a central angle of

\frac{\pi}{3}

radians. Using the radian arc length formula,

s = r\theta

, substituting

r = 12

and

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

yields

s = 12 \times \frac{\pi}{3} = 4\pi

feet.

Step-by-Step Solution

1

Identify the given values from the problem statement.

The radius of the circular path is

r = 12

feet, and the central angle is

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

radians.

These parameters are required to calculate the arc length of the path.

2

Recall the formula for the arc length of a circle when the angle is in radians.

The formula is

s = r\theta

.

Since the angle is given in radians, the arc length is directly the product of the radius and the angle.

3

Substitute the values into the formula and calculate the result.

s = 12 \times \frac{\pi}{3} = 4\pi

.

Multiplying the radius by the angle in radians gives the length of the path traveled.

Key Concept

Calculating the arc length of a circle when the central angle is measured in radians using the formula

s = r\theta

.

Question 2625Question →

In the standard $(x, y)$ coordinate plane, quadrilateral $ABCD$ is a kite that is not a rhombus, with $AB = BC$ and $AD = CD$ . The vertices $A$ and $C$ are located at $(2, 5)$ and $(8, 13)$ , respectively. If the vertex $B$ is located at $(9, 6)$ , which of the following could be the coordinates of vertex $D$ ?

(1, 12)

(-1, 17)

(-3, 15)

(-3, 3)

(-5, 10)

Show answer & explanation

Answer: $(-3, 15)$

Answer

The coordinates

(-3, 15)

The correct answer is the coordinates

(-3, 15)

. The diagonals of a kite are perpendicular, and the diagonal

BD

perpendicularly bisects the diagonal

AC

. The midpoint of

AC

is

M(5, 9)

and the slope of

AC

is

\frac{4}{3}

. Therefore, the line containing diagonal

BD

must pass through

M(5, 9)

with a perpendicular slope of

-\frac{3}{4}

. The equation of this line is

3x + 4y = 51

. Testing the coordinates

(-3, 15)

shows that the point lies on this line. Since the distance from

(-3, 15)

to

M(5, 9)

is

10

units while the distance from

B(9, 6)

to

M(5, 9)

is

5

units, the diagonals do not bisect each other, confirming the kite is not a rhombus.

Step-by-Step Solution

1

Find the midpoint

M

of diagonal

AC

.

M = \left(\frac{2 + 8}{2}, \frac{5 + 13}{2}\right) = (5, 9)

In a kite where

AB = BC

and

AD = CD

, the diagonal

BD

perpendicularly bisects the diagonal

AC

. Therefore, the line containing

BD

must pass through the midpoint of

AC

.

2

Calculate the slope of diagonal

AC

.

m_{AC} = \frac{13 - 5}{8 - 2} = \frac{8}{6} = \frac{4}{3}

The slope is needed to find the perpendicular slope of the line containing diagonal

BD

.

3

Determine the slope of the line containing diagonal

BD

.

m_{BD} = -\frac{1}{m_{AC}} = -\frac{3}{4}

Since the diagonals of a kite are perpendicular, the slope of the line containing diagonal

BD

is the negative reciprocal of the slope of diagonal

AC

.

4

Write the equation of the line containing diagonal

BD

and test the coordinates of vertex

D

.

y - 9 = -\frac{3}{4}(x - 5) \Rightarrow 3x + 4y = 51

Using the point-slope formula with the midpoint

M(5, 9)

and slope

-\frac{3}{4}

gives the line equation. Testing the correct option

(-3, 15)

gives

3(-3) + 4(15) = -9 + 60 = 51

, which lies on this line. Furthermore, the distance from

(-3, 15)

to

M(5, 9)

is

\sqrt{(-3-5)^2 + (15-9)^2} = 10

, which is different from the distance of

5

between

(9, 6)

and

M(5, 9)

, ensuring the kite is not a rhombus.

Key Concept

Diagonals of a kite are perpendicular, and the diagonal connecting the vertices between the unequal adjacent sides perpendicularly bisects the other diagonal.

Question 2626Question →

For each angle or terminal ray described on the left, match it to the correct coordinates $(x, y)$ of its intersection with the unit circle in the standard coordinate plane on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches associate the angle of 7pi/6 radians with coordinates (-sqrt(3)/2, -1/2); the angle of 5pi/4 radians with coordinates (-sqrt(2)/2, -sqrt(2)/2); the angle of pi/3 radians with coordinates (1/2, sqrt(3)/2); and the y-axis reflected angle of 4pi/3 radians with coordinates (1/2, -sqrt(3)/2).

