All practice questions

Question 221Question →

In the standard $(x,y)$ coordinate plane, a region in the first quadrant is bounded by the $x$ -axis, the $y$ -axis, and the line with equation $ax + by = c$ , where $a$ , $b$ , and $c$ are positive constants. The line passes through the point $(8, 18)$ . If the area of this region is minimized when $a = 3$ , what is the value of $c$ ?

Show answer & explanation

Answer: 48

Answer

48

Substituting the given point and

a = 3

into the equation yields

c = 24 + 18b

. The area of the triangle formed by the intercepts is

A = \frac{c^2}{6b}

. Substituting

c

gives

A = 6(9b + 24 + \frac{16}{b})

. Using AM-GM, the minimum occurs when

9b = \frac{16}{b}

, resulting in

b = \frac{4}{3}

. Using this value, we find

c = 48

.

Step-by-Step Solution

1

Substitute the point

(8, 18)

and

a = 3

into the equation

ax + by = c

.

24 + 18b = c

This establishes a relationship between the constants

b

and

c

based on the given point that lies on the line.

2

Calculate the

x

-intercept and

y

-intercept of the line.

x

-intercept is at

x = \frac{c}{3}

, and

y

-intercept is at

y = \frac{c}{b}

.

The boundary of the region in the first quadrant is defined by these coordinate intercepts.

3

Formulate the area

A

of the right triangle bounded by the axes and the line.

A = \frac{c^2}{6b}

The area of a right triangle with vertices at the origin and the intercepts is

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

.

4

Substitute

c = 24 + 18b

into the area formula and simplify.

A = 6\left(9b + 24 + \frac{16}{b}\right)

Expressing the area as a single-variable function of

b

allows us to find its minimum value.

5

Apply the AM-GM inequality to minimize the variable term

9b + \frac{16}{b}

.

b = \frac{4}{3}

minimizes the expression.

The sum of two positive terms is minimized when the terms are equal, so

9b = \frac{16}{b} \implies b^2 = \frac{16}{9} \implies b = \frac{4}{3}

.

6

Calculate the value of

c

using the minimizing value of

b

.

c = 48

Substituting

b = \frac{4}{3}

into the relation

c = 24 + 18b

yields the constant value

c

for the minimum area.

Key Concept

Minimizing the area bounded by a line and the coordinate axes using linear equation forms and algebraic minimization.

Alternative Method

Instead of using the AM-GM inequality, you can find the minimum by taking the derivative of the area function

A(b) = 54b + 144 + \frac{96}{b}

with respect to

b

. Setting the derivative

A'(b) = 54 - \frac{96}{b^2} = 0

yields

b^2 = \frac{96}{54} = \frac{16}{9}

, which gives

b = \frac{4}{3}

for

b > 0

.

Estimated Time:3m 0s

Question 222Question →

In the standard $(x, y)$ coordinate plane, an ellipse is centered at the origin $(0, 0)$ and has vertices at $(-5, 0)$ and $(5, 0)$ . If the distance between the two foci of the ellipse is $8$ , what is the length of the minor axis of the ellipse?

Show answer & explanation

Answer: 6

Answer

The correct answer is 6.

The correct answer is 6 because the ellipse has a horizontal major axis with

a = 5

and focal distance

c = 4

. Using the relationship

c^2 = a^2 - b^2

, we solve for the semi-minor axis

b

to get

b = 3

. The total length of the minor axis is

2b = 6

.

Step-by-Step Solution

1

Determine the semi-major axis length

a

.

a = 5

The vertices are at

(\pm 5, 0)

, which are

5

units from the center

(0, 0)

along the major axis.

2

Determine the distance from the center to each focus

c

.

c = 4

The distance between the two foci is

2c = 8

, so the distance from the center to a focus is

c = 4

.

3

Find the semi-minor axis length

b

.

b = 3

Using the relation

c^2 = a^2 - b^2

for ellipses, we get

4^2 = 5^2 - b^2 \implies b^2 = 9 \implies b = 3

.

4

Calculate the full length of the minor axis.

6

The length of the minor axis is

2b = 2(3) = 6

.

Key Concept

The relationship between the semi-major axis, semi-minor axis, and focal distance of an ellipse.

Estimated Time:1m 30s

Question 223Question →

Given the functions $f(x) = \sqrt{3x + 4}$ and $g(x) = x^2 - 5$ , what is the value of $g(f(7))$ ?

Show answer & explanation

Answer: 20

Answer

20

Evaluating the inner function first gives

f(7) = \sqrt{3(7) + 4} = 5

. Substituting this value into the outer function yields

g(5) = 5^2 - 5 = 20

.

