All practice questions

Question 301Question →

In a triangle, the lengths of the sides are $x$ , $y$ , and $z$ , where $x$ , $y$ , and $z$ are integers such that $x < y < z$ . If $x = 7$ and the perimeter of the triangle is $32$ , what is the number of possible integer values for $z$ ?

Show answer & explanation

Answer: 3

Answer

There are exactly 3 possible integer values for the side length z.

The correct answer is 3. By expressing the second side as

y = 25 - z

and applying the ordering constraint

7 < 25 - z < z

, we determine that

12.5 < z < 18

. Applying the Triangle Inequality Theorem (

7 + y > z

) yields the restriction

z < 16

. Combining these conditions restricts the integer values of

z

to

\{13, 14, 15\}

, which counts to exactly 3 possible values.

Step-by-Step Solution

1

Express the side length

y

in terms of

z

.

y = 25 - z

The perimeter of the triangle is the sum of the side lengths:

x + y + z = 32

. Substituting

x = 7

gives

7 + y + z = 32

, which simplifies to

y = 25 - z

.

2

Apply the given inequality constraint

x < y < z

to find initial bounds for

z

.

12.5 < z < 18

Substituting

x = 7

and

y = 25 - z

into

x < y < z

yields

7 < 25 - z < z

. The left inequality

7 < 25 - z

simplifies to

z < 18

. The right inequality

25 - z < z

simplifies to

25 < 2z

, or

z > 12.5

.

3

Apply the Triangle Inequality Theorem to establish the final constraint on

z

.

z < 16

Since

z

is the longest side, the sum of the two shorter sides must be strictly greater than

z

:

x + y > z

. Substituting

x = 7

and

y = 25 - z

gives

7 + 25 - z > z

, which simplifies to

32 > 2z

, or

z < 16

.

4

Combine all constraints and count the valid integer values for

z

.

3 possible values (

13, 14, 15

)

Combining the bounds from the steps gives

12.5 < z < 16

. The integers satisfying this inequality are

13

,

14

, and

15

, which gives a total of 3 possible integer values.

Key Concept

Triangle Inequality Theorem and algebraic constraints on side lengths

Question 302Question →

A shipping company charges a rate based on the weight of a package. The total cost $C$ , in dollars, to ship a package of weight $w$ pounds is given by the formula $C = \frac{5}{8}(w - 2) + 6.50$ for packages weighing more than $2$ pounds. If the shipping cost for a certain package is $\$14.00$ , what is the weight of the package, in pounds?

Show answer & explanation

Answer: 14

Answer

The weight of the package is 14 pounds.

Substituting

C = 14.00

into the formula gives

14.00 = \frac{5}{8}(w - 2) + 6.50

. Subtracting

6.50

from both sides yields

7.50 = \frac{5}{8}(w - 2)

. Multiplying both sides by the reciprocal

\frac{8}{5}

yields

12 = w - 2

. Finally, adding

2

to both sides gives the weight

w = 14

pounds.

Step-by-Step Solution

1

Substitute the total shipping cost into the formula.

14.00 = \frac{5}{8}(w - 2) + 6.50

Since the shipping cost

C

is given as

14.00

, we substitute this value into the formula to solve for the unknown weight

w

.

2

Subtract

6.50

from both sides of the equation.

7.50 = \frac{5}{8}(w - 2)

Subtracting

6.50

isolates the term containing the variable

w

on the right side of the equation.

3

Multiply both sides of the equation by the reciprocal of the fraction.

12 = w - 2

Multiplying by

\frac{8}{5}

eliminates the fractional coefficient of

\frac{5}{8}

on the right side.

4

Add

2

to both sides of the equation to solve for

w

.

w = 14

Adding

2

isolates the variable

w

, giving the final weight of the package.

Key Concept

Solving linear equations with fractional and decimal terms

Estimated Time:1m 30s

Question 303Question →

A line segment in the standard $(x, y)$ coordinate plane has endpoints at $A(-1, 10)$ and $B(7, -2)$ . If point $M$ is the midpoint of this segment, what is the distance from $M$ to the origin $(0, 0)$ ?

Show answer & explanation

Answer: 5

Answer

The distance from the midpoint

M

to the origin is 5.

To find the distance from the midpoint

M

to the origin

(0,0)

, we first determine the coordinates of

M

by taking the average of the coordinates of

A(-1, 10)

and

B(7, -2)

. This results in

M\left(\frac{-1+7}{2}, \frac{10+(-2)}{2}\right) = (3, 4)

. Next, we apply the distance formula between

M(3,4)

and the origin

(0,0)

to get

\sqrt{(3-0)^2 + (4-0)^2} = \sqrt{9+16} = \sqrt{25} = 5

.

Step-by-Step Solution

1

Calculate the coordinates of the midpoint

M

of the segment

AB

.

The midpoint is

M(3, 4)

.