Each description on the left simplifies to a unique angle in standard position on the unit circle. The coordinates on the right represent the exact

(\cos\theta, \sin\theta)

values for each corresponding angle: the angle of

\frac{7\pi}{6}

radians corresponds to coordinates

\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)

; the angle of

\frac{5\pi}{4}

radians corresponds to coordinates

\left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)

; the angle of

\frac{\pi}{3}

radians corresponds to coordinates

\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)

; and the terminal ray symmetric to the terminal ray of

\frac{4\pi}{3}

radians across the

y

-axis corresponds to coordinates

\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)

.

Step-by-Step Solution

1

Analyze the first angle description by converting all angle measures to a common unit (radians) and computing the net rotation.

For the first angle,

570^\circ = \frac{19\pi}{6}

radians, and the net rotation is

\frac{13\pi}{3} - \frac{19\pi}{6} = \frac{7\pi}{6}

radians.

Converting degrees to radians and subtracting the clockwise rotation from the counterclockwise rotation simplifies the angle to a standard radian value.

2

For the second angle, express the starting point as an angle in radians and add the given counterclockwise rotation, then find the coterminal angle in

[0, 2\pi)

.

Starting at

(0, -1)

is equivalent to

\frac{3\pi}{2}

radians. Total rotation is

\frac{3\pi}{2} + \frac{23\pi}{4} = \frac{29\pi}{4}

radians, which is coterminal with

\frac{29\pi}{4} - 6\pi = \frac{5\pi}{4}

radians.

Using the standard position angle of the starting point allows us to sum the rotations and determine the final terminal ray position.

3

Determine the coterminal angle for

-1020^\circ

in the interval

[0, 360^\circ)

and convert it to radians.

-1020^\circ + 1080^\circ = 60^\circ

, which equals

\frac{\pi}{3}

radians.

Adding multiples of

360^\circ

finds the equivalent positive angle within one full revolution.

4

Find the coordinates of the terminal ray of

\frac{4\pi}{3}

radians on the unit circle, reflect the point across the

y

-axis, and determine the coordinates of the resulting point.

The coordinate point of

\frac{4\pi}{3}

is

\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)

. Reflecting this point across the

y

-axis yields

\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)

.

Symmetry across the

y

-axis negates the

x

-coordinate of the point on the unit circle.

5

Match each of the simplified angles to their corresponding standard coordinates

(x, y) = (\cos\theta, \sin\theta)

on the unit circle.

The angle

\frac{7\pi}{6}

matches

\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)

;

\frac{5\pi}{4}

matches

\left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)

;

\frac{\pi}{3}

matches

\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)

; and the reflected terminal ray matches

\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)

.

Evaluating the sine and cosine functions at each angle yields the final coordinates.

Key Concept

Identifying terminal coordinates of angles on the unit circle by converting between degrees and radians, calculating coterminal angles, and applying coordinate symmetries.

Estimated Time:3m 0s

Question 2627Question →

In the standard $(x, y)$ coordinate plane, a right triangle has vertices $A(-2, -3)$ , $B(6, -3)$ , and $C(6, y)$ , where $y > 0$ . If the hypotenuse of the triangle has a length of 17 units, what is the value of $y$ ?

6

12

15

18

14

Show answer & explanation

Answer: 12

Answer

The correct value of

y

is

12

.

The horizontal leg

AB

has a length of

6 - (-2) = 8

units. By the Pythagorean theorem, the vertical leg

BC

has a length of

\sqrt{17^2 - 8^2} = 15

units. Since the vertex

B

is at

(6, -3)

and

C

is at

(6, y)

with

y > 0

, the vertical distance is

y - (-3) = 15

, which yields

y = 12

.

Step-by-Step Solution

1

Determine the length of the horizontal leg

AB

.

The length of

AB

is

8

units.

Since vertices

A(-2, -3)

and

B(6, -3)

share the same

y

-coordinate, the segment is horizontal. The length is the positive difference between their

x

-coordinates:

6 - (-2) = 8

.

2

Use the Pythagorean theorem to calculate the length of the vertical leg

BC

.

The length of

BC

is

15

units.

Since

AB

is horizontal and

BC

is vertical (vertices

B

and

C

share the same

x

-coordinate of

6

), the angle at

B

is a right angle. The hypotenuse is

AC = 17

. By the Pythagorean theorem,

AB^2 + BC^2 = AC^2 \implies 8^2 + BC^2 = 17^2 \implies 64 + BC^2 = 289 \implies BC^2 = 225 \implies BC = 15

.

3

Set up an equation using the coordinates to find

y

.

y = 12

The length of the vertical segment

BC

is the difference in

y

-coordinates:

|y - (-3)| = 15

, which simplifies to

y + 3 = 15

since

y > 0

. Solving for

y

yields

y = 12

.

Key Concept

Using coordinate differences and the Pythagorean theorem to determine unknown vertices of geometric figures in the coordinate plane.

