All practice questions

A line passes through the points $(-3, 2)$ and $(5, -4)$ in a coordinate plane. What is the slope of this line?

\frac{4}{3}

\frac{3}{4}

-\frac{3}{4}

-\frac{4}{3}

-\frac{1}{4}

Show answer & explanation

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

Answer

-\frac{3}{4}

The correct answer shows a slope of

-\frac{3}{4}

. Applying the slope formula

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

to the points

(-3, 2)

and

(5, -4)

yields

\frac{-4 - 2}{5 - (-3)} = \frac{-6}{8}

, which simplifies to

-\frac{3}{4}

.

Step-by-Step Solution

1

Identify the coordinates of the two points and state the slope formula.

The points are

(x_1, y_1) = (-3, 2)

and

(x_2, y_2) = (5, -4)

. The slope formula is

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

.

To find the slope of a line, we calculate the ratio of vertical change (rise) to horizontal change (run).

2

Substitute the coordinate values into the slope formula.

m = \frac{-4 - 2}{5 - (-3)}

This sets up the subtraction of the

y

-coordinates in the numerator and the

x

-coordinates in the denominator.

3

Simplify the numerator and denominator to calculate the final slope.

m = \frac{-6}{8} = -\frac{3}{4}

Subtracting a negative number in the denominator is equivalent to addition:

5 - (-3) = 8

. Simplifying the fraction yields the slope of

-\frac{3}{4}

.

Key Concept

The slope

m

of a line passing through points

(x_1, y_1)

and

(x_2, y_2)

is defined as the change in

y

divided by the change in

x

:

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

.

Question 2282Question →

An equation of a circle is given by $x^2 + y^2 - 8x + 6y + k = 0$ , where $k$ is a constant. A second circle has a center that is the reflection of the first circle's center across the line $y = x$ . If the second circle is tangent to the $x$ -axis and has the same radius as the first circle, what is the value of $k$ ?

9

16

21

34

41

Show answer & explanation

Answer: 9

Answer

The value of the constant

k

is 9.

The correct answer is 9. Completing the square for the first circle's equation gives

(x - 4)^2 + (y + 3)^2 = 25 - k

, which identifies the center as

(4, -3)

and the radius squared as

r^2 = 25 - k

. Reflecting

(4, -3)

across the line

y = x

swaps the coordinates to give the new center

(-3, 4)

. Because the second circle is tangent to the

x

-axis, its radius is the absolute value of the

y

-coordinate of its center, which is

|4| = 4

. Since both circles have the same radius, we set the radius squared equal to

4^2

:

25 - k = 16

, which yields

k = 9

.

Step-by-Step Solution

1

Complete the square for the first circle's equation

x^2 + y^2 - 8x + 6y + k = 0

to identify its center and radius.

(x - 4)^2 + (y + 3)^2 = 25 - k

, which represents a circle with center

(4, -3)

and radius squared

r^2 = 25 - k

.

This puts the equation into the standard form

(x - h)^2 + (y - k)^2 = r^2

to find the center and radius.

2

Reflect the center of the first circle across the line

y = x

.

Reflecting the point

(4, -3)

across the line

y = x

swaps the coordinates, yielding the new center

(-3, 4)

.

To find the center of the second circle.

3

Determine the radius of the second circle using the tangency condition.

Since the second circle is tangent to the

x

-axis, its radius is equal to the absolute value of the

y

-coordinate of its center, which is

|4| = 4

.

The distance from a circle's center

(h, k)

to the line of tangency

y = 0

(the

x

-axis) is equal to its radius.

4

Equate the radius squared of the first circle to the square of the radius of the second circle.

25 - k = 4^2 \implies 25 - k = 16 \implies k = 9

.

Both circles are stated to have the same radius.

Key Concept

Converting the general form of a circle's equation to standard form by completing the square, and using coordinate transformations and geometric tangency conditions to solve for unknowns.

Estimated Time:3m 0s

Question 2283Question →

A triangle has two sides of lengths $5$ centimeters and $11$ centimeters. Which of the following could be the perimeter of the triangle, in centimeters?

16

22

27

32

34

Show answer & explanation

Answer: 27

Answer

27

The correct answer is

27

centimeters. According to the Triangle Inequality Theorem, the length of the third side,

x

, of a triangle must be strictly greater than the difference of the other two sides (

11 - 5 = 6

) and strictly less than their sum (

11 + 5 = 16

). This gives the inequality range

6 < x < 16

. The perimeter is the sum of all three sides, which is

P = 5 + 11 + x = 16 + x

. Applying the bounds of

x

, we find that the perimeter must satisfy

16 + 6 < P < 16 + 16

, which simplifies to

22 < P < 32

. Among the options,

27

is the only value that is strictly within this range.

Step-by-Step Solution

1

Apply the Triangle Inequality Theorem to find the limits for the third side.