The midpoint formula is

M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)

. Substituting the coordinates of

A(-1, 10)

and

B(7, -2)

gives the x-coordinate as

\frac{-1 + 7}{2} = 3

and the y-coordinate as

\frac{10 + (-2)}{2} = 4

.

2

Calculate the distance from the midpoint

M(3, 4)

to the origin

(0, 0)

.

The distance is 5.

The distance formula is

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

. Substituting the coordinates of

M(3, 4)

and the origin

(0, 0)

gives

d = \sqrt{(3 - 0)^2 + (4 - 0)^2} = \sqrt{9 + 16} = \sqrt{25} = 5

.

Key Concept

Using the midpoint formula to find the center point of a line segment, and then using the distance formula to find the length between that point and another specified coordinate.

Question 304Question →

In a right triangle, the length of the side opposite the $60^\circ$ angle is $6\sqrt{3}$ centimeters. What is the length, in centimeters, of the hypotenuse of this triangle?

Show answer & explanation

Answer: 12

Answer

The length of the hypotenuse is

12

centimeters.

In a

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

special right triangle, the sides opposite the

30^\circ

angle, the

60^\circ

angle, and the

90^\circ

(hypotenuse) angle are in the ratio

x : x\sqrt{3} : 2x

. Given that the side opposite the

60^\circ

angle is

6\sqrt{3}

centimeters, we have

x\sqrt{3} = 6\sqrt{3}

, which means

x = 6

. The hypotenuse is

2x = 2(6) = 12

centimeters.

Step-by-Step Solution

1

Determine the type of special right triangle.

A

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

right triangle.

Since the triangle is a right triangle and has a

60^\circ

angle, the remaining angle must be

180^\circ - 90^\circ - 60^\circ = 30^\circ

.

2

Set up the relation for the side lengths using the ratio of a

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

triangle.

The side opposite the

60^\circ

angle is

x\sqrt{3}

centimeters, where

x

is the length of the side opposite the

30^\circ

angle.

In any

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

triangle, the side lengths are in the ratio

1 : \sqrt{3} : 2

.

3

Solve for the base variable

x

.

x = 6

We are given that the side opposite the

60^\circ

angle is

6\sqrt{3}

centimeters, so

x\sqrt{3} = 6\sqrt{3}

.

4

Calculate the length of the hypotenuse.

The hypotenuse is

2x = 2(6) = 12

centimeters.

The hypotenuse of a

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

triangle is twice the length of the shorter leg, which is

2x

.

Key Concept

Using the side length ratios of a

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

special right triangle to find missing lengths.

Question 305Question →

A straight line graphed on a coordinate grid has a slope of $1.25$ and passes through the point $(4, 11)$ . If the point $(k, -4)$ also lies on this line, what is the value of $k$ ?

Show answer & explanation

Answer: -8

Answer

The value of

k

is

-8

.

The correct answer is

-8

. By using the point-slope formula of a line, we determine the equation to be

y - 11 = 1.25(x - 4)

. Substituting the coordinates of the point

(k, -4)

yields the equation

-4 - 11 = 1.25(k - 4)

. Solving this equation results in

k = -8

.

Step-by-Step Solution

1

Use the point-slope formula of a line,

y - y_1 = m(x - x_1)

, to write the equation of the line.

y - 11 = 1.25(x - 4)

We are given the slope

m = 1.25

and a point

(4, 11)

on the line.

2

Substitute the coordinates of the second point,

(k, -4)

, into the line's equation.

-4 - 11 = 1.25(k - 4)

Since the point

(k, -4)

lies on the line, its coordinates must satisfy the line's equation.

3

Solve the equation for

k

.

k = -8

Simplify the left side to get

-15 = 1.25(k - 4)

, divide both sides by

1.25

to get

-12 = k - 4

, and add

4

to both sides to find

k = -8

.

Key Concept

Writing and solving linear equations using point-slope form.

Question 306Question →

A regular hexagon $ABCDEF$ has a side length of $8$ centimeters. A point $P$ lies on the side $CD$ such that the ratio of the length of $CP$ to the length of $PD$ is $1:3$ . What is the length, in centimeters, of the segment $AP$ ?

Show answer & explanation

Answer: 14

Answer

The length of the segment

AP

is

14

centimeters.

The correct answer is

14

. Dropping a perpendicular from

P

to the main diagonal

AD

creates a

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

triangle

\triangle PHD

with hypotenuse

PD = 6

. The legs are

DH = 3

and

PH = 3\sqrt{3}

. This leaves

AH = 13

. Applying the Pythagorean Theorem to the right triangle

\triangle AHP

with legs

13

and

3\sqrt{3}

yields

AP = \sqrt{13^2 + (3\sqrt{3})^2} = 14

.

Step-by-Step Solution

1

Determine the length of the main diagonal

AD

of the regular hexagon.

AD = 16

cm

In a regular hexagon with side length

s

, the main diagonal connecting opposite vertices has a length of

2s

. Given

s = 8

, we find

AD = 2 \times 8 = 16

.