Step-by-Step Solution

1

Evaluate the inner function

f(x)

at

x = 7

.

f(7) = 5

To evaluate a composite function of the form

g(f(x))

, we first calculate the value of the inner function

f(x)

at the given input.

2

Substitute the output of the inner function as the input for the outer function

g(x)

.

g(5) = 20

Since

f(7) = 5

, evaluating

g(f(7))

is equivalent to evaluating

g(5)

.

Key Concept

Function composition involves evaluating an inner function and then using that result as the input for an outer function.

Question 224Question →

If $x$ and $y$ are non-zero real numbers such that the expression $\frac{(x^2 y^a)^3}{x^5 y^{-2}}$ is equivalent to $x^b y^{14}$ for some integers $a$ and $b$ , what is the value of $a + b$ ?

Show answer & explanation

Answer: 5

Answer

The value of

a + b

is

5

.

Simplifying the expression

\frac{(x^2 y^a)^3}{x^5 y^{-2}}

using the exponent rules yields

x^{6-5} y^{3a-(-2)} = x^1 y^{3a+2}

. Equating this to

x^b y^{14}

shows that

b = 1

and

3a + 2 = 14

. Solving for

a

gives

a = 4

. Thus, the sum

a + b

is

4 + 1 = 5

.

Step-by-Step Solution

1

Simplify the numerator of the given expression.

x^6 y^{3a}

Applying the power of a product rule

(uv)^n = u^n v^n

and the power of a power rule

(u^m)^n = u^{mn}

to

(x^2 y^a)^3

results in

(x^2)^3 (y^a)^3 = x^6 y^{3a}

.

2

Simplify the quotient by subtracting exponents with the same base.

x^1 y^{3a+2}

Using the quotient rule

\frac{u^m}{u^n} = u^{m-n}

, the base

x

term becomes

x^{6-5} = x^1

, and the base

y

term becomes

y^{3a - (-2)} = y^{3a+2}

.

3

Equate the exponents of like bases to find the values of

a

and

b

.

b = 1

and

a = 4

Comparing

x^1 y^{3a+2}

to

x^b y^{14}

gives

b = 1

and

3a + 2 = 14

. Solving

3a + 2 = 14

yields

3a = 12

, which simplifies to

a = 4

.

4

Sum the values of

a

and

b

.

5

Adding

a

and

b

yields

4 + 1 = 5

.

Key Concept

Applying properties of exponents, including the power of a product, power of a power, and quotient rules, to simplify algebraic expressions.

Estimated Time:1m 30s

Question 225Question →

In the standard $(x, y)$ coordinate plane, a line segment has endpoints at $(3, -4)$ and $(9, 8)$ . What is the $y$ -coordinate of the midpoint of this line segment?

Show answer & explanation

Answer: 2

Answer

The

y

-coordinate of the midpoint is

2

.

The

y

-coordinate of the midpoint is calculated by finding the average of the

y

-coordinates of the endpoints:

\frac{y_1 + y_2}{2}

. Substituting

y_1 = -4

and

y_2 = 8

gives

\frac{-4 + 8}{2} = \frac{4}{2} = 2

.

Step-by-Step Solution

1

Identify the

y

-coordinates of the two given endpoints

(3, -4)

and

(9, 8)

.

y_1 = -4

and

y_2 = 8

The midpoint formula relies on the coordinates of the endpoints.

2

Calculate the average of the

y

-coordinates using the formula

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

.

y_m = \frac{-4 + 8}{2} = 2

The

y

-coordinate of a midpoint is the arithmetic mean of the

y

-coordinates of the endpoints.

Key Concept

Midpoint Formula

Estimated Time:45s

Question 226Question →

A botanist is monitoring the heights of two bamboo plants. Plant A is $20$ inches tall and grows at a constant rate of $1.5$ inches per day. Plant B is $12$ inches tall and grows at a constant rate of $2.5$ inches per day. After how many days will Plant B be exactly $6$ inches taller than Plant A?

Show answer & explanation

Answer: 14

Answer

The correct answer is 14 days.

The correct answer is

14

days because when we translate the relationship, we get the equation

12 + 2.5d = 20 + 1.5d + 6

. Solving this equation yields

d = 14

.

Step-by-Step Solution

1

Define the variable

d

as the number of days and write the expressions for the heights of both plants.

Plant A's height is

20 + 1.5d

inches, and Plant B's height is

12 + 2.5d

inches.

To represent the growth of each plant algebraically over time.