Let the third side be

x

. The length of

x

must satisfy:

11 - 5 < x < 11 + 5 \implies 6 < x < 16

.

The length of any side of a triangle must be strictly between the positive difference and the sum of the lengths of the other two sides.

2

Set up the equation for the perimeter of the triangle.

Perimeter

P = 5 + 11 + x = 16 + x

.

The perimeter of a triangle is defined as the sum of the lengths of all three of its sides.

3

Find the range of possible values for the perimeter

P

by applying the inequality bounds of

x

.

Add

16

to all parts of the inequality

6 < x < 16

:

16 + 6 < 16 + x < 16 + 16 \implies 22 < P < 32

.

We must shift the inequality bounds of the third side by the sum of the two known sides to find the range of the perimeter.

4

Identify the option that falls strictly inside the range

22 < P < 32

.

The value

27

is the only option that satisfies

22 < 27 < 32

.

Only a value strictly between

22

and

32

can represent a mathematically valid perimeter for this triangle.

Key Concept

Triangle Inequality Theorem and Perimeter

Estimated Time:1m 0s

Question 2284Question →

A system of equations consists of the linear equation $y = x + 5$ and the quadratic equation $y = (x - 3)^2 + 2$ . If the graphs of these two equations intersect at two distinct points, $(x_1, y_1)$ and $(x_2, y_2)$ , in the standard $(x, y)$ coordinate plane, what is the value of $y_1 + y_2$ ?

3

7

11

15

17

Show answer & explanation

Answer: 17

Answer

17

To find the points of intersection, we set the two equations equal:

x + 5 = (x - 3)^2 + 2

. Expanding the right side results in

x + 5 = x^2 - 6x + 9 + 2

, which simplifies to

x + 5 = x^2 - 6x + 11

. Subtracting

x + 5

from both sides gives the quadratic equation

x^2 - 7x + 6 = 0

. Factoring this equation yields

(x - 1)(x - 6) = 0

, giving

x

-coordinates of

1

and

6

. Substituting these values into the linear equation

y = x + 5

gives the

y

-coordinates:

y = 1 + 5 = 6

and

y = 6 + 5 = 11

. The sum of these

y

-coordinates is

6 + 11 = 17

.

Step-by-Step Solution

1

Expand the quadratic equation to its standard polynomial form.

y = x^2 - 6x + 11

Expanding

(x - 3)^2 + 2

allows for easier substitution and algebraic manipulation.

2

Equate the linear and quadratic expressions for

y

to find the intersection points.

x + 5 = x^2 - 6x + 11

Setting the two equations equal to each other identifies the

x

-coordinates where the graphs intersect.

3

Rearrange the equation into standard quadratic form and solve for

x

by factoring.

x^2 - 7x + 6 = 0 \implies (x - 1)(x - 6) = 0

, which gives

x = 1

and

x = 6

.

Factoring allows us to find the roots of the quadratic equation representing the intersection

x

-coordinates.

4

Substitute the

x

-coordinates back into the linear equation

y = x + 5

to determine the corresponding

y

-coordinates.

For

x = 1

,

y = 6

. For

x = 6

,

y = 11

.

Finding the

y

-coordinates corresponding to the intersection points.

5

Calculate the sum of the two

y

-coordinates.

6 + 11 = 17

The question asks specifically for the value of

y_1 + y_2

.

Key Concept

Solving systems of linear and quadratic equations by substitution and factoring

Question 2285Question →

A temperature control system in a chemical reactor regulates target output temperature, in degrees Celsius, based on the reactor's pressure coefficient. The pressure coefficient $P$ is determined by the elapsed time $t$ , in minutes, according to the function $P(t) = (t - 3)^2 - 5$ . The target temperature $T$ is determined by the pressure coefficient $P$ according to the function $T(P) = 3|P - 2| + 1$ . What is the target output temperature of the system, in degrees Celsius, at time $t = 1$ minute?

-8

12

10

-4

46

Show answer & explanation

Answer: 10

Answer

10

The correct temperature is 10. By first finding the pressure at

t = 1

to be

P(1) = (1 - 3)^2 - 5 = 4 - 5 = -1

, and then substituting this pressure into the temperature function, we get

T(-1) = 3|-1 - 2| + 1 = 3(3) + 1 = 10

.

Step-by-Step Solution

1

Calculate the pressure coefficient

P(1)

at time

t = 1

minute by substituting

t = 1

into

P(t) = (t - 3)^2 - 5

.

P(1) = -1

Before calculating the temperature, we must find the input value

P

corresponding to

t = 1

.

2

Evaluate the target temperature

T(P)

at the pressure coefficient

P = -1

by substituting

P = -1

into

T(P) = 3|P - 2| + 1

.

T(-1) = 10

Applying the absolute value definition

|-3| = 3

, we find

3(3) + 1 = 10

.

Key Concept

Evaluating composite functions in a real-world scenario.