Alternative Method

Alternatively, you can apply the distance formula directly between the vertices

A(-2, -3)

and

C(6, y)

with a distance of 17 units:

\sqrt{(6 - (-2))^2 + (y - (-3))^2} = 17 \implies 8^2 + (y + 3)^2 = 289 \implies (y + 3)^2 = 225

, which yields

y + 3 = 15 \implies y = 12

since

y > 0

.

Estimated Time:1m 30s

Question 2628Question →

A vertical flagpole is secured by two straight guide wires anchored to the flat ground on opposite sides of the pole. The first guide wire is $13$ feet long and its anchor is $5$ feet from the base of the pole. The second guide wire is anchored such that it makes a $30^\circ$ angle of elevation with the ground. If both guide wires are attached to the flagpole at the same height, what is the length, in feet, of the second guide wire?

Show answer & explanation

Answer: 24

Answer

The length of the second guide wire is

24

feet.

The correct answer is

24

. First, the height of the attachment point is found using the Pythagorean theorem:

h = \sqrt{13^2 - 5^2} = 12

feet. Since the second wire makes a

30^\circ

angle of elevation with the ground, it forms a 30-60-90 right triangle where the flagpole height of

12

feet is the leg opposite the

30^\circ

angle. The length of the wire is the hypotenuse of this triangle, which is twice the length of the opposite leg:

2 \times 12 = 24

feet.

Step-by-Step Solution

1

Use the Pythagorean theorem to calculate the height of the flagpole where the guide wires are attached.

The flagpole height is

12

feet.

The first guide wire, the flagpole, and the ground form a right triangle with a hypotenuse of

13

feet and a horizontal leg of

5

feet. Thus,

h = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = 12

.

2

Apply the properties of a 30-60-90 special right triangle to find the length of the second guide wire.

The length of the second guide wire is

24

feet.

The second wire forms a 30-60-90 right triangle with the flagpole and the ground. The angle of elevation is

30^\circ

, which means the side opposite this angle is the vertical height of the flagpole (

12

feet). In a 30-60-90 triangle, the hypotenuse (the wire length) is twice the length of the shorter leg (opposite the

30^\circ

angle), so the length is

2 \times 12 = 24

.

Key Concept

Applying the Pythagorean Theorem and the ratio properties of 30-60-90 special right triangles to solve multi-step geometry problems.

Question 2629Question →

A goat is tethered to one of the outer corners of a flat, rectangular shed that measures $6\text{ meters}$ by $8\text{ meters}$ . The tether is $10\text{ meters}$ long. Assuming the goat remains outside the shed, what is the total area, in square meters, of the region the goat can graze?

75\pi

83\pi

80\pi

90\pi

100\pi

Show answer & explanation

Answer: $80\pi$

Answer

The correct grazing area is

80\pi

square meters.

The total grazing area is the sum of three sectors: a sector of radius

10\text{ meters}

with central angle

270^\circ

(area

75\pi

), a sector of radius

4\text{ meters}

with central angle

90^\circ

(area

4\pi

), and a sector of radius

2\text{ meters}

with central angle

90^\circ

(area

\pi

). Adding these yields

80\pi

square meters.

Step-by-Step Solution

1

Determine the area of the main sector.

The main sector has a radius of

10\text{ meters}

and a central angle of

360^\circ - 90^\circ = 270^\circ

. The area is

\frac{270}{360} \times \pi \times 10^2 = 75\pi

square meters.

The corner of the rectangular shed blocks

90^\circ

of a full circle, leaving a

270^\circ

sector.

2

Determine the area of the sector at the corner adjacent to the

6

-meter side.

The tether wraps around the corner, leaving a remaining length of

10 - 6 = 4\text{ meters}

. It sweeps through an angle of

90^\circ

. The area is

\frac{90}{360} \times \pi \times 4^2 = 4\pi

square meters.

When the tether extends past the adjacent corner along the

6

-meter side, the pivot point becomes that corner and the tether length decreases by the side length.

3

Determine the area of the sector at the corner adjacent to the

8

-meter side.

The tether wraps around the corner, leaving a remaining length of

10 - 8 = 2\text{ meters}

. It sweeps through an angle of

90^\circ

. The area is

\frac{90}{360} \times \pi \times 2^2 = \pi

square meters.

When the tether extends past the adjacent corner along the

8

-meter side, the pivot point becomes that corner and the tether length decreases by the side length.

4

Check for overlap and calculate the total grazing area.

The total area is

75\pi + 4\pi + \pi = 80\pi

square meters.

The two smaller sectors are at different corners of the rectangular shed and do not overlap. Summing the three sector areas gives the total grazing region.

Key Concept

Calculating sector areas by determining the correct radii and central angles based on geometric constraints.

Question 2630Question →

An angle in standard position measures $-468^\circ$ . What is the radian measure of the coterminal angle that lies in the interval $[0, 2\pi)$ ?

