All practice questions

Question 1441Question →

If $a$ is a real number greater than $1$ such that $a^x - a^{-x} = 2$ , what is the value of the expression $\frac{a^{3x} + a^{-3x}}{a^{2x} + a^{-2x}}$ ?

\frac{5\sqrt{2}}{3}

\frac{4}{3}

2

\frac{5\sqrt{6}}{6}

\frac{5}{3}

Show answer & explanation

Answer: $\frac{5\sqrt{2}}{3}$

Answer

\frac{5\sqrt{2}}{3}

Squaring the given relationship

a^x - a^{-x} = 2

gives

a^{2x} - 2 + a^{-2x} = 4

, which simplifies to

a^{2x} + a^{-2x} = 6

. Adding 4 to both sides gives

(a^x + a^{-x})^2 = 8

, so

a^x + a^{-x} = 2\sqrt{2}

. By the sum of cubes identity,

a^{3x} + a^{-3x} = (a^x + a^{-x})(a^{2x} + a^{-2x} - 1) = 2\sqrt{2}(6 - 1) = 10\sqrt{2}

. Dividing

10\sqrt{2}

by

6

gives the simplified result

\frac{5\sqrt{2}}{3}

.

Step-by-Step Solution

1

Square both sides of the given equation

a^x - a^{-x} = 2

.

(a^x - a^{-x})^2 = a^{2x} - 2(a^x)(a^{-x}) + a^{-2x} = 4 \implies a^{2x} + a^{-2x} = 6

.

Expanding the binomial square allows us to find the denominator

a^{2x} + a^{-2x}

directly.

2

Determine the value of

a^x + a^{-x}

.

(a^x + a^{-x})^2 = a^{2x} + 2 + a^{-2x} = 6 + 2 = 8 \implies a^x + a^{-x} = \sqrt{8} = 2\sqrt{2}

.

Since

a > 1

,

a^x > 0

and

a^{-x} > 0

, their sum must be positive.

3

Use the sum of cubes identity

u^3 + v^3 = (u + v)(u^2 - uv + v^2)

to evaluate

a^{3x} + a^{-3x}

.

a^{3x} + a^{-3x} = (a^x + a^{-x})(a^{2x} - 1 + a^{-2x}) = (2\sqrt{2})(6 - 1) = 10\sqrt{2}

.

Factoring the numerator breaks it into terms whose numerical values are known.

4

Compute the ratio of numerator to denominator.

\frac{a^{3x} + a^{-3x}}{a^{2x} + a^{-2x}} = \frac{10\sqrt{2}}{6} = \frac{5\sqrt{2}}{3}.

Simplify the fraction by dividing numerator and denominator by 2.

Key Concept

Algebraic transformations of exponential expressions using polynomial identities.

Estimated Time:2m 30s

Question 1442Question →

In a circle centered at point $O$ with a radius of $6$ units, points $P$ and $Q$ lie on the circle such that central angle $\angle POQ$ measures $120^\circ$ . Which of the following statements are true? Select all such statements.

Select all that apply

The length of minor arc

PQ

is

4\pi

.

The area of minor sector

POQ

is

12\pi

.

The perimeter of sector

POQ

is

4\pi

.

The area of sector

POQ

is

36\pi

.

The length of chord

PQ

is

6\sqrt{3}

.

Show answer & explanation

Answer: The length of minor arc $PQ$ is $4\pi$ .; The area of minor sector $POQ$ is $12\pi$ .; The length of chord $PQ$ is $6\sqrt{3}$ .

Answer

The true statements are that the length of minor arc

PQ

is

4\pi

, the area of minor sector

POQ

is

12\pi

, and the length of chord

PQ

is

6\sqrt{3}

.

The arc length of a

120^\circ

central angle is

\frac{1}{3}

of the total circumference

12\pi

, which gives

4\pi

. The sector area is

\frac{1}{3}

of the total area

36\pi

, which gives

12\pi

. Furthermore, by geometry of an isosceles triangle with vertex angle

120^\circ

and congruent sides

6

, the base chord length is

2 \times 6 \sin(60^\circ) = 6\sqrt{3}

.

Step-by-Step Solution

1

Calculate the arc length of minor arc

PQ

Arc length =

4\pi

Arc length equals

\frac{\theta}{360^\circ} \times 2\pi r = \frac{120^\circ}{360^\circ} \times 2\pi(6) = 4\pi

.

2

Calculate the area of sector

POQ

Sector area =

12\pi

Sector area equals

\frac{\theta}{360^\circ} \times \pi r^2 = \frac{120^\circ}{360^\circ} \times \pi(6^2) = 12\pi

.

3

Determine the perimeter of sector

POQ

Perimeter =

4\pi + 12

The boundary of a sector consists of the curved arc length plus two straight radii:

4\pi + 6 + 6 = 4\pi + 12

.

4

Calculate the length of chord

PQ

Chord

PQ = 6\sqrt{3}

Using triangle

OPQ

with sides

OP = OQ = 6

and

\angle POQ = 120^\circ

, law of cosines or bisecting

\angle POQ

into two

30^\circ

-

60^\circ

-

90^\circ

triangles yields chord length

2 \times (6 \sin 60^\circ) = 2 \times 3\sqrt{3} = 6\sqrt{3}

.

Key Concept

Arc Length, Sector Area, and Chord Relationships in Circles

Estimated Time:1m 30s

Question 1443Question →

A car rental agency charges a flat daily rate of $45 plus$ 0.20 per mile driven. If a customer rented a car for one day and the total rental cost before taxes was $75, how many miles did the customer drive?

150

225

375

600

60

Show answer & explanation

Answer: 150

Answer

150 miles

Subtracting the

45 fixed fee from the total charge of

75 leaves

30 attributable to mileage. Dividing

30 by the variable rate of $0.20 per mile gives 150 miles.