2

Set up an equation representing that Plant B's height is

6

inches more than Plant A's height.

12 + 2.5d = (20 + 1.5d) + 6

To translate the verbal relationship into a mathematical equation.

3

Simplify the equation and solve for

d

.

d = 14

To find the number of days that satisfies the given condition.

Key Concept

Translating and Solving Algebraic Word Problems

Estimated Time:1m 15s

Question 227Question →

A hyperbola in the standard $(x, y)$ coordinate plane is defined by the equation $9x^2 - 16y^2 - 36x - 32y - 124 = 0$ . One of the foci of this hyperbola is located at the point $(f, -1)$ , where $f > 0$ . What is the value of $f$ ?

Show answer & explanation

Answer: 7

Answer

7

Completing the square transforms the equation into the standard form of a horizontal hyperbola,

\frac{(x - 2)^2}{16} - \frac{(y + 1)^2}{9} = 1

, which has its center at

(2, -1)

with

a^2 = 16

and

b^2 = 9

. The distance to the foci is

c = \sqrt{a^2 + b^2} = \sqrt{16 + 9} = 5

. Since the transverse axis is horizontal, the foci are located at

(2 \pm 5, -1)

, which are

(-3, -1)

and

(7, -1)

. Given the constraint that

f > 0

, the positive x-coordinate of the focus is 7.

Step-by-Step Solution

1

Group the terms and prepare to complete the square.

9(x^2 - 4x) - 16(y^2 + 2y) = 124

Grouping the variables helps isolate the quadratic expressions for completing the square.

2

Complete the square for both the

x

and

y

terms.

9(x - 2)^2 - 16(y + 1)^2 = 144

To complete the square for

x^2 - 4x

, add 4 inside the first parentheses, adding

9 \times 4 = 36

to the right side. To complete the square for

y^2 + 2y

, add 1 inside the second parentheses, which subtracts

16 \times 1 = 16

from the right side because of the leading negative coefficient. This leaves the right side as

124 + 36 - 16 = 144

.

3

Divide both sides of the equation by the constant to find the standard form.

\frac{(x - 2)^2}{16} - \frac{(y + 1)^2}{9} = 1

Dividing by 144 puts the equation in the standard horizontal hyperbola form:

\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1

.

4

Find the distance

c

from the center to the foci.

c = 5

For a hyperbola, the relationship between the semi-axes and the focal distance is

c = \sqrt{a^2 + b^2}

. Substituting

a^2 = 16

and

b^2 = 9

gives

c = \sqrt{16 + 9} = 5

.

5

Determine the coordinates of the foci and extract the value of

f

.

f = 7

The center of the hyperbola is

(h, k) = (2, -1)

. The foci are located at

(h \pm c, k) = (2 \pm 5, -1)

, which corresponds to the points

(-3, -1)

and

(7, -1)

. Since the problem states

f > 0

, the target focus must be

(7, -1)

, meaning

f = 7

.

Key Concept

Converting a general hyperbola equation into standard form to calculate focal points

Question 228Question →

A certain radioactive isotope decays according to the formula $N(t) = N_0 \cdot 2^{-t/8}$ , where $N_0$ is the initial amount of the isotope and $t$ is the time in years. If a sample initially contains $120$ grams of the isotope, how many years will it take for the amount of the isotope to decay to $15$ grams?

Show answer & explanation

Answer: 24

Answer

24

The correct answer is 24 because substituting the initial value of 120 and the final value of 15 into the equation yields

15 = 120 \cdot 2^{-t/8}

. Dividing both sides by 120 gives

\frac{1}{8} = 2^{-t/8}

, which can be rewritten as

2^{-3} = 2^{-t/8}

. Setting the exponents equal to each other gives

-3 = -t/8

, and solving for

t

yields 24.

Step-by-Step Solution

1

Substitute the given values into the decay formula.

15 = 120 \cdot 2^{-t/8}

The initial amount

N_0

is

120

grams and the final amount

N(t)

is

15

grams.

2

Isolate the exponential term.

\frac{1}{8} = 2^{-t/8}

Divide both sides by

120

. Since

\frac{15}{120}

reduces to

\frac{1}{8}

, this isolates the base

2

term.

3

Write the fraction as a power with base 2.

2^{-3} = 2^{-t/8}

Using exponent rules,

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

.

4

Equate the exponents and solve for

t

.

t = 24

Since the bases are equal, the exponents must be equal, so

-3 = -\frac{t}{8}

which gives

t = 24

.

Key Concept

Solving exponential equations using a common base.