\frac{5\pi}{7}

\frac{7\pi}{5}

\frac{\pi}{6}

\frac{3\pi}{5}

\frac{9\pi}{5}

Show answer & explanation

Answer: $\frac{7\pi}{5}$

Answer

The correct radian measure of the coterminal angle is

\frac{7\pi}{5}

.

To find the correct radian measure, first determine the positive coterminal angle in degrees by adding multiples of

360^\circ

until the angle lies in the interval

[0^\circ, 360^\circ)

. Adding

720^\circ

to

-468^\circ

results in

252^\circ

. Convert this angle to radians by multiplying by

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

, which yields

\frac{252\pi}{180}

. Dividing the numerator and denominator by their greatest common divisor,

36

, simplifies the expression to

\frac{7\pi}{5}

radians.

Step-by-Step Solution

1

Find a positive coterminal angle in degrees by adding multiples of

360^\circ

to the initial angle.

-468^\circ + 360^\circ = -108^\circ

, and

-108^\circ + 360^\circ = 252^\circ

.

Adding

720^\circ

(two full rotations) to

-468^\circ

shifts the angle into the standard positive range

[0^\circ, 360^\circ)

while maintaining the same terminal ray.

2

Convert the coterminal angle from degrees to radians by multiplying by the conversion factor

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

.

252^\circ \times \frac{\pi}{180^\circ} = \frac{252\pi}{180}

radians.

Since

\pi

radians corresponds to

180^\circ

, multiplying by

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

changes the unit of measure from degrees to radians.

3

Simplify the fraction

\frac{252\pi}{180}

by dividing both the numerator and the denominator by their greatest common divisor.

Dividing

252

and

180

by their greatest common divisor of

36

yields

\frac{7\pi}{5}

radians.

Simplifying the fraction expresses the final radian measure in its standard, reduced form.

Key Concept

Finding positive coterminal angles and converting degree measures to radian measures on the unit circle.

Alternative Method

Convert the initial angle of

-468^\circ

directly to radians first by multiplying by

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

, yielding

-\frac{13\pi}{5}

radians. To find the positive coterminal angle in the interval

[0, 2\pi)

, add multiples of

2\pi

radians (which is

\frac{10\pi}{5}

):

-\frac{13\pi}{5} + \frac{10\pi}{5} = -\frac{3\pi}{5}

, and then

-\frac{3\pi}{5} + \frac{10\pi}{5} = \frac{7\pi}{5}

radians.

Estimated Time:2m 0s

Question 2631Question →

Passage

The Settlement Movement and the Foundations of Hull House

In the late nineteenth century, rapid industrialization and urbanization in the United States led to overcrowded cities and widespread poverty among working-class immigrant populations. In Chicago, Illinois, this reality was particularly acute, as thousands of new arrivals sought work in the city's factories and meatpacking plants. It was against this backdrop of social inequality that Jane Addams, a young woman from a wealthy Illinois family, embarked on a mission to bridge the gap between different social classes.

Addams’s journey toward social reform began in earnest in 1883 during a tour of Europe. During her travels, she witnessed the stark contrast between the lives of the wealthy elite and the urban poor. However, the pivotal moment of her journey occurred in 1887 when she visited Toynbee Hall in London's East End. Toynbee Hall was the world's first university settlement house, a place where wealthy university graduates lived and worked alongside local residents in an effort to alleviate poverty and share cultural resources. Inspired by this model, Addams resolved to establish a similar institution in the United States.

Upon returning to Chicago in 1889, Addams and her close friend Ellen Gates Starr began searching for a suitable location in the city's crowded West Side. They eventually located a large, run-down mansion that had been built in 1856 by a wealthy real estate developer named Charles Hull. The mansion, located at the corner of Halsted and Polk Streets, was surrounded by tenements, factories, and saloons, making it the ideal location for a settlement project. Addams and Starr leased the home, which they named Hull House in honor of its original owner, and moved in to begin their work.

In September 1889, Hull House officially opened its doors to the public. Initially, Addams and Starr focused on addressing the immediate practical needs of the neighborhood's immigrant residents. They established a nursery and a kindergarten to care for the young children of working mothers, and they organized social clubs and reading groups for older children and adults. The response from the community was overwhelming, and the house quickly became a bustling center of neighborhood life.

As the needs of the neighborhood grew, so did the facilities of Hull House. In 1891, the settlement house expanded its physical footprint with the construction of the Butler Art Gallery, which provided local residents with access to art exhibitions and lectures. A few years later, in 1893, Hull House added a public playground, which was the first of its kind in the city of Chicago. By 1895, the complex had grown to include a cooperative residence for young working women, known as the Jane Club, which offered affordable housing and a supportive community.