Alternative Method

Instead of evaluating step-by-step numerically, we can write the composite function symbolically:

T(P(t)) = 3|((t - 3)^2 - 5) - 2| + 1 = 3|(t - 3)^2 - 7| + 1

. Plugging in

t = 1

yields

T(P(1)) = 3|(1 - 3)^2 - 7| + 1 = 3|4 - 7| + 1 = 3|-3| + 1 = 10

.

Estimated Time:1m 30s

Question 2286Question →

A landscaping company charges a one-time equipment mobilization fee of $45$ plus an hourly labor rate of $32.50$ per worker. A homeowner hires a crew of $3$ workers to clear their yard. If the total bill for the job is $435$ dollars, for how many hours did the crew work?

Show answer & explanation

Answer: 4

Answer

4

The total cost of

435

dollars is the sum of the one-time

45

dollar mobilization fee and the hourly labor cost of

32.50

dollars per worker for

3

workers over

h

hours. This translates to the equation

45 + 3(32.50)h = 435

, which simplifies to

45 + 97.50h = 435

. Subtracting

45

from both sides gives

97.50h = 390

. Dividing by

97.50

yields

h = 4

hours.

Step-by-Step Solution

1

Set up the algebraic expression representing the total cost based on the number of hours worked,

h

. The total cost consists of a fixed mobilization fee of

45

dollars and a variable labor cost of

32.50

dollars per worker per hour.

45 + 3(32.50)h

To represent the relation between the hours worked and the total charge.

2

Equate the expression for the total cost to the actual total bill of

435

dollars and simplify the labor rate coefficient.

45 + 97.50h = 435

To form a solvable linear equation in terms of the unknown number of hours,

h

.

3

Subtract the fixed mobilization fee of

45

from both sides of the equation.

97.50h = 390

To isolate the term containing the variable

h

.

4

Divide both sides of the equation by

97.50

to solve for

h

.

h = 4

To determine the number of hours the crew worked.

Key Concept

Formulating and solving linear equations from real-world contexts

Estimated Time:1m 30s

Question 2287Question →

In the standard $(x, y)$ coordinate plane, line segment $PQ$ has endpoints $P(-2, 3)$ and $Q(4, 1)$ . First, segment $PQ$ is rotated $90^\circ$ counterclockwise about the origin to form segment $P'Q'$ . Next, segment $P'Q'$ is reflected across the line $y = x$ to form segment $P''Q''$ . What are the coordinates of the midpoint of segment $P''Q''$ ?

(1, -2)

(-1, 2)

(-2, 1)

(2, -1)

(-2, -1)

Show answer & explanation

Answer: $(1, -2)$

Answer

The coordinates of the midpoint of segment

P''Q''

are

(1, -2)

.

The midpoint of the original segment

PQ

is

(1, 2)

. Under a

90^\circ

counterclockwise rotation about the origin, the point

(x, y)

maps to

(-y, x)

, so

(1, 2)

maps to

(-2, 1)

. Under a reflection across the line

y = x

, the point

(x, y)

maps to

(y, x)

, so

(-2, 1)

maps to

(1, -2)

. Since rigid transformations preserve midpoints, the midpoint of the final segment is

(1, -2)

.

Step-by-Step Solution

1

Calculate the midpoint of the original segment

PQ

.

Midpoint

M = \left(\frac{-2+4}{2}, \frac{3+1}{2}\right) = (1, 2)

Since rotation and reflection are rigid transformations (isometries), the midpoint of the transformed segment is the transformed midpoint of the original segment.

2

Apply a

90^\circ

counterclockwise rotation about the origin to the midpoint coordinate

(1, 2)

.

Intermediate midpoint

M' = (-2, 1)

The coordinate rule for a

90^\circ

counterclockwise rotation about the origin is

(x, y) \rightarrow (-y, x)

.

3

Apply a reflection across the line

y = x

to the intermediate midpoint

(-2, 1)

.

Final midpoint

M'' = (1, -2)

The coordinate rule for reflection across the line

y = x

is

(x, y) \rightarrow (y, x)

.

Key Concept

Applying composite transformations (rotations and reflections) to geometric figures on the coordinate plane, utilizing the property that the midpoint of a transformed segment is the transformed midpoint of the original segment.

Estimated Time:1m 30s

Question 2288Question →

A marketing firm charges a flat setup fee of $160$ plus $3.20$ for each promotional brochure printed. For a large order, the firm applies a $15\%$ discount to the total cost (the sum of the setup fee and the printing cost). If the final discounted bill is $680$ , how many brochures were printed?

170

20

200

250

300

Show answer & explanation

Answer: 200

Answer

200

The correct answer of 200 brochures is found by formulating the equation

0.85(160 + 3.2b) = 680

. Dividing both sides by

0.85

yields

160 + 3.2b = 800

. Subtracting

160

gives

3.2b = 640

, and dividing by

3.2

results in

b = 200

.