Question 229Question →

In the standard $(x, y)$ coordinate plane, the line defined by the equation $3x - 4y = 24$ intersects the $x$ -axis at a certain point. What is the $x$ -coordinate of this intersection point?

Show answer & explanation

Answer: 8

Answer

The

x

-coordinate of the intersection point is

8

.

To find the

x

-intercept, set

y = 0

in the equation

3x - 4y = 24

. This simplifies to

3x = 24

. Dividing both sides of the equation by

3

gives

x = 8

.

Step-by-Step Solution

1

Set

y = 0

in the equation.

3x - 4(0) = 24 \implies 3x = 24

The intersection of any graph with the

x

-axis occurs where the

y

-coordinate is

0

.

2

Solve for

x

.

x = 8

Dividing both sides of

3x = 24

by

3

isolates the variable

x

.

Key Concept

Finding the

x

-intercept of a line by setting

y = 0

Alternative Method

Convert the standard form equation

3x - 4y = 24

into slope-intercept form:

-4y = -3x + 24 \implies y = \frac{3}{4}x - 6

. To find the

x

-intercept, set

y = 0

and solve the equation

0 = \frac{3}{4}x - 6 \implies 6 = \frac{3}{4}x \implies x = 8

.

Estimated Time:45s

Question 230Question →

In $\triangle PQR$ , the measure of $\angle P$ is $50^\circ$ , and the measure of the exterior angle at vertex $Q$ is $110^\circ$ . If the bisector of $\angle PRQ$ intersects side $PQ$ at point $S$ , what is the measure, in degrees, of $\angle PRS$ ?

Show answer & explanation

Answer: 30

Answer

The measure of

\angle PRS

is

30^\circ

.

By using the linear pair relationship, the interior angle

\angle PQR

is found to be

70^\circ

. Applying the triangle angle sum theorem, the third interior angle

\angle PRQ

is

180^\circ - (50^\circ + 70^\circ) = 60^\circ

. The angle bisector

RS

divides this angle into two equal parts, resulting in a measure of

30^\circ

for

\angle PRS

.

Step-by-Step Solution

1

Find the interior angle at vertex

Q

\angle PQR = 70^\circ

The interior and exterior angles at a vertex are supplementary, summing to

180^\circ

.

2

Find the measure of interior angle

\angle PRQ

\angle PRQ = 60^\circ

The sum of the interior angles in any triangle is

180^\circ

.

3

Calculate the measure of the bisected angle

\angle PRS

\angle PRS = 30^\circ

An angle bisector divides the angle into two equal measures.

Key Concept

Triangle Angle Sum Theorem and Exterior Angle Relationships

Estimated Time:1m 30s

Question 231Question →

The daily revenue $R$ , in dollars, of a manufacturing company is modeled by the quadratic function $R(x) = -0.2x^2 + kx - 1,200$ , where $x$ is the number of units produced and sold, and $k$ is a positive constant. If the maximum daily revenue the company can achieve is $800$ dollars, what is the value of $k$ ?

Show answer & explanation

Answer: 40

Answer

The value of the constant

k

is 40.

Setting the daily revenue function equal to 800 and rewriting it in standard form yields

-0.2x^2 + kx - 2000 = 0

. For a quadratic equation to have exactly one real solution, which represents the maximum vertex of the parabola, the discriminant must be equal to 0. Setting the discriminant

b^2 - 4ac = 0

gives

k^2 - 4(-0.2)(-2000) = 0

, which simplifies to

k^2 - 1600 = 0

. Solving for the positive constant

k

gives

k = 40

.

Step-by-Step Solution

1

Set the revenue function equal to the maximum daily revenue of 800 dollars.

-0.2x^2 + kx - 1,200 = 800

The maximum revenue is the highest point (vertex) on the parabola, where the line

y = 800

is tangent to the curve.

2

Convert the equation into standard quadratic form

ax^2 + bx + c = 0

.

-0.2x^2 + kx - 2,000 = 0

Standard form is required to identify the coefficients

a

,

b

, and

c

for the discriminant formula.

3

Set the discriminant equal to zero.

k^2 - 4(-0.2)(-2,000) = 0

Since the maximum daily revenue is achieved at exactly one point, the quadratic equation must have exactly one real solution, meaning its discriminant (

b^2 - 4ac

) must be zero.

4

Solve for the positive constant

k

.

k^2 - 1,600 = 0 \implies k = 40

Solving the equation yields

k = \pm 40

. Since the problem specifies that

k

is a positive constant, we select

k = 40

.