Through its rapid growth, Hull House demonstrated the effectiveness of the settlement house model. Addams and her colleagues did not merely provide charity; they lived as neighbors to those they served, working together to advocate for child labor laws, housing reform, and sanitation improvements. By the early twentieth century, Hull House had evolved from a single rented mansion into a sprawling thirteen-building complex, serving as a national model for social reform and community empowerment.

Based on the passage, arrange the following events in the chronological order in which they occurred, from the earliest to the latest.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct chronological order of the events is: Jane Addams visits Toynbee Hall in London's East End, followed by Jane Addams and Ellen Gates Starr leasing the mansion built by Charles Hull, then the construction of the Butler Art Gallery, and finally the addition of a public playground to the Hull House complex.

The correct chronological order of the events is explicitly provided in the passage with specific dates: Jane Addams visited Toynbee Hall in 1887; she and Ellen Gates Starr leased the Charles Hull mansion in 1889; the Butler Art Gallery was constructed in 1891; and the public playground was added in 1893.

Step-by-Step Solution

1

Locate the mention of Jane Addams visiting Toynbee Hall in the text.

The text states she visited Toynbee Hall in 1887.

This establishes the year for the first event.

2

Locate the details about leasing the Charles Hull mansion.

The text states she and Ellen Gates Starr returned to Chicago and leased the mansion in 1889.

This establishes the year for the second event.

3

Locate the details regarding the Butler Art Gallery.

The text explicitly states the gallery was constructed in 1891.

This establishes the year for the third event.

4

Locate the details regarding the public playground.

The text explicitly states the playground was added in 1893.

This establishes the year for the final event.

Key Concept

Identifying Explicit Details

Question 2632Question →

A projectile is launched vertically upward from an initial height of $5$ meters. Its height, $h(t)$ in meters, $t$ seconds after launch is given by the function $h(t) = -4.9t^2 + 19.6t + 5$ . To the nearest tenth of a second, how many seconds after launch does the projectile reach a height of $15$ meters on its way down?

Show answer & explanation

Answer: 3.4

Answer

To the nearest tenth of a second, the projectile reaches a height of

15

meters on its way down at

3.4

seconds.

The correct answer is

3.4

seconds. Setting the height equation

h(t) = 15

yields

-4.9t^2 + 19.6t - 10 = 0

. Solving this quadratic equation via the quadratic formula gives two solutions:

t \approx 0.6

seconds and

t \approx 3.4

seconds. The projectile travels upward first, passing the

15

-meter mark at

0.6

seconds, and then descends, passing the

15

-meter mark again at

3.4

seconds.

Step-by-Step Solution

1

Set up the quadratic equation by setting the height function

h(t)

equal to

15

.

-4.9t^2 + 19.6t + 5 = 15

This allows us to find the specific values of time

t

when the height of the projectile is exactly

15

meters.

2

Rearrange the quadratic equation into the standard form

at^2 + bt + c = 0

by subtracting

15

from both sides.

-4.9t^2 + 19.6t - 10 = 0

Writing the equation in standard form is necessary before applying the quadratic formula.

3

Substitute the coefficients

a = -4.9

,

b = 19.6

, and

c = -10

into the quadratic formula.

t = \frac{-19.6 \pm \sqrt{(19.6)^2 - 4(-4.9)(-10)}}{2(-4.9)}

Since the quadratic equation has non-integer decimal coefficients, using the quadratic formula is the most reliable method to solve for the roots.

4

Simplify the discriminant and calculate the two values of

t

.

t \approx 0.6

and

t \approx 3.4

The discriminant is

19.6^2 - 196 = 188.16

. Taking the square root gives

\sqrt{188.16} \approx 13.72

, resulting in two real roots.

5

Determine which root corresponds to the projectile's motion on the way down.

t \approx 3.4

seconds

The smaller root (

0.6

seconds) represents the first time the projectile reaches

15

meters while ascending. The larger root (

3.4

seconds) represents the time the projectile passes

15

meters while descending.

Key Concept

Solving quadratic equations with decimal coefficients using the quadratic formula and interpreting the physical context of the roots.

Question 2633Question →

A circle has a radius of $12\text{ centimeters}$ . What is the length, in centimeters, of the arc intercepted by a central angle of $30^\circ$ ?

\pi

2\pi

12\pi

24\pi

2

Show answer & explanation

Answer: $2\pi$

Answer

2\pi

centimeters

To find the arc length, multiply the total circumference of the circle,

2\pi r

, by the fraction of the circle represented by the central angle,

\frac{\theta}{360^\circ}

. With a radius of

12

centimeters and a central angle of

30^\circ

, this calculation yields

2\pi (12) \times \frac{30}{360} = 24\pi \times \frac{1}{12} = 2\pi

centimeters.

Step-by-Step Solution

1

Identify the formula for arc length:

s = 2\pi r \left(\frac{\theta}{360^\circ}\right)

, where

r

is the radius and

\theta

is the central angle in degrees.