2

Calculate the length of the segment

PD

on the side

CD

.

PD = 6

cm

The point

P

divides the side

CD

of length

8

in the ratio

CP:PD = 1:3

. Thus,

PD = \frac{3}{1+3} \times 8 = 6

.

3

Identify the angles and type of triangle formed by dropping a perpendicular from

P

to diagonal

AD

.

\triangle PHD

is a

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

right triangle.

The diagonal

AD

bisects the interior angle

\angle CDE = 120^\circ

of the regular hexagon, making

\angle ADC = 60^\circ

. Since

PH \perp AD

, the triangle

\triangle PHD

has angles

90^\circ

,

60^\circ

, and

30^\circ

.

4

Find the lengths of the legs

DH

and

PH

of the special right triangle

\triangle PHD

.

DH = 3

cm and

PH = 3\sqrt{3}

cm

Using the ratios of a

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

triangle with hypotenuse

PD = 6

, the leg adjacent to the

60^\circ

angle is

DH = 6 \cos(60^\circ) = 3

, and the leg opposite to the

60^\circ

angle is

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

.

5

Calculate the length of the segment

AH

.

AH = 13

cm

Since

H

lies on the diagonal

AD

, we subtract the length of

DH

from the total length of the diagonal:

AH = AD - DH = 16 - 3 = 13

.

6

Apply the Pythagorean Theorem to the right triangle

\triangle AHP

to find the length of

AP

.

AP = 14

cm

In the right triangle

\triangle AHP

with legs

AH = 13

and

PH = 3\sqrt{3}

, the hypotenuse is

AP = \sqrt{AH^2 + PH^2} = \sqrt{13^2 + (3\sqrt{3})^2} = \sqrt{169 + 27} = \sqrt{196} = 14

.

Key Concept

Applying special right triangle ratios and the Pythagorean Theorem in multi-step geometric figures.

Question 307Question →

A line, $L_1$ , is perpendicular to a second line whose equation is $5x - 2y = 15$ . The line $L_1$ intersects the $y$ -axis at $(0, 12)$ and passes through the midpoint of a line segment with endpoints at $(3, k)$ and $(7, 6)$ in the standard $(x, y)$ coordinate plane. What is the value of $k$ ?

Show answer & explanation

Answer: 14

Answer

The correct value of

k

is 14.

The slope of the line

5x - 2y = 15

is found by solving for

y

, yielding

y = \frac{5}{2}x - 7.5

. The slope of any perpendicular line is the negative reciprocal of

\frac{5}{2}

, which is

-\frac{2}{5}

. Given the

y

-intercept

(0, 12)

, the equation of the perpendicular line

L_1

is

y = -\frac{2}{5}x + 12

. The midpoint of the segment with endpoints

(3, k)

and

(7, 6)

is calculated as

\left(\frac{3+7}{2}, \frac{k+6}{2}\right) = \left(5, \frac{k+6}{2}\right)

. Since the midpoint lies on

L_1

, substituting

x = 5

into the line equation gives

y = -\frac{2}{5}(5) + 12 = 10

. Equating this to the midpoint's

y

-coordinate expression gives

\frac{k+6}{2} = 10

, which solves to

k = 14

.

Step-by-Step Solution

1

Find the slope of the line

5x - 2y = 15

by converting it to slope-intercept form (

y = mx + b

).

The slope of the line is

\frac{5}{2}

.

Rewriting the equation as

-2y = -5x + 15

and dividing by

-2

isolates

y

and reveals the slope.

2

Determine the perpendicular slope for line

L_1

.

The slope of

L_1

is

-\frac{2}{5}

.

Perpendicular lines have slopes that are negative reciprocals of one another.

3

Formulate the equation of line

L_1

using its slope and the given

y

-intercept

(0, 12)

.

The equation of

L_1

is

y = -\frac{2}{5}x + 12

.

The slope-intercept form is

y = mx + b

, where

m

is the slope and

b

is the

y

-coordinate of the

y

-intercept.

4

Find the midpoint of the line segment with endpoints

(3, k)

and

(7, 6)

in terms of

k

.

The midpoint is

\left(5, \frac{k+6}{2}\right)

.

The midpoint formula calculates the average of the

x

-coordinates and the average of the

y

-coordinates.

5

Substitute the midpoint coordinates into the equation of

L_1

and solve for

k

.

k = 14

Since the midpoint lies on line

L_1

, substituting its

x

and

y

values into the equation must satisfy the equality.

Key Concept

Using perpendicular slopes and the midpoint formula to determine unknown coordinate values.

Question 308Question →

One of the solutions to the quadratic equation $2x^2 - kx + 18 = 0$ is exactly four times the other solution. If $k$ is a positive constant, what is the value of $k$ ?