Key Concept

Using the discriminant of a quadratic equation to find the value of a parameter when there is exactly one real solution.

Estimated Time:2m 0s

Question 232Question →

In the standard $(x, y)$ coordinate plane, a line segment $AB$ has midpoint $M(3, 4)$ . If endpoint $A$ lies on the line $y = 2x - 7$ , and the length of segment $AB$ is $10$ units, what is the product of all possible $x$ -coordinates of endpoint $A$ ?

Show answer & explanation

Answer: 21

Answer

The product of all possible

x

-coordinates of endpoint

A

is

21

.

By determining that the distance from endpoint

A(x, 2x - 7)

to the midpoint

M(3, 4)

is half of the segment length

AB

(which is

5

), we set up the distance formula equation:

(x - 3)^2 + (2x - 11)^2 = 25

. Simplifying this equation leads to the quadratic expression

x^2 - 10x + 21 = 0

, which factors into

(x - 3)(x - 7) = 0

. The two possible

x

-coordinates are

3

and

7

. Multiplying these values yields the product

21

.

Step-by-Step Solution

1

Find the distance between endpoint

A

and midpoint

M

.

The distance

AM

is

5

.

Since

M

is the midpoint of segment

AB

of length

10

, the distance from either endpoint to the midpoint is half of the total length:

10 \div 2 = 5

.

2

Express the coordinates of endpoint

A

in terms of a single variable.

Endpoint

A

is represented as

(x, 2x - 7)

.

Endpoint

A

lies on the line

y = 2x - 7

.

3

Apply the distance formula to find the relationship for

x

.

(x - 3)^2 + (2x - 11)^2 = 25

The distance between

A(x, 2x - 7)

and

M(3, 4)

is

5

, so the square of the distance is

5^2 = 25

.

4

Simplify the quadratic equation.

x^2 - 10x + 21 = 0

Expanding

(x - 3)^2 + (2x - 11)^2 = 25

yields

x^2 - 6x + 9 + 4x^2 - 44x + 121 = 25

. Combining like terms gives

5x^2 - 50x + 130 = 25

. Subtracting

25

from both sides gives

5x^2 - 50x + 105 = 0

. Dividing the entire equation by

5

yields

x^2 - 10x + 21 = 0

.

5

Solve for the possible values of

x

.

x = 3

or

x = 7

Factoring the quadratic equation yields

(x - 3)(x - 7) = 0

, which gives the roots

x = 3

and

x = 7

.

6

Calculate the product of the possible

x

-coordinates.

21

The product of the two possible

x

-coordinates is

3 \times 7 = 21

.

Key Concept

Distance and Midpoint Formulas

Alternative Method

Instead of expanding the quadratic equation, one can use the geometric interpretation. The points

A

are the intersections of the circle

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

and the line

y = 2x - 7

. Substituting

y = 2x - 7

directly into the circle equation and simplifying to

x^2 - 10x + 21 = 0

is the most direct approach.

Estimated Time:2m 0s

Question 233Question →

In the standard $(x, y)$ coordinate plane, a line with a negative slope passes through the point $(4, 3)$ and has an $x$ -intercept that is twice its $y$ -intercept. If the equation of this line is written in the form $Ax + By = C$ , where $A$ , $B$ , and $C$ are integers with no common factor greater than 1, and $A > 0$ , what is the value of $C$ ?

Show answer & explanation

Answer: 10

Answer

The value of

C

is 10.

The correct answer is 10. By setting the intercepts as

(2b, 0)

and

(0, b)

, we find the slope is

m = -\frac{1}{2}

. Applying the point-slope formula with

(4, 3)

gives

y - 3 = -\frac{1}{2}(x - 4)

, which simplifies to

y = -\frac{1}{2}x + 5

. Converting this to standard form with

A > 0

yields

x + 2y = 10

, where

C = 10

.

Step-by-Step Solution

1

Represent the coordinates of the intercepts using a variable.

The

y

-intercept is

(0, b)

and the

x

-intercept is

(2b, 0)

.

The problem states that the

x

-intercept is twice the

y

-intercept.

2

Calculate the slope (

m

) of the line using the intercepts.

m = \frac{b - 0}{0 - 2b} = -\frac{1}{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}

.

3

Find the equation of the line using the point-slope form with point

(4, 3)

.

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

The point-slope form of a linear equation is

y - y_1 = m(x - x_1)

.

4

Convert the equation to the standard form

Ax + By = C

.

x + 2y = 10

Multiplying the equation by 2 and moving the

x

term to the left side results in integer coefficients where the coefficient of

x

is positive (

A = 1 > 0

).