Formula established:

s = 2\pi r \left(\frac{\theta}{360^\circ}\right)

Arc length is the fraction of the total circumference determined by the central angle.

2

Substitute the given radius

r = 12\text{ cm}

and central angle

\theta = 30^\circ

into the formula.

Equation set up:

s = 2\pi (12) \left(\frac{30}{360}\right)

This sets up the specific calculation for the given circle.

3

Simplify the expression to find the final arc length.

s = 24\pi \left(\frac{1}{12}\right) = 2\pi

Simplification yields the exact length of the arc in terms of

\pi

.

Key Concept

The length of an arc is proportional to the fraction of the circle's circumference represented by the central angle.

Estimated Time:1m 0s

Question 2634Question →

In rhombus $ABCD$ , the perimeter is $100$ and the length of diagonal $BD$ is $30$ . Point $P$ lies on diagonal $AC$ such that the ratio of the length of segment $AP$ to the length of segment $PC$ is $3:7$ . What is the length of segment $BP$ ?

Show answer & explanation

Answer: 17

Answer

The length of segment

BP

is 17.

The correct answer is found by utilizing the properties of a rhombus. A rhombus has four congruent sides, meaning each side of a rhombus with perimeter

100

has a length of

25

. The diagonals of a rhombus are perpendicular bisectors of one another. Letting

O

be the intersection of the diagonals, we find

BO = 15

since diagonal

BD = 30

. Using the Pythagorean theorem on right triangle

AOB

, we determine that the other half-diagonal is

AO = \sqrt{25^2 - 15^2} = 20

, which means the full diagonal

AC = 40

. Point

P

divides

AC

in the ratio

3:7

, meaning

AP = 12

and

PC = 28

. The distance from

P

to the intersection point

O

is

OP = AO - AP = 20 - 12 = 8

. Finally, applying the Pythagorean theorem to right triangle

BOP

with legs

BO = 15

and

OP = 8

yields

BP = \sqrt{15^2 + 8^2} = 17

.

Step-by-Step Solution

1

Calculate the side length of rhombus

ABCD

from its perimeter.

Each side length is

25

.

A rhombus has four equal sides, so the side length is the perimeter divided by four:

\frac{100}{4} = 25

.

2

Find the length of half of diagonal

BD

.

BO = 15

, where

O

is the intersection of diagonals

AC

and

BD

.

The diagonals of a rhombus bisect each other.

3

Calculate the half-diagonal length

AO

and full diagonal length

AC

.

AO = 20

and

AC = 40

.

The diagonals of a rhombus are perpendicular, forming right triangle

AOB

. By the Pythagorean theorem,

AO = \sqrt{AB^2 - BO^2} = \sqrt{25^2 - 15^2} = 20

. Since the diagonals bisect each other, the total length of diagonal

AC

is

2 \times 20 = 40

.

4

Determine the length of segment

AP

.

AP = 12

.

Point

P

lies on diagonal

AC

such that the ratio of segment

AP

to

PC

is

3:7

. Therefore,

AP = \frac{3}{3+7} \times AC = \frac{3}{10} \times 40 = 12

.

5

Find the distance

OP

between point

P

and the intersection point

O

.

OP = 8

.

Since

AO = 20

and

P

is

12

units from

A

,

P

lies on the segment

AO

. Thus, the distance from

P

to

O

is

OP = AO - AP = 20 - 12 = 8

.

6

Calculate the length of segment

BP

.

BP = 17

.

Because the diagonals of a rhombus are perpendicular,

\triangle BOP

is a right triangle with legs

BO = 15

and

OP = 8

. Using the Pythagorean theorem,

BP = \sqrt{BO^2 + OP^2} = \sqrt{15^2 + 8^2} = 17

.

Key Concept

Properties of Rhombuses (perpendicular bisecting diagonals, equal side lengths) and the Pythagorean Theorem

Estimated Time:2m 0s

Question 2635Question →

A circular garden plot has a radius of $10\text{ meters}$ . A sector of the garden with a central angle of $72^\circ$ is planted with roses. What is the area, in square meters, of the sector planted with roses?

4\pi

40\pi

20\pi

72\pi

100\pi

Show answer & explanation

Answer: $20\pi$

Answer

The correct answer is

20\pi

, representing the area of the sector in square meters.

The area of a sector of a circle is calculated using the formula

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

. Substituting

\theta = 72^\circ

and

r = 10\text{ meters}

gives

\frac{72^\circ}{360^\circ} \pi (10)^2 = \frac{1}{5} \cdot 100\pi = 20\pi

.

Step-by-Step Solution

1

Identify the formula for the area of a sector.

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

The area of a sector is a proportional fraction of the total area of the circle.

2

Substitute the given values into the formula.

\text{Area} = \frac{72^\circ}{360^\circ} \pi (10)^2

The central angle is

72^\circ

and the radius is

10\text{ meters}

.