Show answer & explanation

Answer: 15

Answer

15

By representing the roots as

r

and

4r

, we can use Vieta's formulas to find that the product of the roots is

4r^2 = 9

, which gives

r = 1.5

. The sum of the roots is

5r = k/2

, which gives

k = 10(1.5) = 15

.

Step-by-Step Solution

1

Represent the roots of the quadratic equation.

Let the two solutions be

r

and

4r

, where

r

is a real number.

We are given that one solution is exactly four times the other.

2

Apply Vieta's formulas for the product of the roots.

r \cdot 4r = 4r^2 = \frac{18}{2} = 9

For a quadratic equation

ax^2 + bx + c = 0

, the product of the roots is equal to

\frac{c}{a}

.

3

Solve for the root

r

.

r^2 = \frac{9}{4} \Rightarrow r = 1.5

(since

k

is positive,

r

must be positive)

Solving the equation

4r^2 = 9

gives

r = \pm 1.5

. Since the sum of the roots is positive, we select the positive root.

4

Apply Vieta's formulas for the sum of the roots to find

k

.

r + 4r = 5r = -\frac{-k}{2} = \frac{k}{2} \Rightarrow k = 10r = 10(1.5) = 15

For a quadratic equation

ax^2 + bx + c = 0

, the sum of the roots is equal to

-\frac{b}{a}

.

Key Concept

Using Vieta's formulas to relate the roots of a quadratic equation to its coefficients.

Question 309Question →

A landscape architect is designing a park layout. The total area of the park is represented by the polynomial $A(t) = (2t - 3)(3t^2 + 4t - 1)$ square meters, while the area allocated for a parking lot is represented by $P(t) = 2t^2(3t - 5)$ square meters, where $t$ represents a scaling factor. The remaining green space in the park is modeled by the polynomial $g(t) = A(t) - P(t)$ . When $g(t)$ is written in the standard form $at^3 + bt^2 + ct + d$ , where $a$ , $b$ , $c$ , and $d$ are constants, what is the value of $b$ ?

Show answer & explanation

Answer: 9

Answer

The value of

b

, the coefficient of the

t^2

term in the standard form of

g(t)

, is

9

.

Expanding the polynomials gives

A(t) = 6t^3 - t^2 - 14t + 3

and

P(t) = 6t^3 - 10t^2

. Subtracting

P(t)

from

A(t)

yields

g(t) = (6t^3 - t^2 - 14t + 3) - (6t^3 - 10t^2) = 9t^2 - 14t + 3

. In the standard form

at^3 + bt^2 + ct + d

, the coefficient

b

of the

t^2

term is

9

.

Step-by-Step Solution

1

Expand the polynomial

A(t) = (2t - 3)(3t^2 + 4t - 1)

6t^3 - t^2 - 14t + 3

To represent the total area as a single polynomial in standard form before subtraction.

2

Expand the polynomial

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

6t^3 - 10t^2

To represent the parking lot area as a simplified polynomial in standard form.

3

Subtract

P(t)

from

A(t)

to find the green space polynomial

g(t)

9t^2 - 14t + 3

Subtracting

P(t)

from

A(t)

requires distributing the negative sign to both terms, yielding

-t^2 - (-10t^2) = 9t^2

.

4

Identify the coefficient

b

of the

t^2

term in

at^3 + bt^2 + ct + d

9

Comparing

g(t) = 9t^2 - 14t + 3

to the standard cubic form reveals that

a = 0

and

b = 9

.

Key Concept

Polynomial operations including expansion of products and subtraction with negative sign distribution

Question 310Question →

A line in the standard $(x, y)$ coordinate plane has a $y$ -intercept that is $4$ units greater than its slope. If the line passes through the point $(5, -2)$ , what is the $y$ -intercept of the line?

Show answer & explanation

Answer: 3

Answer

The correct answer is 3.

The

y

-intercept is found by setting up the equation using the given point and the relationship between the slope and the

y

-intercept, yielding

b = 3

.

Step-by-Step Solution

1

Define the relationship between the slope

m

and the

y

-intercept

b

.

m = b - 4

The problem states the

y

-intercept is

4

units greater than the slope, so

b = m + 4

.

2

Substitute the point

(5, -2)

and the slope expression into the slope-intercept equation

y = mx + b

.

-2 = (b - 4)(5) + b

The line passes through the point

(5, -2)

, so its coordinates must satisfy the line's equation.

3

Solve the algebraic equation for

b

.

b = 3

Expanding and simplifying the equation yields

-2 = 6b - 20

, which gives

6b = 18

, and dividing by

6

results in

b = 3

.

Key Concept

Linear Equations and Graphing

Question 311Question →

In the standard $(x, y)$ coordinate plane, a triangle has vertices at $A(-2, -3)$ , $B(6, 1)$ , and $C(1, 6)$ . The length of the altitude from vertex $C$ to side $AB$ can be written in the simplified radical form $a\sqrt{b}$ , where $a$ and $b$ are positive integers. What is the value of $a + b$ ?