Step-by-Step Solution

1

Translate the word problem into a linear equation where

b

represents the number of brochures. The total cost before discount is

160 + 3.2b

. Applying a

15\%

discount means paying

85\%

of this total.

0.85(160 + 3.2b) = 680

This sets up the relationship between the total discounted bill and the number of brochures.

2

Divide both sides of the equation by

0.85

to isolate the expression within parentheses.

160 + 3.2b = 800

Dividing by

0.85

simplifies the equation by calculating the total cost before the discount was applied.

3

Subtract

160

from both sides to isolate the term containing the variable

b

.

3.2b = 640

Subtracting the flat setup fee isolates the printing cost of the brochures.

4

Divide both sides by

3.2

to solve for

b

.

b = 200

Dividing by the per-brochure printing cost yields the total number of brochures printed.

Key Concept

Solving multi-step linear equations involving percentage discounts and decimal coefficients.

Alternative Method

Instead of dividing by

0.85

first, you can distribute

0.85

to both terms inside the parentheses to get

136 + 2.72b = 680

. Subtract

136

from both sides to get

2.72b = 544

, then divide by

2.72

to find

b = 200

.

Estimated Time:1m 30s

Question 2289Question →

In the standard $(x, y)$ coordinate plane, a rhombus $ABCD$ has vertices $A(-2, 3)$ and $C(4, 1)$ . If vertex $B$ lies on the line $y = 2x - 5$ , what is the area of the rhombus $ABCD$ ?

20

40

80

100

200

Show answer & explanation

Answer: 100

Answer

100

The correct answer is 100. By finding the midpoint of diagonal

AC

as

(1, 2)

and using the perpendicular slope of

3

, the line containing diagonal

BD

is

y = 3x - 1

. Intersecting this with

y = 2x - 5

yields vertex

B(-4, -13)

, which gives vertex

D(6, 17)

by midpoint symmetry. The diagonal lengths are

2\sqrt{10}

and

10\sqrt{10}

, and their half-product is 100.

Step-by-Step Solution

1

Find the midpoint

M

of diagonal

AC

.

M(1, 2)

Since

ABCD

is a rhombus, its diagonals bisect each other at their midpoint.

2

Calculate the slope of diagonal

AC

.

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

The slope is calculated using the formula

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

with vertices

A(-2, 3)

and

C(4, 1)

.

3

Find the equation of the line containing diagonal

BD

.

y = 3x - 1

The diagonals of a rhombus are perpendicular. The slope of

BD

is the negative reciprocal of

-\frac{1}{3}

, which is

3

. The line passes through

M(1, 2)

.

4

Find the coordinates of vertex

B

.

B(-4, -13)

Vertex

B

lies at the intersection of the diagonal line

y = 3x - 1

and the given line

y = 2x - 5

.

5

Find the coordinates of vertex

D

.

D(6, 17)

Since

M(1, 2)

is the midpoint of diagonal

BD

, we solve

\frac{-4 + x_D}{2} = 1

and

\frac{-13 + y_D}{2} = 2

.

6

Calculate the lengths of the diagonals

AC

and

BD

.

AC = 2\sqrt{10}

and

BD = 10\sqrt{10}

Use the distance formula

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

on the pairs of vertices.

7

Calculate the area of the rhombus.

100

The area of a rhombus is given by

\text{Area} = \frac{1}{2} \times d_1 \times d_2

.

Key Concept

Using perpendicular bisector properties of rhombus diagonals on a coordinate plane to find vertices and calculate area.

Question 2290Question →

A straight road on a map is modeled by the linear equation $2x - 5y = 10$ in the standard $(x, y)$ coordinate plane. A second road, which is parallel to the first road, is being constructed. What is the slope of the second road?

2

-\frac{5}{2}

\frac{2}{5}

-\frac{2}{5}

\frac{5}{2}

Show answer & explanation

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

Answer

The slope of the second road is

\frac{2}{5}

Parallel lines have the same slope. Rewriting the line's equation

2x - 5y = 10

into slope-intercept form (

y = mx + b

) yields

y = \frac{2}{5}x - 2

. The slope of this line is the coefficient of

x

, which is

\frac{2}{5}

. Therefore, any line parallel to it must also have a slope of

\frac{2}{5}

.

Step-by-Step Solution

1

Determine the relationship between the slopes of parallel lines.

Parallel lines have identical slopes.

Since the second road is parallel to the first, its slope must be equal to the slope of the first road.

2

Convert the equation of the first road,

2x - 5y = 10

, into slope-intercept form (

y = mx + b

).

Subtract

2x

from both sides to get

-5y = -2x + 10

. Then, divide both sides by

-5

to get

y = \frac{2}{5}x - 2

.

In slope-intercept form, the coefficient of

x

(represented by

m

) is the slope of the line.