Step-by-Step Solution

1

Set up the linear equation representing total cost

45 + 0.20m = 75

, where

m

represents the number of miles driven.

The total cost consists of a fixed fee plus the variable per-mile charge.

2

Isolate the variable term by subtracting the fixed fee from both sides

0.20m = 75 - 45 \implies 0.20m = 30

This determines the portion of the total cost accrued strictly from mileage.

3

Solve for

m

by dividing by the per-mile rate

m = \frac{30}{0.20} = 150

Dividing total mileage cost by the cost per mile gives the total miles driven.

Key Concept

Linear Modeling and Single-Variable Equations

Estimated Time:45s

Question 1444Question →

What is the minimum integer value of $x$ that satisfies the inequality $|2x - 7| \le 5$ ?

Show answer & explanation

Answer: 1

Answer

The minimum integer value of

x

that satisfies the inequality is 1.

To solve

|2x - 7| \le 5

, write it as the compound inequality

-5 \le 2x - 7 \le 5

. Adding 7 across all parts gives

2 \le 2x \le 12

. Dividing by 2 yields

1 \le x \le 6

. The integer solutions are 1, 2, 3, 4, 5, and 6. The minimum integer among these is 1.

Step-by-Step Solution

1

Convert the absolute value inequality into a compound inequality.

-5 \le 2x - 7 \le 5

An inequality of the form

|u| \le a

(where

a \ge 0

) is equivalent to

-a \le u \le a

.

2

Add 7 to all three parts of the inequality.

2 \le 2x \le 12

Adding a constant to an inequality preserves the direction of the inequality signs.

3

Divide all three parts by 2.

1 \le x \le 6

Dividing by a positive constant isolates

x

without reversing the inequality signs.

4

Determine the minimum integer within the solution set

[1, 6]

.

1

The solution set contains integers

\{1, 2, 3, 4, 5, 6\}

, making 1 the smallest integer value.

Key Concept

Solving absolute value inequalities using equivalent compound linear inequalities

Question 1445Question →

Let $S$ be the set of all real numbers $x$ that satisfy the nested absolute value inequality $|3 - |2x - 1|| \le 4$ . Which of the following inequalities MUST be satisfied by every value of $x$ in $S$ ? Select all such inequalities.

Select all that apply

|x| \le 4

(x + 3)(x - 4) \le 0

|2x - 1| \le 4

-1 \le x \le 7

x \ge -2

Show answer & explanation

Answer: $|x| \le 4$ ; $(x + 3)(x - 4) \le 0$

Answer

The inequalities that must be satisfied by every value of

x

in

S

are

|x| \le 4

and

(x + 3)(x - 4) \le 0

.

The solution set to the nested inequality is

S = [-3, 4]

. For any value of

x

in

[-3, 4]

, the absolute value

|x|

ranges from

0

to

4

, so the inequality stating that the magnitude of

x

is at most 4 is satisfied. Additionally, the quadratic expression factored as

(x + 3)(x - 4)

has roots at

x = -3

and

x = 4

and opens upward, meaning it takes on values less than or equal to zero for all

x

between

-3

and

4

.

Step-by-Step Solution

1

Unfold the outer absolute value inequality

-4 \le 3 - |2x - 1| \le 4

By definition,

|u| \le k

(with

k \ge 0

) is equivalent to

-k \le u \le k

.

2

Isolate the inner absolute value expression by subtracting 3 and multiplying by -1

-1 \le |2x - 1| \le 7

Subtracting 3 yields

-7 \le -|2x - 1| \le 1

. Multiplying by

-1

flips the inequality signs, giving

-1 \le |2x - 1| \le 7

.

3

Simplify the compound absolute value bound and solve for

x

x \in [-3, 4]

Since an absolute value is non-negative,

|2x - 1| \ge -1

is satisfied for all real

x

. Thus, we only need

|2x - 1| \le 7

, which gives

-7 \le 2x - 1 \le 7 \implies -6 \le 2x \le 8 \implies -3 \le x \le 4

.

4

Test the solution set

S = [-3, 4]

against each given statement

Statements

|x| \le 4

and

(x + 3)(x - 4) \le 0

hold for all

x \in [-3, 4]

.

For

x \in [-3, 4]

, the extreme values of

x

yield

|x| \le 4

. Furthermore, a quadratic with roots at

-3

and

4

is non-positive on

[-3, 4]

.

Key Concept

Solving nested absolute value inequalities by systematic expansion and isolating valid intervals.

Question 1446Question →

For all real numbers $x$ that satisfy the absolute value inequality $|3 - 2x| - 4 \le x$ , the rational expression $y = \frac{|x - 6|}{2 - x}$ is defined. Which of the following intervals represents the complete set of all possible real values of $y$ ?

(-\infty, 0] \cup [\frac{19}{7}, \infty)

(-\infty, -\frac{1}{5}] \cup [\frac{19}{7}, \infty)

(-\infty, 0) \cup (1, \infty)

[-\frac{1}{5}, \frac{19}{7}]

(-\infty, 1] \cup [\frac{12}{7}, \infty)

Show answer & explanation

Answer: $(-\infty, 0] \cup [\frac{19}{7}, \infty)$

Answer

The complete set of all possible real values of

y

is

(-\infty, 0] \cup [\frac{19}{7}, \infty)

.

The correct answer is

(-\infty, 0] \cup [\frac{19}{7}, \infty)

. Solving

|3 - 2x| - 4 \le x

gives

-\frac{1}{3} \le x \le 7

. Excluding

x = 2

where the denominator is zero, evaluating

y = \frac{|x - 6|}{2 - x}

on

[-\frac{1}{3}, 2)

yields

y \ge \frac{19}{7}

, on

(2, 6)

yields

y < 0

, and on

[6, 7]

yields

-\frac{1}{5} \le y \le 0

. Taking the union of these intervals gives

(-\infty, 0] \cup [\frac{19}{7}, \infty)

.