5

Identify the value of

C

from the standard form equation.

C = 10

Comparing

x + 2y = 10

with

Ax + By = C

shows that

A=1

,

B=2

, and

C=10

. These integers share no common factors greater than 1.

Key Concept

Converting a linear equation to standard form using given geometric features and a coordinate point.

Alternative Method

Instead of using the point-slope formula, substitute the point

(4, 3)

directly into the intercept form of a linear equation,

\frac{x}{a} + \frac{y}{b} = 1

. Since

a = 2b

, this becomes

\frac{4}{2b} + \frac{3}{b} = 1

. Simplifying this gives

\frac{2}{b} + \frac{3}{b} = 1 \Rightarrow \frac{5}{b} = 1 \Rightarrow b = 5

. Thus,

a = 10

. The intercept form is

\frac{x}{10} + \frac{y}{5} = 1

. Multiplying by 10 to clear denominators gives

x + 2y = 10

, where

C = 10

.

Estimated Time:1m 30s

Question 234Question →

In the standard $(x, y)$ coordinate plane, the perpendicular bisector of the line segment with endpoints $(1, 2)$ and $(5, 10)$ intersects the curve $y = x^2$ at two points. If one of these intersection points lies in the second quadrant, what is the $y$ -coordinate of this point?

Show answer & explanation

Answer: 9

Answer

The correct answer is

9

.

The perpendicular bisector of the segment connecting

(1, 2)

and

(5, 10)

passes through their midpoint

(3, 6)

and has a slope of

-\frac{1}{2}

(the negative reciprocal of

2

). The equation of this line is

x + 2y = 15

. Substituting

y = x^2

yields the quadratic equation

2x^2 + x - 15 = 0

, which factors into

(2x - 5)(x + 3) = 0

. The solution in the second quadrant corresponds to the negative

x

-coordinate,

x = -3

. Squaring this value gives a

y

-coordinate of

9

.

Step-by-Step Solution

1

Find the midpoint of the segment with endpoints

(1, 2)

and

(5, 10)

.

The midpoint is

(3, 6)

.

The perpendicular bisector of a segment passes through its midpoint.

2

Calculate the slope of the segment and the slope of the perpendicular bisector.

The slope of the segment is

2

, and the perpendicular slope is

-\frac{1}{2}

.

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

3

Write the equation of the perpendicular bisector.

The equation of the line is

x + 2y = 15

.

Using the point-slope form with point

(3, 6)

and slope

-\frac{1}{2}

yields

y - 6 = -\frac{1}{2}(x - 3)

, which simplifies to

x + 2y = 15

.

4

Find the intersection points of the line and the curve

y = x^2

.

The

x

-coordinates of the intersection points are

2.5

and

-3

.

Substituting

y = x^2

into

x + 2y = 15

gives the quadratic equation

2x^2 + x - 15 = 0

, which factors as

(2x - 5)(x + 3) = 0

.

5

Identify the point in the second quadrant and find its

y

-coordinate.

The point is

(-3, 9)

, so the

y

-coordinate is

9

.

A point in the second quadrant must have a negative

x

-coordinate (

x = -3

) and a positive

y

-coordinate (

y = 9

).

Key Concept

Perpendicular bisectors and systems of linear-quadratic equations in coordinate geometry

Question 235Question →

A delivery drone starts a flight with a battery charge of $95\%$ . The battery charge decreases at a constant rate of $1.5\%$ per minute of flight time. During the flight, the drone lands on a charging station for $15$ minutes, during which its battery charge increases at a constant rate of $2.4\%$ per minute. After this charging period, the drone resumes its flight. If the drone ends its flight with a battery charge of $65\%$ , and its total flight time (excluding the time spent charging) was $t$ minutes, what is the value of $t$ ?

Show answer & explanation

Answer: 44

Answer

44

The correct answer is found by setting up a linear equation that models the drone's battery charge changes. The drone starts with

95\%

battery, loses

1.5\%

per minute for

t

minutes, and gains

36\%

from charging (

15 \times 2.4\%

). Setting this equal to the final charge of

65\%

gives the equation

95 - 1.5t + 36 = 65

. Simplifying gives

131 - 1.5t = 65

, which leads to

-1.5t = -66

and

t = 44

.

Step-by-Step Solution

1

Calculate the total percentage of battery charge gained while charging.

36\%

The drone charges at a rate of

2.4\%

per minute for

15

minutes, so the total gain is

15 \times 2.4\% = 36\%

.

2

Set up a linear equation for the final battery charge.