Show answer & explanation

Answer: 8

Answer

The final answer is 8.

The altitude length of the triangle is

3\sqrt{5}

. In this simplified radical form, the coefficient

a

is 3 and the radicand

b

is 5. Summing these values gives

3 + 5 = 8

.

Step-by-Step Solution

1

Calculate the area of triangle

ABC

using the shoelace formula or by bounding the triangle in a rectangle.

Area = 30

The area is needed to determine the altitude length using the area formula of a triangle.

2

Calculate the length of the base side

AB

using the distance formula.

AB = 4\sqrt{5}

The base length is required to solve for the height perpendicular to it.

3

Set up the triangle area formula

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

to find the height

h

.

h = 3\sqrt{5}

This solves for the length of the altitude from vertex

C

to side

AB

.

4

Identify

a

and

b

from the simplified radical form

a\sqrt{b} = 3\sqrt{5}

and calculate

a + b

.

8

To provide the final requested sum.

Key Concept

Calculating the altitude of a triangle in the coordinate plane by relating its area and side lengths.

Question 312Question →

In right triangle $ABC$ , the hypotenuse $AC$ has a length of $15$ centimeters, and leg $AB$ has a length of $9$ centimeters. What is the length, in centimeters, of leg $BC$ ?

Show answer & explanation

Answer: 12

Answer

The length of leg

BC

is

12

centimeters.

The length of leg

BC

is found using the Pythagorean Theorem,

AB^2 + BC^2 = AC^2

. Substituting the given values gives

9^2 + BC^2 = 15^2

, which simplifies to

81 + BC^2 = 225

. Subtracting 81 from both sides yields

BC^2 = 144

. Taking the square root of 144 gives the correct length of 12 centimeters.

Step-by-Step Solution

1

Identify the given dimensions and apply the Pythagorean Theorem.

AB^2 + BC^2 = AC^2

For any right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.

2

Substitute the known values

AB = 9

and

AC = 15

into the equation.

9^2 + BC^2 = 15^2

The hypotenuse

AC

is the side opposite the right angle, and

AB

is one of the legs.

3

Simplify the squared terms.

81 + BC^2 = 225

Squaring 9 yields 81, and squaring 15 yields 225.

4

Isolate the unknown term by subtracting 81 from both sides.

BC^2 = 144

Subtracting 81 from both sides isolates

BC^2

on the left side of the equation.

5

Take the square root of both sides to solve for the leg length.

BC = 12

Taking the square root of 144 gives the side length, which must be positive.

Key Concept

Pythagorean Theorem

Question 313Question →

A circle in the standard $(x, y)$ coordinate plane has a diameter with one endpoint at $(-3, 5)$ and its center at $(2, -7)$ . What is the $y$ -coordinate of the other endpoint of the diameter?

Show answer & explanation

Answer: -19

Answer

The y-coordinate of the other endpoint of the diameter is -19.

Because the center of a circle is the midpoint of any diameter, the midpoint formula applies. For the y-coordinate, the equation is

y_m = \frac{y_1 + y_2}{2}

. Substituting the y-coordinate of the given endpoint (

5

) and the center (

-7

) yields

-7 = \frac{5 + y_2}{2}

. Multiplying both sides by 2 gives

-14 = 5 + y_2

. Subtracting 5 from both sides results in

-19

.

Step-by-Step Solution

1

Identify the relationship between the circle's center and its diameter.

The center of the circle,

(2, -7)

, is the midpoint of the diameter.

By definition, the center of a circle bisects any diameter, making it the midpoint of the diameter's endpoints.

2

Apply the midpoint formula to set up an equation for the y-coordinate.

y_m = \frac{y_1 + y_2}{2}

The y-coordinate of a midpoint is the average of the y-coordinates of the two endpoints.

3

Substitute the given values into the formula and solve for the unknown y-coordinate.

y_2 = -19

Substituting the given y-coordinates yields

-7 = \frac{5 + y_2}{2}

. Multiplying by 2 gives

-14 = 5 + y_2

, and subtracting 5 from both sides yields

-19

.

Key Concept

Finding a missing endpoint given the midpoint and one endpoint

Estimated Time:1m 0s

Question 314Question →

A marathon runner plans to run a total of $d$ miles this week. The runner schedules $\frac{2}{3}(d - 6)$ miles of the run on soft dirt trails and $0.4(d + 15)$ miles on asphalt roads. If the runner completes only these two segments for a total of $34$ miles, what is the value of $d$ ?

Show answer & explanation

Answer: 30

Answer

The value of

d

is 30.

The value of

d

that satisfies the equation is

30

because substituting

30

back into the original equation yields a true statement:

\frac{2}{3}(30 - 6) + 0.4(30 + 15) = 16 + 18 = 34

.

Step-by-Step Solution

1

Set up the equation based on the given information: the trail running distance plus the road running distance equals the total distance.

\frac{2}{3}(d - 6) + 0.4(d + 15) = 34

This relates the individual segment distances to the total planned mileage of

34

miles.