Step-by-Step Solution

1

Isolate the absolute value expression in the inequality.

|3 - 2x| \le x + 4

Adding

4

to both sides prepares the inequality for standard double-inequality solving.

2

Set up the compound inequality and solve for

x

.

-(x + 4) \le 3 - 2x \le x + 4

, yielding

x \le 7

from

-x - 4 \le 3 - 2x

, and

x \ge -\frac{1}{3}

from

3 - 2x \le x + 4

(remembering to flip the inequality sign when dividing by

-3

). Thus,

x \in [-\frac{1}{3}, 7]

.

An absolute value inequality

|A| \le B

(with

B \ge 0

) is equivalent to

-B \le A \le B

.

3

Identify domain restrictions for

y = \frac{|x - 6|}{2 - x}

.

The expression is undefined at

x = 2

. Therefore, the domain of

x

is divided into three sub-intervals:

[-\frac{1}{3}, 2)

,

(2, 6)

, and

[6, 7]

.

The denominator cannot be zero, and the absolute value

|x - 6|

changes definition at

x = 6

.

4

Analyze

y

on the first interval

[-\frac{1}{3}, 2)

.

For

x < 6

,

|x - 6| = 6 - x

. So

y = \frac{6 - x}{2 - x} = 1 + \frac{4}{2 - x}

. As

x

increases from

-\frac{1}{3}

towards

2

,

2 - x

decreases from

\frac{7}{3}

to

0^+

, so

y

increases from

1 + \frac{4}{7/3} = \frac{19}{7}

to

+\infty

. Hence

y \in [\frac{19}{7}, \infty)

.

As the positive denominator approaches zero from above, the positive fraction grows without bound towards

+\infty

.

5

Analyze

y

on the second interval

(2, 6)

.

Here

x < 6

, so

|x - 6| = 6 - x

and

y = 1 + \frac{4}{2 - x}

. As

x

increases from

2^+

to

6

,

2 - x

increases from

0^-

to

-4

. Thus

y

increases from

-\infty

up to

1 + \frac{4}{-4} = 0

. Hence

y \in (-\infty, 0)

.

As the negative denominator moves away from zero towards

-4

, the expression increases from

-\infty

to

0

.

6

Analyze

y

on the third interval

[6, 7]

.

For

x \ge 6

,

|x - 6| = x - 6

. So

y = \frac{x - 6}{2 - x} = -1 - \frac{4}{2 - x}

. At

x = 6

,

y = 0

. At

x = 7

,

y = \frac{1}{-5} = -\frac{1}{5}

. As

x

increases from

6

to

7

,

y

decreases continuously from

0

down to

-\frac{1}{5}

. Hence

y \in [-\frac{1}{5}, 0]

.

Combining

(-\infty, 0)

from the second interval and

[-\frac{1}{5}, 0]

from the third interval gives

(-\infty, 0]

.

7

Combine the ranges from all intervals.

y \in (-\infty, 0] \cup [\frac{19}{7}, \infty)

.

Taking the union of all output values across the valid domain yields the total range.

Key Concept

Solving absolute value inequalities and finding the range of rational expressions with absolute values over restricted domains.

Estimated Time:3m 0s

Question 1447Question →

Which of the following inequalities represents all real values of $x$ that satisfy $-3x + 14 \le 2$ ?

x \ge 4

x \le 4

x \ge -4

x \le -4

x \ge \frac{16}{3}

Show answer & explanation

Answer: $x \ge 4$

Answer

x \ge 4

Subtracting 14 from both sides of

-3x + 14 \le 2

yields

-3x \le -12

. Dividing both sides by

-3

requires reversing the inequality sign from

\le

to

\ge

, giving

x \ge 4

. Thus, the inequality

x \ge 4

correctly represents all solution values.

Step-by-Step Solution

1

Isolate the variable term on the left side of the inequality.

Subtract 14 from both sides:

-3x \le 2 - 14

, which simplifies to

-3x \le -12

.

To solve for

x

, constant terms must first be removed from the variable side using inverse operations.

2

Divide both sides by the coefficient of

x

and apply the inequality rule for negative multipliers.

Divide by

-3

and reverse the inequality sign:

x \ge \frac{-12}{-3}

, which simplifies to

x \ge 4

.

Dividing or multiplying an inequality by a negative quantity changes the direction of the inequality sign.

Key Concept

Solving linear inequalities with negative coefficients

Estimated Time:45s

Question 1448Question →

A solid rectangular box has a length of $6$ centimeters, a width of $4$ centimeters, and a height of $5$ centimeters. What is the total surface area, in square centimeters, of the box?

74

120

148

240

288

Show answer & explanation

Answer: $148$

Answer

The total surface area of the rectangular box is

148

square centimeters.

The total surface area of a rectangular solid with length

l

, width

w

, and height

h

is given by

2(lw + lh + wh)

. Substituting

l = 6

,

w = 4

, and

h = 5

yields

2(6 \cdot 4 + 6 \cdot 5 + 4 \cdot 5) = 2(24 + 30 + 20) = 2(74) = 148

square centimeters.

Step-by-Step Solution

1

Identify the given dimensions of the rectangular box.

Length

l = 6

cm, width

w = 4

cm, and height

h = 5

cm.

These dimensions are required to compute the area of each face.

2

Calculate the surface area of the three distinct face pairs.

Top and bottom faces:

6 \times 4 = 24

; Front and back faces:

6 \times 5 = 30

; Left and right side faces:

4 \times 5 = 20

.

A rectangular solid has six rectangular faces grouped into three identical pairs.

3

Apply the surface area formula

A = 2(lw + lh + wh)

.

A = 2(24 + 30 + 20) = 2(74) = 148

square centimeters.