3

Identify the slope of the first road and match it to the parallel road.

The slope of the first road is

\frac{2}{5}

, so the slope of the parallel road is also

\frac{2}{5}

.

The slope is the coefficient of

x

, which is

\frac{2}{5}

.

Key Concept

Parallel lines in a coordinate plane have equal slopes.

Question 2291Question →

In the standard $(x, y)$ coordinate plane, point $P$ undergoes two transformations. First, it is reflected across the $y$ -axis. Second, it is translated $3$ units down and $4$ units to the right, resulting in the image point $(1, -2)$ . What are the coordinates of the original point $P$ ?

(-3, -1)

(3, -5)

(-5, -5)

(3, 1)

(1, -3)

Show answer & explanation

Answer: $(3, 1)$

Answer

The coordinates of the original point

P

are

(3, 1)

.

To find the coordinates of the original point, we must work backward from the final image

(1, -2)

. First, we undo the translation (which was

3

units down and

4

units to the right) by performing the opposite actions: moving

3

units up and

4

units to the left. This shifts

(1, -2)

to

(1 - 4, -2 + 3) = (-3, 1)

. Next, we undo the reflection across the

y

-axis. Reflecting a point across the

y

-axis negates its

x

-coordinate. The reflection of

(-3, 1)

across the

y

-axis yields the original point

(3, 1)

.

Step-by-Step Solution

1

Identify the transformations in reverse order to work backward from the final image

(1, -2)

to the original point

P

.

The final image is

(1, -2)

. We must first undo the translation (

3

units down,

4

units right) and then undo the reflection across the

y

-axis.

To find the pre-image, we apply the inverse transformations in the reverse order of the original operations.

2

Undo the translation by applying the opposite operations: move

3

units up and

4

units to the left.

The intermediate point is

(1 - 4, -2 + 3) = (-3, 1)

.

Undoing a translation of

+4

in the

x

-direction and

-3

in the

y

-direction requires subtracting

4

from the

x

-coordinate and adding

3

to the

y

-coordinate.

3

Undo the reflection across the

y

-axis by reflecting the intermediate point

(-3, 1)

across the

y

-axis.

The original point

P

is

(3, 1)

.

Reflecting a point across the

y

-axis negates its

x

-coordinate. Since reflection is its own inverse, applying it again returns the original coordinates:

-(-3) = 3

.

Key Concept

Working backward through composite transformations in the coordinate plane.

Alternative Method

Instead of working backward step-by-step, write the transformation equations. If the original point is

P(x, y)

, the reflection across the

y

-axis gives

P'(-x, y)

. The subsequent translation of

3

units down and

4

units right gives

P''(-x + 4, y - 3)

. Set this expression equal to the final coordinates:

-x + 4 = 1 \implies x = 3

and

y - 3 = -2 \implies y = 1

, which gives the original point

(3, 1)

.

Estimated Time:1m 30s

Question 2292Question →

In the standard $(x, y)$ coordinate plane, a triangle is first reflected across the line $y = -x$ and then translated 4 units to the right and 3 units down. If the vertex $A$ of the triangle is mapped to the final image point $A''(1, -5)$ under this composite transformation, what were the coordinates of the original vertex $A$ ?

(8, -5)

(-2, -3)

(2, 3)

(1, 2)

(9, -4)

Show answer & explanation

Answer: $(2, 3)$

Answer

The original coordinates of vertex

A

are

(2, 3)

.

To find the original coordinates of the vertex, the transformations must be undone in reverse order. First, reverse the translation of 4 units right and 3 units down by moving the final image point

A''(1, -5)

4 units left and 3 units up, yielding the intermediate point

A'(-3, -2)

. Second, reverse the reflection across

y = -x

by applying the rule

(x, y) \rightarrow (-y, -x)

to

A'

, resulting in the original coordinates

(2, 3)

.

Step-by-Step Solution

1

Identify the inverse of each transformation and their correct reverse order. The composite transformation consists of (1) reflection across the line

y = -x

, followed by (2) translation of 4 units to the right and 3 units down. To backtrack from the final image, we must apply the inverse transformations in reverse order: first (1) reverse the translation, then (2) reverse the reflection.

The reverse process is: first translate 4 units left and 3 units up, then reflect across the line

y = -x

.

Reversing the composite transformation requires reversing both the individual operations and the order of their application.

2

Apply the inverse translation to the image point

A''(1, -5)

. Reversing a translation of 4 units right and 3 units down means translating 4 units left and 3 units up, represented by the rule

(x, y) \rightarrow (x - 4, y + 3)

.

The intermediate point is

A' = (1 - 4, -5 + 3) = (-3, -2)

.

This step determines the coordinates of the vertex after the reflection but before the translation.

3

Apply the inverse reflection to

A'(-3, -2)

. Since a reflection is its own inverse, we reflect

A'(-3, -2)

across the line

y = -x

using the rule

(x, y) \rightarrow (-y, -x)

.