3

Simplify the fraction and the squared term.

\text{Area} = \frac{1}{5} \pi (100)

72/360

simplifies to

1/5

and

10^2 = 100

.

4

Perform the final multiplication.

20\pi

One-fifth of

100

is

20

.

Key Concept

Calculating the area of a circle sector given the radius and the central angle in degrees.

Estimated Time:1m 0s

Question 2636Question →

For a certain convex polygon, the ratio of the sum of the interior angles to the sum of the exterior angles is $5:1$ . How many sides does this polygon have?

7

8

10

12

14

Show answer & explanation

Answer: 12

Answer

The correct answer is 12 sides.

The sum of the interior angles of any convex polygon with

n

sides is given by

(n-2) \times 180^\circ

. The sum of the exterior angles of any convex polygon is always

360^\circ

. Since the ratio of the sum of the interior angles to the sum of the exterior angles is

5:1

, we can set up the equation:

\frac{(n-2) \times 180^\circ}{360^\circ} = \frac{5}{1}

. Simplifying the fraction on the left gives

\frac{n-2}{2} = 5

. Multiplying both sides by 2 gives

n - 2 = 10

, and adding 2 to both sides gives

n = 12

. Therefore, the polygon has 12 sides.

Step-by-Step Solution

1

Write the formulas for the sum of the interior angles and the sum of the exterior angles of a convex polygon.

The sum of the interior angles is

(n-2) \times 180^\circ

, where

n

is the number of sides. The sum of the exterior angles is always

360^\circ

for any convex polygon.

These formulas represent the components of the given ratio.

2

Set up the ratio equation using the given information.

\frac{(n-2) \times 180^\circ}{360^\circ} = \frac{5}{1}

The problem states that the ratio of the sum of the interior angles to the sum of the exterior angles is

5:1

.

3

Simplify the equation and solve for the number of sides,

n

.

\frac{n-2}{2} = 5 \implies n-2 = 10 \implies n = 12

Simplifying

\frac{180}{360}

to

\frac{1}{2}

makes it easier to solve the algebraic equation for

n

.

Key Concept

The sum of the interior angles of a convex polygon with

n

sides is

(n-2) \times 180^\circ

, while the sum of the exterior angles is always

360^\circ

.

Estimated Time:1m 30s

Question 2637Question →

Match each of the trigonometric functions listed on the left with the correct description of its amplitude and period listed on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The function

y = 3\sin(2x)

matches the description stating 'Amplitude is 3 and period is

\pi

'. The function

y = 2\cos(3x)

matches the description stating 'Amplitude is 2 and period is

\frac{2\pi}{3}

'. The function

y = 4\sin(\pi x)

matches the description stating 'Amplitude is 4 and period is 2'.

Each trigonometric function of the form

y = a\sin(bx)

or

y = a\cos(bx)

has an amplitude equal to the absolute value of the coefficient of the trigonometric term (

|a|

) and a period equal to

2\pi

divided by the absolute value of the coefficient of the angle variable (

|b|

). Applying these formulas gives the correct properties for each function.

Step-by-Step Solution

1

Identify the general form of the trigonometric functions

The functions are in the form

y = a\sin(bx)

or

y = a\cos(bx)

, where the amplitude is given by the absolute value of the vertical stretch coefficient (

|a|

), and the period is calculated as

\frac{2\pi}{|b|}

.

This establishes the formulas needed to determine the amplitude and period for each equation.

2

Calculate the properties for

y = 3\sin(2x)

The vertical stretch coefficient is 3, so the amplitude is 3. The frequency coefficient is 2, so the period is

\frac{2\pi}{2} = \pi

.

To find the amplitude and period for the first function.

3

Calculate the properties for

y = 2\cos(3x)

The vertical stretch coefficient is 2, so the amplitude is 2. The frequency coefficient is 3, so the period is

\frac{2\pi}{3}

.

To find the amplitude and period for the second function.

4

Calculate the properties for

y = 4\sin(\pi x)

The vertical stretch coefficient is 4, so the amplitude is 4. The frequency coefficient is

\pi

, so the period is

\frac{2\pi}{\pi} = 2

.

To find the amplitude and period for the third function.

Key Concept

Identifying the amplitude and calculating the period of trigonometric functions from their equations

Estimated Time:1m 30s

Question 2638Question →

In the standard $(x, y)$ coordinate plane, three vertices of a rectangle are $A(-4, 1)$ , $B(2, 9)$ , and $C(6, 6)$ . What is the area of the rectangle, in square units?

Show answer & explanation

Answer: 50

Answer

50

The length of side

AB

is calculated as

\sqrt{(2 - (-4))^2 + (9 - 1)^2} = \sqrt{36 + 64} = 10

. The length of the adjacent side

BC

is calculated as

\sqrt{(6 - 2)^2 + (6 - 9)^2} = \sqrt{16 + 9} = 5

. The area of the rectangle is the product of these two perpendicular side lengths, which is

10 \times 5 = 50

.