Summing the area of all six faces gives the total surface area.

Key Concept

Surface Area of a Rectangular Solid

Question 1449Question →

If $x$ is a real number such that $|2x - 7| \le 5$ , what is the minimum possible value of $|x - 8|$ ?

Show answer & explanation

Answer: 2

Answer

2

Solving the given inequality

|2x - 7| \le 5

yields the compound inequality

-5 \le 2x - 7 \le 5

. Adding 7 across the inequality gives

2 \le 2x \le 12

, which simplifies to

1 \le x \le 6

. Geometrically,

|x - 8|

represents the distance between

x

and 8 on the real number line. To minimize this distance for any

x

in the closed interval

[1, 6]

, we select the point in

[1, 6]

closest to 8, which is

x = 6

. Evaluating at

x = 6

produces

|6 - 8| = 2

.

Step-by-Step Solution

1

Unpack the absolute value inequality

1 \le x \le 6

The inequality

|2x - 7| \le 5

is equivalent to

-5 \le 2x - 7 \le 5

. Adding 7 gives

2 \le 2x \le 12

, and dividing by 2 yields

1 \le x \le 6

.

2

Determine the value in the domain

[1, 6]

that minimizes

|x - 8|

x = 6

The expression

|x - 8|

measures the distance from

x

to 8 on the number line. The value within

[1, 6]

nearest to 8 is

x = 6

.

3

Evaluate the expression at

x = 6

2

Substituting

x = 6

into

|x - 8|

gives

|6 - 8| = |-2| = 2

.

Key Concept

Properties of Linear Inequalities and Absolute Value as Distance

Estimated Time:1m 30s

Question 1450Question →

The table below shows the frequency distribution of the weights, in grams, for a sample of 80 manufactured components.

Weight (grams)	Frequency
100–109	12
110–119	24
120–129	28
130–139	16

What percentage of the components in the sample have a weight of at least 120 grams?

Show answer & explanation

Answer: 55

Answer

55%

To calculate the percentage of components with a weight of at least 120 grams, sum the frequencies of all weight intervals corresponding to 120 grams or more:

28 + 16 = 44

. Next, divide this count by the total number of components (

80

) and multiply by

100

:

\frac{44}{80} \times 100\% = 55\%

.

Step-by-Step Solution

1

Identify the number of components with weight at least 120 grams

44 components

The weight intervals '120–129' and '130–139' contain 28 and 16 components respectively, giving

28 + 16 = 44

.

2

Calculate the percentage relative to the total sample size of 80 components

55%

Dividing the target count (44) by the total sample size (80) and multiplying by 100 yields

\frac{44}{80} \times 100 = 55\%

.

Key Concept

Calculating class percentages from grouped frequency tables

Question 1451Question →

In a circle with center $O$ and radius $12$ , radii $OA$ and $OB$ are perpendicular. Point $C$ lies on segment $OA$ such that $C$ is the midpoint of $OA$ . A line segment perpendicular to $OA$ is drawn from point $C$ to intersect minor arc $AB$ at point $D$ . What is the area of the region bounded by line segment $CD$ , line segment $CA$ , and minor arc $AD$ ?

12\pi

24\pi - 18

24\pi - 18\sqrt{3}

36\pi - 18\sqrt{3}

24\pi - 36\sqrt{3}

Show answer & explanation

Answer: $24\pi - 18\sqrt{3}$

Answer

24\pi - 18\sqrt{3}

The area of the region bounded by line segment

CD

, segment

CA

, and minor arc

AD

is obtained by subtracting the area of right triangle

OCD

from the area of sector

OAD

. With

OC = 6

and radius

OD = 12

, triangle

OCD

is a

30^\circ-60^\circ-90^\circ

right triangle, giving

CD = 6\sqrt{3}

and central angle

\angle AOD = 60^\circ

. The sector area is

\frac{60^\circ}{360^\circ} \cdot \pi (12^2) = 24\pi

, and the triangle area is

\frac{1}{2} \cdot 6 \cdot 6\sqrt{3} = 18\sqrt{3}

. Subtracting the triangle area from the sector area yields

24\pi - 18\sqrt{3}

.

Step-by-Step Solution

1

Determine the length of

OC

and the height

CD

OC = 6

and

CD = 6\sqrt{3}

Since

C

is the midpoint of radius

OA = 12

,

OC = 6

. Triangle

OCD

is a right triangle at

C

with hypotenuse

OD = 12

(radius of circle). By the Pythagorean theorem,

CD = \sqrt{12^2 - 6^2} = \sqrt{108} = 6\sqrt{3}

.

2

Find the central angle

\angle AOD

\angle AOD = 60^\circ

In right triangle

OCD

,

\cos(\angle AOD) = \frac{OC}{OD} = \frac{6}{12} = \frac{1}{2}

, which implies

\angle AOD = 60^\circ

.

3

Calculate the area of sector

OAD

\text{Area(Sector } OAD) = 24\pi

$

The area of a sector with central angle

60^\circ

and radius

12

is

\frac{60^\circ}{360^\circ} \cdot \pi (12^2) = \frac{1}{6} \cdot 144\pi = 24\pi

.

4

Calculate the area of right triangle

OCD

\text{Area(Triangle } OCD) = 18\sqrt{3}

The area of right triangle

OCD

is

\frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 6 \cdot 6\sqrt{3} = 18\sqrt{3}

.

5

Subtract the area of triangle

OCD

from the area of sector

OAD

\text{Area of shaded region} = 24\pi - 18\sqrt{3}

The bounded region is formed by removing triangle

OCD

from sector

OAD

.

Key Concept

Calculating the area of a region bounded by a circle arc and line segments by subtracting a right triangle area from a sector area.