2

Distribute the coefficients to eliminate the parentheses.

\frac{2}{3}d - 4 + 0.4d + 6 = 34

Distributing

\frac{2}{3}

to

(d-6)

gives

\frac{2}{3}d - 4

, and distributing

0.4

to

(d+15)

gives

0.4d + 6

.

3

Combine the constant terms on the left side of the equation.

\frac{2}{3}d + 0.4d + 2 = 34

Combining the constants

-4

and

6

yields

2

.

4

Subtract 2 from both sides of the equation to isolate the variable terms.

\frac{2}{3}d + 0.4d = 32

Subtracting

2

from both sides simplifies the equation to have variable terms on one side and constant terms on the other.

5

Convert the decimal

0.4

to a fraction to combine the coefficients of

d

.

\frac{2}{3}d + \frac{2}{5}d = 32

Converting

0.4

to

\frac{4}{10} = \frac{2}{5}

allows us to work with a common denominator.

6

Find a common denominator to add the fractions.

\frac{10}{15}d + \frac{6}{15}d = 32 \Rightarrow \frac{16}{15}d = 32

The least common multiple of

3

and

5

is

15

.

7

Multiply both sides of the equation by the reciprocal of the coefficient of

d

to solve for

d

.

d = 32 \times \frac{15}{16} \Rightarrow d = 2 \times 15 \Rightarrow d = 30

Multiplying by

\frac{15}{16}

isolates

d

on the left side of the equation.

Key Concept

Solving linear equations involving fractions and decimals by distributing, combining like terms, and isolating the variable.

Alternative Method

Instead of converting the decimal

0.4

to a fraction, you can multiply the entire equation by a common multiple like

15

to eliminate both the fraction and the decimal:

15 \left[ \frac{2}{3}(d - 6) + 0.4(d + 15) \right] = 15(34) \Rightarrow 10(d - 6) + 6(d + 15) = 510 \Rightarrow 10d - 60 + 6d + 90 = 510 \Rightarrow 16d + 30 = 510 \Rightarrow 16d = 480 \Rightarrow d = 30

. This method avoids working with fractional coefficients.

Estimated Time:1m 30s

Question 315Question →

The volume of a water tank, in cubic feet, is given by the polynomial $V(t) = 3t(2t - 1)(t + 4)$ , where $t$ represents the time in hours since the pump was turned on. When the polynomial is written in standard form, what is the coefficient of the $t^2$ term?

Show answer & explanation

Answer: 21

Answer

21

To find the coefficient of the

t^2

term, we perform polynomial multiplication on the expression

V(t) = 3t(2t - 1)(t + 4)

. First, multiplying the binomials

(2t - 1)(t + 4)

yields

2t^2 + 7t - 4

. Next, distributing the

3t

term to the trinomial yields

6t^3 + 21t^2 - 12t

. The coefficient of the

t^2

term in this standard form polynomial is 21.

Step-by-Step Solution

1

Expand the product of the two binomials

(2t - 1)(t + 4)

.

2t^2 + 7t - 4

Use the distributive property (FOIL) to multiply the binomials:

2t(t) + 2t(4) - 1(t) - 1(4) = 2t^2 + 8t - t - 4 = 2t^2 + 7t - 4

.

2

Multiply the resulting trinomial by the monomial

3t

.

6t^3 + 21t^2 - 12t

Distribute

3t

to each term:

3t(2t^2) + 3t(7t) + 3t(-4) = 6t^3 + 21t^2 - 12t

.

3

Identify the coefficient of the

t^2

term.

21

The coefficient is the numerical factor of the

t^2

term, which is 21.

Key Concept

Operations on Polynomials

Question 316Question →

In acute triangle $PQR$ , an altitude $PS$ is drawn from vertex $P$ perpendicular to side $QR$ at point $S$ . The measure of $\angle PQS$ is $60^\circ$ , the length of segment $PQ$ is $24$ , and the length of segment $PR$ is $39$ . What is the length of side $QR$ ?

Show answer & explanation

Answer: 45

Answer

The length of side

QR

is

45

.

The altitude divides the acute triangle into two right triangles. In the first right triangle,

\triangle PQS

, the angles are

30^\circ

,

60^\circ

, and

90^\circ

, with a hypotenuse of

24

. This makes the adjacent leg

QS = 12

and the shared altitude

PS = 12\sqrt{3}

. In the second right triangle,

\triangle PRS

, the hypotenuse is

39

and one leg is

12\sqrt{3}

. Using the Pythagorean Theorem, we find the other leg

SR = \sqrt{39^2 - (12\sqrt{3})^2} = \sqrt{1521 - 432} = \sqrt{1089} = 33

. Summing the two segments gives the total length of side

QR = 12 + 33 = 45

.

Step-by-Step Solution

1

Identify the two right triangles formed by the altitude.