95 - 1.5t + 36 = 65

The final battery charge (

65\%

) is the initial charge (

95\%

) minus the battery consumed during flight (

1.5\%

per minute for

t

minutes) plus the charge gained (

36\%

).

3

Combine constant terms on the left side of the equation.

131 - 1.5t = 65

Adding

95

and

36

simplifies the expression on the left side.

4

Isolate the variable term by subtracting

131

from both sides.

-1.5t = -66

Subtracting

131

from both sides leaves only the variable term on the left.

5

Divide both sides by

-1.5

to solve for

t

.

t = 44

Dividing

-66

by

-1.5

isolates

t

to find the total flight time.

Key Concept

Solving linear equations in a real-world context by setting up an algebraic equation.

Estimated Time:1m 30s

Question 236Question →

In $\triangle PQR$ , the lengths of sides $PQ$ and $PR$ are equal. If the measure of $\angle P$ is $70^\circ$ , what is the measure, in degrees, of $\angle Q$ ?

Show answer & explanation

Answer: 55

Answer

The measure of

\angle Q

is

55^\circ

.

Since

PQ = PR

,

\triangle PQR

is an isosceles triangle with base

QR

, which means the base angles

\angle Q

and

\angle R

have equal measures. The sum of the interior angles of a triangle is

180^\circ

. Setting up the equation gives

70^\circ + 2(\angle Q) = 180^\circ

. Subtracting

70^\circ

from both sides yields

2(\angle Q) = 110^\circ

. Dividing by 2, we find that the measure of

\angle Q

is

55^\circ

.

Step-by-Step Solution

1

Identify the properties of the given triangle.

\triangle PQR

is an isosceles triangle with base

QR

and base angles

\angle Q = \angle R

.

A triangle with two equal sides is isosceles, and the angles opposite those sides are equal in measure.

2

Apply the triangle angle sum theorem.

\angle P + \angle Q + \angle R = 180^\circ

The sum of the measures of the interior angles of any triangle is always

180^\circ

.

3

Substitute the known values and solve for

\angle Q

.

70^\circ + 2(\angle Q) = 180^\circ \implies 2(\angle Q) = 110^\circ \implies \angle Q = 55^\circ

Substituting

\angle P = 70^\circ

and

\angle R = \angle Q

allows us to solve the linear equation for the unknown angle measure.

Key Concept

Isosceles Triangle Properties and Triangle Angle Sum Theorem

Estimated Time:45s

Question 237Question →

For the functions $f(x) = |4x - 17|$ and $g(x) = 3 - 2x$ , what is the value of $f(g(5))$ ?

Show answer & explanation

Answer: 45

Answer

The correct answer is 45.

Evaluating the inner function first yields

g(5) = 3 - 2(5) = -7

. Substituting this result into the outer function gives

f(-7) = |4(-7) - 17| = |-45| = 45

.

Step-by-Step Solution

1

Evaluate the inner function

g(5)

g(5) = 3 - 2(5) = -7

In a composite function of the form

f(g(x))

, the inner function

g(x)

must be evaluated first at the given input.

2

Evaluate the outer function

f(x)

at the output of the inner function

f(-7) = |4(-7) - 17| = |-28 - 17| = |-45| = 45

The output of the inner function,

-7

, becomes the input for the outer function

f(x)

.

Key Concept

Function Evaluation and Composition

Question 238Question →

On a coordinate map, two tracking stations are located at the points $(2, -3)$ and $(-6, 12)$ . What is the straight-line distance, in map units, between the two stations?

Show answer & explanation

Answer: 17

Answer

The straight-line distance between the two stations is 17 map units.

Applying the coordinate distance formula to the coordinates

(2, -3)

and

(-6, 12)

yields a distance of

\sqrt{(-6 - 2)^2 + (12 - (-3))^2} = \sqrt{(-8)^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17

.

Step-by-Step Solution

1

Identify coordinates of the two stations.

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

and

(x_2, y_2) = (-6, 12)

Defining the coordinate variables is necessary to apply the formula correctly.

2

Apply the distance formula.

d = \sqrt{(-6 - 2)^2 + (12 - (-3))^2}

The distance formula calculates the straight-line distance between two coordinates.

3

Simplify the arithmetic terms.

d = \sqrt{(-8)^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289}

Simplify the differences and squares under the radical.

4

Calculate the final square root.

d = 17

The square root of 289 is 17.