Question 1452Question →

In a plane, lines $l_1$ and $l_2$ are parallel ( $l_1 \parallel l_2$ ). A transversal line $t$ intersects $l_1$ at point $P$ and $l_2$ at point $Q$ . The measures of two consecutive interior angles on the same side of transversal $t$ are $(4x - 10)^\circ$ on line $l_1$ and $(3x + 50)^\circ$ on line $l_2$ . A third line $l_3$ passes through point $P$ and is perpendicular to line $l_2$ .

Which of the following statements MUST be true? Select all that apply.

Select all that apply

x = 20

The acute angle formed between line

t

and line

l_3

at point

P

measures

20^\circ

.

The obtuse angle formed at point

Q

between line

t

and line

l_2

measures

120^\circ

.

Line

l_3

is parallel to transversal line

t

.

The sum of all four interior angles located between lines

l_1

and

l_2

along transversal line

t

is

360^\circ

.

Show answer & explanation

Answer: $x = 20$ ; The acute angle formed between line $t$ and line $l_3$ at point $P$ measures $20^\circ$ .; The sum of all four interior angles located between lines $l_1$ and $l_2$ along transversal line $t$ is $360^\circ$ .

Answer

The correct statements are that

x = 20

, the acute angle formed between line

t

and line

l_3

at point

P

measures

20^\circ

, and the sum of all four interior angles located between lines

l_1

and

l_2

along transversal line

t

is

360^\circ

.

The statements confirming

x = 20

, the

20^\circ

acute angle between lines

t

and

l_3

, and the

360^\circ

interior angle sum are all correct because consecutive interior angles sum to

180^\circ

, perpendicular lines create

90^\circ

angles, and the four interior angles sum to

360^\circ

.

Step-by-Step Solution

1

Set up an equation for consecutive interior angles.

(4x - 10) + (3x + 50) = 180^\circ

Consecutive interior angles on the same side of a transversal intersecting parallel lines are supplementary.

2

Solve the algebraic equation for

x

.

7x + 40 = 180 \implies 7x = 140 \implies x = 20

Combine like terms and isolate

x

.

3

Calculate the interior angle measures at points

P

and

Q

.

Angle at

P = 4(20) - 10 = 70^\circ

; Angle at

Q = 3(20) + 50 = 110^\circ

Substitute

x = 20

back into the original expressions.

4

Determine perpendicular relationships and angle measure between line

t

and line

l_3

.

Angle between

t

and

l_3

at

P

is

90^\circ - 70^\circ = 20^\circ

Line

l_3 \perp l_2

implies

l_3 \perp l_1

because

l_1 \parallel l_2

. Thus

l_3

forms a

90^\circ

angle with

l_1

at

P

.

5

Sum the four interior angles between lines

l_1

and

l_2

.

70^\circ + 110^\circ + 70^\circ + 110^\circ = 360^\circ

The four interior angles comprise two pairs of interior angles measuring

70^\circ

and

110^\circ

each.

Key Concept

Parallel line angle theorems (consecutive interior angles, alternate interior angles) and perpendicular line relationships.

Question 1453Question →

An isosceles triangle has a base of length $10$ inches and a perimeter of $36$ inches. What is the area of the triangle, in square inches?

30

60

65

120

130

Show answer & explanation

Answer: $60$

Answer

60

square inches

Subtracting the base length of

10

inches from the total perimeter of

36

inches leaves

26

inches for the remaining two congruent sides, giving

13

inches each. The perpendicular height divides the isosceles triangle into two right triangles with base

5

inches and hypotenuse

13

inches. By the Pythagorean theorem, the height is

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

inches. Substituting base

10

and height

12

into the area formula

\frac{1}{2} \times \text{base} \times \text{height}

gives

\frac{1}{2} \times 10 \times 12 = 60

square inches.

Step-by-Step Solution

1

Determine the length of the two congruent sides of the isosceles triangle.

Each side has a length of

13

inches.

Since the perimeter is

36

inches and the base is

10

inches, the sum of the two equal sides is

36 - 10 = 26

inches. Dividing by

2

yields

13

inches per side.

2

Calculate the perpendicular height of the triangle using the Pythagorean theorem.

The height of the triangle is

12

inches.

Dropping an altitude from the top vertex to the base bisects the base into two segments of

5

inches. This creates two right triangles with a base of

5

inches and hypotenuse of

13

inches. Thus,

h = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12

inches.

3

Calculate the area of the isosceles triangle.

The area is

60

square inches.

Using the triangle area formula

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}

, we compute

\frac{1}{2} \times 10 \times 12 = 60

.

Key Concept

Properties of Isosceles Triangles and Area Formula

Question 1454Question →

If

x

satisfies the linear equation

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

what is the value of

4x - 3

?

5

13

17

57

77

Show answer & explanation

Answer: 17

Answer

17

Solving the equation by grouping terms with like denominators gives

\frac{2x + 2}{4} = \frac{2x + 5}{5}

. Simplifying the left side to

\frac{x + 1}{2}

and cross-multiplying yields

5(x + 1) = 2(2x + 5)

, which simplifies to

5x + 5 = 4x + 10

, giving

x = 5

. Evaluating

4x - 3

at

x = 5

produces

4(5) - 3 = 17

, making this answer correct.

Step-by-Step Solution

1

Group like fractional terms with denominator 4 on one side and denominator 5 on the other side of the equation.

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

Grouping terms with common denominators simplifies algebraic combination.

2

Combine the numerators over their common denominators, carefully distributing signs.

\frac{(3x - 1) - (x - 3)}{4} = \frac{(2x + 3) + 2}{5} \implies \frac{2x + 2}{4} = \frac{2x + 5}{5}

Subtracting

(x - 3)

requires distributing the negative sign to yield

-x + 3

.

3

Simplify the left side fraction and cross-multiply to eliminate denominators.