The original point is

A = (-(-2), -(-3)) = (2, 3)

.

This step yields the final pre-image coordinates of vertex

A

.

Key Concept

Backtracking a composite transformation in the coordinate plane by applying inverse transformations in reverse order.

Question 2293Question →

In the standard $(x, y)$ coordinate plane, a line $L_1$ is perpendicular to the line that contains the points $(3, 5)$ and $(-1, 8)$ . If $L_1$ is also parallel to the line defined by the equation $ax + 6y = 15$ , what is the value of the constant $a$ ?

Show answer & explanation

Answer: -8

Answer

The value of the constant

a

is

-8

.

The slope of the line containing

(3, 5)

and

(-1, 8)

is

-\frac{3}{4}

. The slope of a line perpendicular to it is the negative reciprocal, which is

\frac{4}{3}

. Because line

L_1

is parallel to the line

ax + 6y = 15

, they must have equal slopes. The slope of

ax + 6y = 15

is

-\frac{a}{6}

. Setting the two slopes equal gives

-\frac{a}{6} = \frac{4}{3}

, which yields

a = -8

.

Step-by-Step Solution

1

Calculate the slope of the line containing the points

(3, 5)

and

(-1, 8)

.

The slope is

-\frac{3}{4}

.

Using the slope formula

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

for the points

(3, 5)

and

(-1, 8)

, we get

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

.

2

Determine the slope of line

L_1

using the perpendicular relationship.

The slope of

L_1

is

\frac{4}{3}

.

Since line

L_1

is perpendicular to the line with slope

-\frac{3}{4}

, its slope must be the negative reciprocal, which is

-\frac{1}{-3/4} = \frac{4}{3}

.

3

Express the slope of the line

ax + 6y = 15

in terms of

a

.

The slope is

-\frac{a}{6}

.

Rewriting the equation

ax + 6y = 15

in slope-intercept form (

y = mx + b

) gives

6y = -ax + 15

, which simplifies to

y = -\frac{a}{6}x + \frac{5}{2}

. The slope is the coefficient of

x

, which is

-\frac{a}{6}

.

4

Set the slope of

L_1

equal to the slope of the parallel line to solve for

a

.

a = -8

Because line

L_1

is parallel to the line

ax + 6y = 15

, their slopes are equal:

-\frac{a}{6} = \frac{4}{3}

. Multiplying both sides by

-6

gives

a = -8

.

Key Concept

Parallel lines have equal slopes, and perpendicular lines have slopes that are negative reciprocals of each other.

Estimated Time:1m 30s

Question 2294Question →

A linear equation in the standard $(x, y)$ coordinate plane is given by $y = -4$ . Is the statement that the slope of this line is undefined true or false?

True

False

Show answer & explanation

Answer: False

Answer

False

The correct answer is false because the line

y = -4

is horizontal, and all horizontal lines have a slope of

0

. A slope is only undefined for vertical lines, which are written in the form

x = c

.

Step-by-Step Solution

1

Identify the orientation of the line given by the equation

y = -4

.

The equation is of the form

y = c

, where

c

is a constant. This represents a horizontal line where every point on the line has a

y

-coordinate of

-4

.

Since the

y

-value never changes regardless of the

x

-value, the line is parallel to the

x

-axis.

2

Calculate the slope using two points on the line.

Let the two points be

(0, -4)

and

(1, -4)

. The slope

m

is given by

m = \frac{-4 - (-4)}{1 - 0} = \frac{0}{1} = 0

.

The slope formula is

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

.

3

Compare the calculated slope to the statement in the question.

The calculated slope is

0

, which is a defined number. The statement claims the slope is undefined, which is only true for vertical lines. Therefore, the statement is false.

An undefined slope occurs when dividing by zero, which is not the case for a horizontal line.

Key Concept

The slope of any horizontal line is

0

, whereas the slope of any vertical line is undefined.

Question 2295Question →

If $u$ and $v$ are positive real numbers such that $\frac{(u^{-2} v^3)^3}{u^k v^{-1}} = \frac{v^{10}}{u^{10}}$ , what is the value of the exponent $k$ ?

Show answer & explanation

Answer: 4

Answer

The value of the exponent is 4.

By applying the exponent rules systematically, the expression on the left simplifies to

u^{-6-k} v^{10}

, and the expression on the right is

u^{-10} v^{10}

. Equating the exponents of

u

gives

-6-k = -10

, which solves to

k = 4

.

Step-by-Step Solution

1

Apply the power of a product and power of a power properties to the numerator

(u^{-2} v^3)^3

.

u^{-6} v^9

According to the power of a product rule,

(xy)^a = x^a y^a

, and the power of a power rule,

(x^a)^b = x^{ab}

.