The altitude

PS

divides

\triangle PQR

into two adjacent right triangles:

\triangle PQS

and

\triangle PRS

, which share the side

PS

.

Establishing these right triangles allows us to apply right-triangle trigonometric ratios and the Pythagorean Theorem.

2

Use the properties of the

30^\circ\text{-}60^\circ\text{-}90^\circ

triangle

\triangle PQS

to find

QS

and

PS

.

QS = 12

and

PS = 12\sqrt{3}

.

In a

30^\circ\text{-}60^\circ\text{-}90^\circ

triangle, the leg opposite the

30^\circ

angle is half the hypotenuse, and the leg opposite the

60^\circ

angle is

\sqrt{3}

times the shorter leg. Here, hypotenuse

PQ = 24

, so

QS = 12

and

PS = 12\sqrt{3}

.

3

Apply the Pythagorean Theorem to

\triangle PRS

to find

SR

.

SR = 33

.

In right triangle

\triangle PRS

, the hypotenuse is

PR = 39

. By the Pythagorean Theorem,

PS^2 + SR^2 = PR^2

. Squaring the sides gives

(12\sqrt{3})^2 + SR^2 = 39^2 \implies 432 + SR^2 = 1521

. Solving for

SR

gives

SR^2 = 1089 \implies SR = 33

.

4

Sum the segments

QS

and

SR

to find the total length of

QR

.

QR = 45

.

Because

\triangle PQR

is an acute triangle, the altitude

PS

lands at a point

S

on the segment

QR

, meaning

QR = QS + SR

. Adding the lengths gives

12 + 33 = 45

.

Key Concept

Applying properties of

30^\circ\text{-}60^\circ\text{-}90^\circ

special right triangles and the Pythagorean Theorem in multi-step geometric figures.

Estimated Time:2m 30s

Question 317Question →

In the standard $(x, y)$ coordinate plane, a circle is defined by the equation $x^2 + y^2 - 10x - 4y = 0$ . The line $y = x$ intersects the circle at the origin $(0, 0)$ and at a second point $P$ . What is the $x$ -coordinate of point $P$ ?

Show answer & explanation

Answer: 7

Answer

The correct answer is 7.

To find the points of intersection between the circle and the line, we substitute the line's equation

y = x

into the circle's equation

x^2 + y^2 - 10x - 4y = 0

. This substitution yields

2x^2 - 14x = 0

. Factoring out

2x

gives

2x(x - 7) = 0

, which has solutions

x = 0

and

x = 7

. The solution

x = 0

corresponds to the origin

(0, 0)

. Therefore, the

x

-coordinate of the second point

P

is

7

.

Step-by-Step Solution

1

Substitute

y = x

into the equation of the circle.

x^2 + x^2 - 10x - 4x = 0

Since the line is

y = x

, any point of intersection must satisfy both equations.

2

Combine like terms to simplify the quadratic equation.

2x^2 - 14x = 0

Grouping the quadratic and linear terms yields a single equation in terms of

x

.

3

Factor the quadratic equation.

2x(x - 7) = 0

Factoring allows us to identify the individual roots of the equation.

4

Solve for

x

and identify the value for point

P

.

x = 7

The solution

x = 0

corresponds to the origin. Thus, the

x

-coordinate of the second point

P

must be

7

.

Key Concept

Finding points of intersection between a line and a circle in the coordinate plane.

Question 318Question →

In a right triangle, the measure of one of the acute angles is $30^\circ$ . If the side opposite this $30^\circ$ angle has a length of $6.5$ inches, what is the length, in inches, of the hypotenuse?

Show answer & explanation

Answer: 13

Answer

The length of the hypotenuse is

13

inches.

In a

30^\circ

-

60^\circ

-

90^\circ

special right triangle, the length of the hypotenuse is exactly twice the length of the shorter leg (the side opposite the

30^\circ

angle). Given that the shorter leg has a length of

6.5

inches, the hypotenuse has a length of

2 \times 6.5 = 13

inches.

Step-by-Step Solution

1

Determine the relationship between the given side and the hypotenuse using special right triangle properties.

The triangle is a

30^\circ

-

60^\circ

-

90^\circ

right triangle, meaning the hypotenuse is twice the length of the shorter leg.

By geometric theorem, the sides of a

30^\circ

-

60^\circ

-

90^\circ

triangle are in the ratio

1 : \sqrt{3} : 2

, with the shortest side opposite the

30^\circ

angle and the longest side being the hypotenuse.

2

Multiply the length of the side opposite the

30^\circ

angle by 2.

6.5 \text{ inches} \times 2 = 13 \text{ inches}

Doubling the shorter leg length of

6.5

inches gives the length of the hypotenuse.

Key Concept

In a

30^\circ

-

60^\circ

-

90^\circ

special right triangle, the length of the hypotenuse is always twice the length of the shorter leg (the side opposite the

30^\circ

angle).