\frac{x + 1}{2} = \frac{2x + 5}{5} \implies 5(x + 1) = 2(2x + 5)

Simplifying

\frac{2x+2}{4}

to

\frac{x+1}{2}

reduces computation before cross-multiplication.

4

Expand both sides and solve for

x

.

5x + 5 = 4x + 10 \implies 5x - 4x = 10 - 5 \implies x = 5

Subtracting

4x

and

5

from both sides isolates

x

.

5

Substitute

x = 5

into the required expression

4x - 3

.

4(5) - 3 = 20 - 3 = 17

The question asks for the value of

4x - 3

, not

x

alone.

Key Concept

Solving linear equations in one variable involving fractional terms and evaluating targeted algebraic expressions.

Estimated Time:2m 0s

Question 1455Question →

A solid right circular cone has a base radius of $r$ and a height of $h = \frac{4}{3}r$ . A solid sphere has a radius of $R$ . If the total surface area of the cone is equal to the total surface area of the sphere, which of the following statements must be true? Select all that apply.

Select all that apply

The base radius of the cone,

r

, is strictly greater than the radius of the sphere,

R

.

The volume of the cone is strictly less than the volume of the sphere.

The ratio of the volume of the cone to the volume of the sphere is

\frac{\sqrt{6}}{4}

.

The volume of the cone is strictly greater than the volume of the sphere.

The ratio of the base radius of the cone to the radius of the sphere,

\frac{r}{R}

, is equal to

\frac{3}{2}

.

Show answer & explanation

Answer: The base radius of the cone, $r$ , is strictly greater than the radius of the sphere, $R$ .; The volume of the cone is strictly less than the volume of the sphere.; The ratio of the volume of the cone to the volume of the sphere is $\frac{\sqrt{6}}{4}$ .

Answer

The statements confirming that the base radius of the cone is strictly greater than the radius of the sphere, that the volume of the cone is strictly less than the volume of the sphere, and that the ratio of the volume of the cone to the volume of the sphere is

\frac{\sqrt{6}}{4}

are all correct.

Equating the total surface area of the cone

\frac{8}{3}\pi r^2

with the surface area of the sphere

4\pi R^2

yields

\frac{r^2}{R^2} = \frac{3}{2}

, which simplifies to

\frac{r}{R} = \sqrt{\frac{3}{2}}

. Because

\sqrt{1.5} > 1

, the base radius of the cone is strictly greater than the radius of the sphere. Furthermore, evaluating the ratio of their volumes gives

\frac{V_{\text{cone}}}{V_{\text{sphere}}} = \frac{\frac{4}{9}\pi r^3}{\frac{4}{3}\pi R^3} = \frac{1}{3}\left(\frac{r}{R}\right)^3 = \frac{\sqrt{6}}{4} \approx 0.612

. Because this ratio is strictly less than

1

, the volume of the cone is strictly less than the volume of the sphere. Thus, the three true statements are those stating

r > R

, that the volume of the cone is strictly less than the sphere's volume, and that their volume ratio is

\frac{\sqrt{6}}{4}

.

Step-by-Step Solution

1

Calculate the slant height and total surface area of the cone in terms of

r

.

Slant height

l = \sqrt{r^2 + \left(\frac{4}{3}r\right)^2} = \sqrt{\frac{25}{9}r^2} = \frac{5}{3}r

. Total surface area

A_{\text{cone}} = \pi r^2 + \pi r l = \pi r^2 + \pi r\left(\frac{5}{3}r\right) = \frac{8}{3}\pi r^2

.

The total surface area of a right circular cone is the sum of its base area

\pi r^2

and lateral area

\pi r l

.

2

Equate the total surface area of the cone to the total surface area of the sphere to find the ratio

\frac{r}{R}

.

\frac{8}{3}\pi r^2 = 4\pi R^2 \implies 2 r^2 = 3 R^2 \implies \frac{r^2}{R^2} = \frac{3}{2} \implies \frac{r}{R} = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}.

The surface area of a sphere of radius

R

is

4\pi R^2

.

3

Compare the linear dimensions

r

and

R

.

Since

\frac{r}{R} = \sqrt{1.5} \approx 1.225 > 1

, it follows that

r > R

.

A ratio greater than

1

implies the numerator is larger than the denominator.

4

Express the volumes of both solids and compute their ratio

\frac{V_{\text{cone}}}{V_{\text{sphere}}}

.

V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 \left(\frac{4}{3}r\right) = \frac{4}{9}\pi r^3

.

V_{\text{sphere}} = \frac{4}{3}\pi R^3

. Therefore,

\frac{V_{\text{cone}}}{V_{\text{sphere}}} = \frac{\frac{4}{9}\pi r^3}{\frac{4}{3}\pi R^3} = \frac{1}{3}\left(\frac{r}{R}\right)^3 = \frac{1}{3}\left(\frac{3}{2}\right)^{3/2} = \frac{1}{3} \cdot \frac{3\sqrt{3}}{2\sqrt{2}} = \frac{\sqrt{6}}{4}

.

Using standard volume formulas for cones and spheres and substituting the known linear dimension ratio.

5

Determine whether the cone's volume is greater than or less than the sphere's volume.

Since

\frac{\sqrt{6}}{4} \approx 0.612 < 1

,

V_{\text{cone}} < V_{\text{sphere}}

.

A volume ratio less than

1

proves the cone has a smaller volume than the sphere.

Key Concept

Analyzing geometric scaling, volume, and total surface area relations between cones and spheres using variable constraints.

Question 1456Question →

In the $xy$ -plane, line $l$ passes through the points $(2, 5)$ and $(6, 13)$ . What is the slope of line $l$ ?

Show answer & explanation

Answer: 2

Answer

The slope of line

l

is 2.

The slope of a line passing through

(x_1, y_1)

and

(x_2, y_2)

is given by

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

. Substituting

(2, 5)

and

(6, 13)

into the formula gives

m = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2

.