2

Use the quotient of powers property to simplify the left side of the equation.

u^{-6-k} v^{10}

The quotient of powers rule states that

\frac{x^a}{x^b} = x^{a-b}

, so the exponents of like bases are subtracted:

-6 - k

for

u

and

9 - (-1) = 10

for

v

.

3

Rewrite the right side of the equation,

\frac{v^{10}}{u^{10}}

, using a negative exponent.

u^{-10} v^{10}

Applying the negative exponent rule,

\frac{1}{x^a} = x^{-a}

.

4

Set the simplified expressions equal and solve for

k

.

k = 4

Since

u^{-6-k} v^{10} = u^{-10} v^{10}

, the exponents of the base

u

must be equal. Therefore,

-6 - k = -10

, which simplifies to

k = 4

.

Key Concept

Properties of Exponents in Algebraic Expressions

Estimated Time:1m 30s

Question 2296Question →

If $b$ is a positive real number unequal to $1$ such that $\log_b 3 = x$ and $\log_b 5 = y$ , what is the value of $\log_b \left( \frac{45}{b^2} \right)$ in terms of $x$ and $y$ ?

2x + y - 2

x^2 + y - 2

\frac{2x + y}{2}

2x + y - 2b

2x - y - 2

Show answer & explanation

Answer: $2x + y - 2$

Answer

2x + y - 2

The correct answer is

2x + y - 2

. By applying the quotient property of logarithms,

\log_b \left( \frac{45}{b^2} \right)

is rewritten as

\log_b 45 - \log_b(b^2)

. Factoring

45

as

3^2 \cdot 5

allows the first term to be expanded using the product and power properties into

2\log_b 3 + \log_b 5

. Simplifying

\log_b(b^2)

to

2

and substituting

x

and

y

yields

2x + y - 2

.

Step-by-Step Solution

1

Apply the quotient property of logarithms.

\log_b \left( \frac{45}{b^2} \right) = \log_b 45 - \log_b(b^2)

The logarithm of a quotient is the difference of the logarithms of the numerator and the denominator:

\log_b(M/N) = \log_b M - \log_b N

.

2

Factor the number 45 and apply the product property of logarithms.

\log_b(3^2 \cdot 5) - \log_b(b^2) = \log_b(3^2) + \log_b 5 - \log_b(b^2)

Since

45 = 9 \cdot 5 = 3^2 \cdot 5

, we can use the product property:

\log_b(M \cdot N) = \log_b M + \log_b N

.

3

Apply the power property to simplify the terms.

2\log_b 3 + \log_b 5 - 2

The power property states that

\log_b(M^k) = k\log_b M

. Also,

\log_b(b^2) = 2

because the base

b

raised to the second power is

b^2

.

4

Substitute the given values

x = \log_b 3

and

y = \log_b 5

.

2x + y - 2

Replacing the logarithmic expressions with

x

and

y

gives the final simplified expression.

Key Concept

Applying logarithmic properties (quotient, product, power) to simplify expressions

Estimated Time:1m 0s

Question 2297Question →

A triangle has two sides of length $8$ and $15$ . If the length of the third side, $s$ , is a prime number, how many possible values are there for $s$ ?

3

4

5

6

7

Show answer & explanation

Answer: $4$

Answer

There are 4 possible values for the third side length s.

The correct answer is

4

. According to the Triangle Inequality Theorem, the length of the third side

s

of a triangle with sides of

8

and

15

must satisfy

15 - 8 < s < 15 + 8

. This simplifies to the open interval

7 < s < 23

. The prime numbers strictly between

7

and

23

are

11

,

13

,

17

, and

19

. Counting these gives exactly

4

possible prime values for

s

.

Step-by-Step Solution

1

Apply the Triangle Inequality Theorem to determine the bounds for the third side

s

.

15 - 8 < s < 15 + 8

, which simplifies to

7 < s < 23

.

The Triangle Inequality Theorem states that the length of any side of a triangle must be strictly greater than the difference of the other two sides and strictly less than their sum.

2

Identify all prime numbers that lie strictly within the range

(7, 23)

.

The prime numbers in this range are

11

,

13

,

17

, and

19

.

A prime number is an integer greater than 1 that has no positive divisors other than 1 and itself.

3

Count the number of identified prime numbers.

There are

4

prime numbers (

11

,

13

,

17

,

19

).

This counts the total number of possible valid lengths for

s

.

Key Concept

Triangle Inequality Theorem and basic number properties

Question 2298Question →

What value of the base $b$ satisfies the equation $\log_b (3b + 10) = 2$ ?

Show answer & explanation

Answer: 5

Answer

The correct answer is 5.

Applying the definition of logarithms to the equation

\log_b(3b + 10) = 2

converts it to the quadratic equation

b^2 = 3b + 10

. Subtracting

3b

and

10

from both sides results in

b^2 - 3b - 10 = 0

. Factoring the quadratic yields

(b - 5)(b + 2) = 0

, giving the potential solutions

b = 5

and

b = -2

. Because the base of a logarithm must be positive (

b > 0

), the negative solution is extraneous. This leaves

5

as the only valid base.