Step-by-Step Solution

1

Calculate the length of side

AB

using the distance formula.

AB = 10

To find one of the side lengths of the rectangle.

2

Calculate the length of side

BC

using the distance formula.

BC = 5

To find the adjacent side length of the rectangle.

3

Multiply the two adjacent side lengths to find the area of the rectangle.

50

The area of a rectangle is equal to the product of its length and width.

Key Concept

Calculating the area of a geometric figure on the coordinate plane by determining its side lengths using the distance formula.

Question 2639Question →

A commercial drone starts at an altitude of $150$ meters and descends at a constant rate. After $12$ seconds of descent, its altitude is $114$ meters. If the altitude, $a$ , in meters, is modeled as a linear function of time, $t$ , in seconds, what is the slope of the line representing this relationship in the standard $(t, a)$ coordinate plane?

-36

-\frac{1}{3}

-3

\frac{1}{3}

3

Show answer & explanation

Answer: $-3$

Answer

The slope of the line representing this relationship is

-3

.

The slope of the line represents the rate of change of the altitude with respect to time. By taking the initial point

(0, 150)

and the point after

12

seconds

(12, 114)

, the change in altitude is

114 - 150 = -36

meters. Dividing this by the change in time of

12 - 0 = 12

seconds yields the correct slope of

-3

.

Step-by-Step Solution

1

Identify the coordinate points

(t, a)

from the given information.

The initial state corresponds to the point

(0, 150)

, and the state after

12

seconds corresponds to

(12, 114)

.

Setting up coordinate pairs allows the direct application of the slope formula.

2

Apply the slope formula

m = \frac{a_2 - a_1}{t_2 - t_1}

to find the rate of change.

m = \frac{114 - 150}{12 - 0}

The slope is the change in the vertical variable (altitude) divided by the change in the horizontal variable (time).

3

Calculate the difference and simplify the fraction.

m = \frac{-36}{12} = -3

Subtracting

150

from

114

gives the net change in altitude, which when divided by the duration of

12

seconds yields the rate of change per second.

Key Concept

The slope of a line represents its constant rate of change, defined as the ratio of the change in the dependent variable to the change in the independent variable.

Question 2640Question →

What are the solutions for $x$ in the quadratic equation $0.2x^2 - 0.6x + 2.2 = 0$ ?

\frac{3 \pm \sqrt{35}}{2}

\frac{3 \pm \sqrt{55}}{2}

\frac{3 \pm i\sqrt{35}}{2}

\pm\frac{i\sqrt{35}}{2}

\frac{15 \pm 2i\sqrt{10}}{10}

Show answer & explanation

Answer: $\frac{3 \pm i\sqrt{35}}{2}$

Answer

The correct answer is

\frac{3 \pm i\sqrt{35}}{2}

.

The correct answer is the pair of complex solutions

\frac{3 \pm i\sqrt{35}}{2}

. By multiplying the equation

0.2x^2 - 0.6x + 2.2 = 0

by

5

, we get the equivalent equation with integer coefficients,

x^2 - 3x + 11 = 0

. Applying the quadratic formula with

a = 1

,

b = -3

, and

c = 11

yields a discriminant of

9 - 44 = -35

. Since the discriminant is negative, the solutions are complex:

\frac{3 \pm i\sqrt{35}}{2}

.

Step-by-Step Solution

1

Multiply both sides of the quadratic equation by

5

to eliminate the decimal coefficients.

The equation

0.2x^2 - 0.6x + 2.2 = 0

becomes

x^2 - 3x + 11 = 0

.

Working with integer coefficients simplifies algebraic manipulation and reduces the risk of calculation errors.

2

Identify the coefficients

a

,

b

, and

c

in the standard quadratic form

ax^2 + bx + c = 0

.

a = 1

,

b = -3

, and

c = 11

.

These values are required to apply the quadratic formula.

3

Substitute the coefficients into the quadratic formula,

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

.

x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(11)}}{2(1)} = \frac{3 \pm \sqrt{9 - 44}}{2} = \frac{3 \pm \sqrt{-35}}{2}

.

The quadratic formula provides the exact solutions for any quadratic equation.

4

Simplify the radical using the imaginary unit,

i = \sqrt{-1}

.

\sqrt{-35} = \sqrt{35} \cdot \sqrt{-1} = i\sqrt{35}

, so the solutions are

x = \frac{3 \pm i\sqrt{35}}{2}

.

Standard mathematical notation represents the square root of a negative number using

i

.

Key Concept

Using the quadratic formula to solve quadratic equations with decimal coefficients and complex roots.

Answer

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