Question 319Question →

Line $A$ passes through the points $(-3, 4)$ and $(5, 8)$ in a coordinate plane. Line $B$ is perpendicular to Line $A$ and passes through the point $(2, -1)$ . What is the $y$ -coordinate of the point where Line $B$ crosses the $y$ -axis?

Show answer & explanation

Answer: 3

Answer

The correct answer is 3.

The correct answer is 3. Finding the slope of the first line gives 0.5. The negative reciprocal of 0.5 is -2, which is the slope of the perpendicular line. Setting up the slope equation between the point (2, -1) and the y-intercept (0, 3) confirms the y-coordinate is indeed 3.

Step-by-Step Solution

1

Calculate the slope of Line A.

The slope of Line A is 0.5.

Using the slope formula

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

for points

(-3, 4)

and

(5, 8)

gives

m_A = \frac{8 - 4}{5 - (-3)} = \frac{4}{8} = 0.5

.

2

Find the slope of Line B.

The slope of Line B is -2.

Because Line B is perpendicular to Line A, its slope must be the negative reciprocal of the slope of Line A:

m_B = -\frac{1}{0.5} = -2

.

3

Determine the y-intercept of Line B.

The y-coordinate of the y-intercept is 3.

Line B passes through

(2, -1)

and

(0, b)

. Using the slope formula:

\frac{b - (-1)}{0 - 2} = -2 \implies \frac{b + 1}{-2} = -2 \implies b + 1 = 4 \implies b = 3

.

Key Concept

The slope of a line measures its steepness, calculated as the change in

y

divided by the change in

x

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

Alternative Method

Instead of using the slope formula directly to solve for the y-intercept, write the equation of Line B in point-slope form:

y - y_1 = m(x - x_1)

. Substituting the point

(2, -1)

and slope

-2

gives

y - (-1) = -2(x - 2)

, which simplifies to

y + 1 = -2x + 4

, or

y = -2x + 3

. To find the y-intercept, set

x = 0

, which yields

y = 3

.

Estimated Time:1m 30s

Question 320Question →

In $\triangle ABC$ , point $D$ lies on side $BC$ such that $AD = BD$ . If the measure of $\angle ADC$ is $112^\circ$ and the measure of $\angle BAC$ is $85^\circ$ , what is the measure of $\angle C$ , in degrees?

Show answer & explanation

Answer: 39

Answer

The measure of

\angle C

is

39^\circ

.

The measure of

\angle C

is found by first calculating the interior angle

\angle ADB = 180^\circ - 112^\circ = 68^\circ

since

BDC

forms a straight line. Because

AD = BD

,

\triangle ABD

is isosceles with

\angle B = \angle BAD

. Using the angle sum of

\triangle ABD

, we have

2(\angle B) + 68^\circ = 180^\circ

, which yields

\angle B = 56^\circ

. Finally, using the angle sum of

\triangle ABC

, we calculate

\angle C = 180^\circ - (85^\circ + 56^\circ) = 39^\circ

.

Step-by-Step Solution

1

Find the measure of

\angle ADB

using the linear pair relationship with

\angle ADC

.

\angle ADB = 68^\circ

Angles on a straight line add up to

180^\circ

. Since

D

lies on

BC

,

\angle ADB + \angle ADC = 180^\circ

.

2

Calculate the measure of

\angle B

using the properties of the isosceles triangle

ABD

.

\angle B = 56^\circ

Since

AD = BD

, the base angles opposite to these sides are equal:

\angle BAD = \angle B

. The sum of angles in

\triangle ABD

is

180^\circ

, so

2(\angle B) + 68^\circ = 180^\circ

.

3

Find the measure of

\angle C

using the triangle angle sum theorem on the large triangle

ABC

.

\angle C = 39^\circ

The sum of the angles in

\triangle ABC

is

180^\circ

, meaning

\angle BAC + \angle B + \angle C = 180^\circ

. Substituting the known values gives

85^\circ + 56^\circ + \angle C = 180^\circ

.

Key Concept

Using the Isosceles Triangle Theorem, the Triangle Angle Sum Theorem, and linear pairs to trace unknown angles in a geometric figure.

Practice More

Try finding the missing angles when a transversal cuts two parallel lines that form a triangle with a third intersecting line.

Alternative Method

Instead of finding

\angle B

first and then solving for

\angle C

in

\triangle ABC

, we can find the angle

\angle DAC

first. Since

\angle ADC = 112^\circ

is an exterior angle to

\triangle ABD

, we have

\angle ADC = \angle B + \angle BAD

. Since

\angle B = \angle BAD

, we get

2(\angle BAD) = 112^\circ \implies \angle BAD = 56^\circ

. Because

\angle BAC = 85^\circ

, we have

\angle DAC = 85^\circ - 56^\circ = 29^\circ

. Now looking at

\triangle ADC

, we can solve for

\angle C

directly:

\angle C = 180^\circ - (112^\circ + 29^\circ) = 39^\circ

.

Estimated Time:1m 30s

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