Step-by-Step Solution

1

Identify the coordinates of the two given points on line

l

.

The points are

(x_1, y_1) = (2, 5)

and

(x_2, y_2) = (6, 13)

.

Two points are required to calculate the slope of a straight line.

2

Apply the slope formula

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

.

m = \frac{13 - 5}{6 - 2} = \frac{8}{4} = 2

.

The slope measures the vertical change (rise) divided by the horizontal change (run).

Key Concept

Slope of a line through two points

Question 1457Question →

Dataset $S$ consists of $80$ distinct numerical scores arranged in ascending order, with an interquartile range denoted by $\text{IQR}_S$ . A new dataset, $T$ , is created by adding four scores to Dataset $S$ : two scores that are strictly smaller than the minimum value of Dataset $S$ , and two scores that are strictly greater than the maximum value of Dataset $S$ . If $\text{IQR}_T$ represents the interquartile range of Dataset $T$ , which of the following statements must be true?

\text{IQR}_T > \text{IQR}_S

\text{IQR}_T = \text{IQR}_S

\text{IQR}_T < \text{IQR}_S

\text{IQR}_T = \text{IQR}_S + 4

The relationship between

\text{IQR}_T

and

\text{IQR}_S

cannot be determined from the given information.

Show answer & explanation

Answer: $\text{IQR}_T > \text{IQR}_S$

Answer

The interquartile range of Dataset

T

is strictly greater than the interquartile range of Dataset

S

(

\text{IQR}_T > \text{IQR}_S

).

The statement that

\text{IQR}_T > \text{IQR}_S

is correct. Inserting two items below the minimum value shifts the rank position of the first quartile to lower-indexed elements of the original dataset, reducing

Q_1

. Inserting two items above the maximum value shifts the rank position of the third quartile to higher-indexed elements of the original dataset, increasing

Q_3

. Since the original numbers are strictly increasing,

Q_3

increases while

Q_1

decreases, resulting in a larger interquartile range for Dataset

T

.

Step-by-Step Solution

1

Determine the quartile positions for the original dataset

S

.

Dataset

S

has

n = 80

distinct elements

x_1 < x_2 < \dots < x_{80}

. Its first quartile

Q_{1,S}

is the average of

x_{20}

and

x_{21}

, and its third quartile

Q_{3,S}

is the average of

x_{60}

and

x_{61}

. Thus,

\text{IQR}_S = Q_{3,S} - Q_{1,S} = \frac{x_{60} + x_{61}}{2} - \frac{x_{20} + x_{21}}{2}

.

Quartiles partition sorted data into four equal parts based on element indices.

2

Determine the new element positions for dataset

T

.

Dataset

T

has

n = 84

elements ordered as

a, b, x_1, x_2, \dots, x_{80}, c, d

, where

a, b < x_1

and

c, d > x_{80}

. The elements of

S

are now shifted right by 2 index positions in

T

.

Inserting two values at the bottom of the ordered list increases the rank index of all original elements by 2.

3

Calculate the first and third quartiles for dataset

T

.

In Dataset

T

, the first quartile

Q_{1,T}

is the average of the 21st and 22nd elements of

T

, which correspond to

x_{19}

and

x_{20}

of

S

. The third quartile

Q_{3,T}

is the average of the 63rd and 64th elements of

T

, which correspond to

x_{61}

and

x_{62}

of

S

.

The 25th and 75th percentile positions in an 84-element dataset correspond to indices 21/22 and 63/64.

4

Compare

\text{IQR}_T

with

\text{IQR}_S

.

Since

S

consists of distinct values in strictly ascending order (

x_{19} < x_{20} < x_{21}

and

x_{60} < x_{61} < x_{62}

),

Q_{1,T} = \frac{x_{19} + x_{20}}{2} < \frac{x_{20} + x_{21}}{2} = Q_{1,S}

and

Q_{3,T} = \frac{x_{61} + x_{62}}{2} > \frac{x_{60} + x_{61}}{2} = Q_{3,S}

. Therefore,

\text{IQR}_T = Q_{3,T} - Q_{1,T} > Q_{3,S} - Q_{1,S} = \text{IQR}_S

.

Subtracting a smaller number from a larger number increases the overall difference.

Key Concept

Interquartile Range and Position Indices under Data Insertion

Question 1458Question →

Working independently at their respective constant rates, Alex and Blair can complete a certain job together. Alex works alone for $4$ hours, after which Blair joins Alex, and together they work for an additional $6$ hours to finish the entire job. If Alex takes strictly less time to complete the job working alone than Blair takes working alone, which of the following statements must be true? Select all that apply.

Select all that apply

Alex working alone would take less than

16

hours to complete the job.

The time required for Alex and Blair to complete the job working together from start to finish is strictly between

8

hours and

10

hours.

Alex completes more than

60\%

of the entire job.

Blair working alone would take more than

24

hours to complete the job.

Blair completes more than

35\%

of the entire job.

Show answer & explanation

Answer: Alex working alone would take less than $16$ hours to complete the job.; The time required for Alex and Blair to complete the job working together from start to finish is strictly between $8$ hours and $10$ hours.; Alex completes more than $60\%$ of the entire job.

Answer

The true statements are that Alex working alone would take less than 16 hours, the combined time to finish the job working together is between 8 and 10 hours, and Alex completes more than 60 percent of the entire job.

By setting up the total work equation

10 r_A + 6 r_B = 1

and using the inequality

r_A > r_B > 0

, we find that

1/16 < r_A < 1/10

. This implies Alex's solo time is strictly less than 16 hours. The combined time

T_{together}

is constrained between 8 and 10 hours because

6/T_{together} = 1 - 4 r_A

. Additionally, Alex's total work contribution is

10 r_A > 10/16 = 62.5\%

, which is strictly greater than 60%.