Step-by-Step Solution

1

Convert the logarithm to exponential form.

b^2 = 3b + 10

By the definition of logarithms,

\log_b(x) = y

is equivalent to

b^y = x

.

2

Rearrange into standard quadratic form.

b^2 - 3b - 10 = 0

Subtracting

3b

and

10

from both sides sets the quadratic expression equal to zero.

3

Factor the quadratic equation.

(b - 5)(b + 2) = 0

Finding two integers that multiply to

-10

and add to

-3

gives

-5

and

2

.

4

Solve for the variable and apply base constraints.

b = 5

Solving the factored equation yields

b = 5

or

b = -2

. Since a logarithmic base must be strictly positive (

b > 0

), we reject

b = -2

as extraneous, leaving

b = 5

.

Key Concept

Converting logarithmic equations to exponential form and verifying base restrictions.

Estimated Time:1m 30s

Question 2299Question →

In a triangle, two of the sides have lengths $13$ and $20$ . The third side has a length of $s$ , where $s$ is an integer. If the side of length $20$ is the longest side of the triangle, and the triangle is obtuse, what is the number of possible values for $s$ ?

Show answer & explanation

Answer: 8

Answer

There are 8 possible integer values for

s

.

To find the number of possible integer values for

s

, we combine the Triangle Inequality Theorem (

13 + s > 20 \implies s > 7

) and the condition for an obtuse triangle with

20

as the longest side (

20^2 > 13^2 + s^2 \implies s^2 < 231 \implies s \leq 15

). This limits

s

to integers in the range

[8, 15]

, which contains exactly

8

values.

Step-by-Step Solution

1

Apply the Triangle Inequality Theorem to find the lower bound for

s

.

s > 7

, so the minimum integer value is

8

.

The sum of the two shorter sides of a triangle must be strictly greater than the longest side.

2

Set up the obtuse triangle inequality with

20

as the longest side.

20^2 > 13^2 + s^2

In any obtuse triangle with longest side

c

, the inequality

c^2 > a^2 + b^2

must hold.

3

Solve the inequality

20^2 > 13^2 + s^2

for

s

.

s^2 < 231 \implies s \leq 15

Simplifying the inequality gives

400 > 169 + s^2 \implies s^2 < 231

. The largest integer whose square is less than

231

is

15

.

4

Determine the number of integers in the range

[8, 15]

.

8 possible values

The integers satisfying both conditions are

\{8, 9, 10, 11, 12, 13, 14, 15\}

, which count to

8

.

Key Concept

Triangle Inequality Theorem and obtuse triangle classification using side lengths

Alternative Method

List the perfect squares and verify which ones satisfy both

s^2 < 231

and the Triangle Inequality Theorem

s > 7

.

Estimated Time:2m 0s

Question 2300Question →

The functions $f$ and $g$ are defined for all permissible real numbers by $f(x) = \frac{x + 3}{x - 1}$ and $g(x) = 2x - 5$ . If $(f \circ g)(x) = 3$ , what is the value of $x$ ?

Show answer & explanation

Answer: 4

Answer

The value of

x

that satisfies the equation

(f \circ g)(x) = 3

is

4

.

To solve for

x

in

(f \circ g)(x) = 3

, we find the composite function

f(g(x))

by substituting

g(x) = 2x - 5

into

f(x)

. This yields

f(g(x)) = \frac{(2x-5)+3}{(2x-5)-1} = \frac{2x-2}{2x-6}

. Setting this equal to

3

gives

\frac{2x-2}{2x-6} = 3

. Multiplying by

2x-6

yields

2x-2 = 6x-18

. Rearranging terms to isolate

x

gives

4x = 16

, which results in

x = 4

.

Step-by-Step Solution

1

Substitute the expression for

g(x)

into

f(x)

to obtain the composite function

(f \circ g)(x)

.

(f \circ g)(x) = \frac{2x - 2}{2x - 6}

By definition of function composition,

(f \circ g)(x) = f(g(x))

. Substituting

g(x) = 2x - 5

into

f(x) = \frac{x+3}{x-1}

yields

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

, which simplifies to

\frac{2x-2}{2x-6}

.

2

Set the composite function expression equal to

3

and solve for

x

.

x = 4

We set

\frac{2x-2}{2x-6} = 3

. Multiplying both sides by the denominator

2x-6

gives

2x-2 = 3(2x-6)

. Expanding the right side gives

2x-2 = 6x-18

. Subtracting

2x

from both sides and adding

18

to both sides results in

16 = 4x

. Dividing by

4

gives

x = 4

.

Key Concept

Function Composition and Evaluation

Estimated Time:1m 30s

Answer

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