Key Concept

Distance Formula

Question 239Question →

A hyperbola is defined by the equation $9x^2 - 4y^2 - 36x - 8y - 4 = 0$ in the standard $(x, y)$ coordinate plane. What is the slope of the asymptote of this hyperbola that has a positive slope?

Show answer & explanation

Answer: 1.5

Answer

The positive slope of the asymptotes is 1.5.

By converting the general form of the hyperbola equation into standard form, we determine that it is a horizontal hyperbola with

a = 2

and

b = 3

. The slopes of the asymptotes for a horizontal hyperbola are

\pm \frac{b}{a}

, making the positive slope equal to

\frac{3}{2} = 1.5

.

Step-by-Step Solution

1

Group x-terms and y-terms, and move the constant to the other side.

9(x^2 - 4x) - 4(y^2 + 2y) = 4

This prepares the equation for completing the square.

2

Complete the square for the quadratic expressions in x and y.

9(x - 2)^2 - 4(y + 1)^2 = 36

Completing the square yields

9[(x-2)^2 - 4] - 4[(y+1)^2 - 1] = 4 \implies 9(x-2)^2 - 36 - 4(y+1)^2 + 4 = 4

.

3

Divide both sides by 36 to format the equation in standard hyperbola form.

\frac{(x-2)^2}{4} - \frac{(y+1)^2}{9} = 1

The standard form of a horizontal hyperbola centered at

(h, k)

is

\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1

.

4

Identify the values of a and b from the denominators.

a = 2

and

b = 3

Since

a^2 = 4

and

b^2 = 9

, taking the square roots gives

a = 2

and

b = 3

.

5

Determine the positive slope of the asymptotes using the formula for a horizontal hyperbola.

m = \frac{b}{a} = 1.5

The asymptotes for a horizontal hyperbola are given by

y - k = \pm \frac{b}{a}(x - h)

, so the positive slope is

\frac{b}{a} = \frac{3}{2} = 1.5

.

Key Concept

Rewriting a hyperbola equation from general form to standard form to find asymptote equations.

Estimated Time:1m 30s

Question 240Question →

A line segment in the standard $(x, y)$ coordinate plane has endpoints at $(1, a)$ and $(5, a^2)$ . The perpendicular bisector of this segment is parallel to the line defined by the equation $x + 3y = 6$ . What is the sum of all possible values of the constant $a$ ?

Show answer & explanation

Answer: 1

Answer

The sum of all possible values of the constant

a

is 1.

To find the sum of all possible values of the constant

a

, we first find the slope of the line

x + 3y = 6

by writing it in slope-intercept form:

y = -\frac{1}{3}x + 2

. Since the perpendicular bisector is parallel to this line, its slope is also

-\frac{1}{3}

. The line segment is perpendicular to its perpendicular bisector, so the slope of the line segment is the negative reciprocal of

-\frac{1}{3}

, which is

3

. Setting the slope of the segment

\frac{a^2 - a}{5 - 1}

equal to

3

gives the equation

\frac{a^2 - a}{4} = 3

, which simplifies to

a^2 - a - 12 = 0

. Solving this quadratic equation yields

(a - 4)(a + 3) = 0

, giving the values

a = 4

and

a = -3

. The sum of these possible values is

4 + (-3) = 1

.

Step-by-Step Solution

1

Find the slope of the given line

x + 3y = 6

.

The slope is

-\frac{1}{3}

.

Rewriting the equation in slope-intercept form (

y = mx + b

) gives

y = -\frac{1}{3}x + 2

, showing the slope is

-\frac{1}{3}

.

2

Determine the slope of the line segment.

The slope is 3.

The line segment is perpendicular to its perpendicular bisector. Because the perpendicular bisector is parallel to the reference line, its slope is also

-\frac{1}{3}

. The line segment's slope is the negative reciprocal of

-\frac{1}{3}

, which is

3

.

3

Write the slope of the segment in terms of

a

and set it equal to 3.

\frac{a^2 - a}{4} = 3

Using the slope formula with endpoints

(1, a)

and

(5, a^2)

gives the expression

\frac{a^2 - a}{5 - 1}

.

4

Solve the quadratic equation for

a

.

a = 4

or

a = -3

Multiplying both sides by 4 yields

a^2 - a = 12

, which simplifies to the quadratic

a^2 - a - 12 = 0

. Factoring gives

(a - 4)(a + 3) = 0

.

5

Sum all possible values of

a

.

1

The sum of the values is

4 + (-3) = 1

.

Key Concept

Understanding that parallel lines have equal slopes, perpendicular lines have slopes that are negative reciprocals of each other, and applying the slope formula to solve for coordinate variables.

Answer

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