Step-by-Step Solution

1

Set up the work equation using individual rates

r_A

and

r_B

.

4 r_A + 6(r_A + r_B) = 1 \implies 10 r_A + 6 r_B = 1

.

Alex works alone for 4 hours and then both Alex and Blair work together for 6 hours to complete 1 unit of work.

2

Apply the condition that Alex takes strictly less time alone than Blair (

r_A > r_B > 0

).

10 r_A + 6 r_A > 10 r_A + 6 r_B = 1 \implies 16 r_A > 1 \implies r_A > \frac{1}{16}

.

Since Alex's rate

r_A

is strictly greater than Blair's rate

r_B

, replacing

r_B

with

r_A

gives an upper bound on Alex's solo time

T_A = 1/r_A < 16

hours.

3

Determine the feasible range for Alex's rate

r_A

.

\frac{1}{16} < r_A < \frac{1}{10}

.

From

6 r_B = 1 - 10 r_A > 0

, we get

r_A < 1/10

. Combined with

r_A > 1/16

, we have

1/16 < r_A < 1/10

.

4

Calculate the combined time

T_{together} = \frac{1}{r_A + r_B}

.

8 < T_{together} < 10

.

Since

6(r_A + r_B) = 1 - 4 r_A

, substituting

1/16 < r_A < 1/10

gives

3/5 < 6(r_A + r_B) < 3/4

, which simplifies to

8 < T_{together} < 10

.

5

Calculate the fraction of total work performed by Alex.

Alex performs

10 r_A > 10 \times \frac{1}{16} = 0.625 = 62.5\%

of the total work.

Alex works for a total of 10 hours (

4 + 6

). Since

r_A > 1/16

, Alex completes over

62.5\%

of the job, which is greater than

60\%

.

Key Concept

Formulating algebraic inequalities for work rate problems when relative individual speeds are given.

Question 1459Question →

In triangle $XYZ$ , the length of side $XY$ is $x + 3$ , the length of side $YZ$ is $2x - 1$ , and the length of side $XZ$ is $12$ , where $x$ is an integer. Which of the following could be the perimeter of triangle $XYZ$ ? Select all such perimeters.

Select all that apply

26

23

44

62

31

Show answer & explanation

Answer: 26; 44

Answer

The perimeters 26 and 44 are the valid values among the choices provided.

The perimeter formula

P = 3x + 14

must yield an integer value corresponding to an integer

x

in the range

4 \le x \le 15

. The perimeters equal to 26 (for

x = 4

) and 44 (for

x = 10

) fall within this valid range and satisfy the triangle inequality.

Step-by-Step Solution

1

Express the perimeter in terms of

x

.

\text{Perimeter } P = (x + 3) + (2x - 1) + 12 = 3x + 14

The perimeter of a triangle is the sum of its three side lengths.

2

Apply the Triangle Inequality Theorem to determine valid bounds for

x

.

Condition 1:

(x + 3) + (2x - 1) > 12 \implies 3x + 2 > 12 \implies 3x > 10 \implies x > 3.33

.
Condition 2:

(x + 3) + 12 > 2x - 1 \implies x + 15 > 2x - 1 \implies x < 16

.
Condition 3:

(2x - 1) + 12 > x + 3 \implies 2x + 11 > x + 3 \implies x > -8

(naturally satisfied for positive

x

).
Thus,

4 \le x \le 15

for integer

x

.

For any non-degenerate triangle, the sum of any two side lengths must be strictly greater than the third side length.

3

Evaluate the allowable perimeters for valid integer values of

x

.

The minimum valid perimeter corresponds to

x = 4

, giving

P = 3(4) + 14 = 26

.
The maximum valid perimeter corresponds to

x = 15

, giving

P = 3(15) + 14 = 59

.
Checking options:
- For

26

:

3x + 14 = 26 \implies x = 4

(valid).
- For

44

:

3x + 14 = 44 \implies x = 10

(valid).

Substituting valid integer

x

values identifies which proposed perimeters satisfy all conditions.

Key Concept

Triangle Inequality Theorem & Algebraic Bounds on Side Lengths

Estimated Time:2m 0s

Question 1460Question →

A sector of a circle with a radius of $10$ units has a total perimeter of $20 + 5\pi$ units. What is the area of this sector?

25\pi

50\pi

5\pi

100\pi

100 + 25\pi

Show answer & explanation

Answer: $25\pi$

Answer

The area of the sector is

25\pi

.

The total perimeter of a sector with radius

r

and arc length

L

is given by

P = 2r + L

. Substituting

r = 10

gives

20 + 5\pi = 20 + L

, so

L = 5\pi

. Using the sector area formula

A = \frac{1}{2} r L

, the area is

\frac{1}{2} \times 10 \times 5\pi = 25\pi

.

Step-by-Step Solution

1

Set up the formula for the perimeter of a sector.

P = 2r + L

, where

r = 10

is the radius and

L

is the arc length.

A sector's perimeter is bounded by two straight radii and one curved arc.

2

Solve for the arc length

L

.

20 + 5\pi = 2(10) + L \implies L = 5\pi

.

Subtracting the combined length of the two radii (

20

) isolates the arc length.

3

Calculate the area of the sector.

Sector Area =

\frac{1}{2} r L = \frac{1}{2} (10)(5\pi) = 25\pi

.

The area of a sector can be directly evaluated using half the product of its radius and arc length.

Key Concept

Perimeter, Arc Length, and Area of a Circular Sector

Alternative Method

Find the central angle

\theta

first: Since

L = 5\pi

and circumference

C = 2\pi(10) = 20\pi

, the fraction of the circle is

\frac{5\pi}{20\pi} = \frac{1}{4}

, which corresponds to

\theta = 90^\circ

. The area is then

\frac{1}{4} \times \pi(10^2) = 25\pi

.

Estimated Time:1m 30s

Answer

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