All practice questions

Question 2121Question →

In the standard $(x,y)$ coordinate plane, the midpoint of a line segment is $(2, 5)$ . If one of the endpoints of the segment is $(-2, 1)$ , what is the $x$ -coordinate of the other endpoint?

Show answer & explanation

Answer: 6

Answer

The

x

-coordinate of the other endpoint is 6.

According to the midpoint formula, the

x

-coordinate of the midpoint is the average of the

x

-coordinates of the two endpoints. Substituting the given values yields the equation

2 = \frac{-2 + x_2}{2}

. Multiplying both sides by 2 gives

4 = -2 + x_2

, and adding 2 to both sides results in

x_2 = 6

.

Step-by-Step Solution

1

Set up the midpoint equation for the

x

-coordinate using the midpoint formula

x_m = \frac{x_1 + x_2}{2}

.

2 = \frac{-2 + x_2}{2}

The

x

-coordinate of the midpoint is the average of the

x

-coordinates of the endpoints.

2

Multiply both sides of the equation by 2 to solve for the numerator.

4 = -2 + x_2

Multiplying by 2 eliminates the denominator on the right side.

3

Add 2 to both sides of the equation to isolate

x_2

.

x_2 = 6

Adding 2 to both sides isolates the variable

x_2

.

Key Concept

Midpoint Formula

Question 2122Question →

An online streaming service offers a basic monthly plan that costs $12$ dollars plus $1.50$ dollars for each premium movie rented. A premium monthly plan costs $20$ dollars and includes $3$ free premium movie rentals, with each additional rental costing $1.00$ dollar. If a subscriber rents $x$ premium movies in a month, where $x > 3$ , which of the following expressions represents the savings, in dollars, of the premium plan compared to the basic plan?

0.50x - 8

5 - 0.50x

0.50x - 5

2.50x - 11

0.50x - 11

Show answer & explanation

Answer: $0.50x - 5$

Answer

0.50x - 5

The cost of the basic plan for

x

rentals is

12 + 1.50x

dollars. Under the premium plan, the first

3

rentals are free, so the subscriber only pays for

x - 3

rentals. The cost of the premium plan is

20 + 1.00(x - 3)

dollars, which simplifies to

17 + x

dollars. The savings of the premium plan compared to the basic plan is the difference between the basic cost and the premium cost:

(12 + 1.50x) - (17 + x) = 0.50x - 5

dollars.

Step-by-Step Solution

1

Write the expression for the cost of the basic monthly plan.

12 + 1.50x

The basic plan has a flat fee of

12

dollars and a rate of

1.50

dollars per movie for all

x

movies rented.

2

Write the expression for the cost of the premium monthly plan.

20 + 1.00(x - 3)

, which simplifies to

17 + x

.

The premium plan has a flat fee of

20

dollars. Since

3

rentals are free, the subscriber only pays

1.00

dollar each for the remaining

x - 3

rentals.

3

Subtract the premium plan cost from the basic plan cost to find the savings.

(12 + 1.50x) - (17 + x) = 0.50x - 5

Savings represents the difference in cost: Basic Cost minus Premium Cost.

Key Concept

Translating verbal descriptions of multi-step costs into linear algebraic expressions and simplifying them.

Estimated Time:1m 30s

Question 2123Question →

In the standard $(x, y)$ coordinate plane, a line $L_1$ with a positive slope $m$ and a negative $y$ -intercept $b$ passes through the point $(4, 3)$ . The region in the fourth quadrant bounded by the line $L_1$ , the $x$ -axis, and the $y$ -axis has an area of exactly $8$ square units. What is the $y$ -coordinate of the $y$ -intercept of line $L_1$ ?

Show answer & explanation

Answer: -6

Answer

The y-coordinate of the y-intercept of line

L_1

is

-6

.

The correct answer is

-6

. Substituting

(4, 3)

into the slope-intercept equation

y = mx + b

gives

3 = 4m + b

, or

m = \frac{3-b}{4}

. The area of the right triangle in the fourth quadrant is

\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} (-\frac{b}{m})(-b) = \frac{b^2}{2m} = 8

. Substituting

m

yields

b^2 = 16\left(\frac{3-b}{4}\right)

, which simplifies to the quadratic equation

b^2 + 4b - 12 = 0

. Factoring gives

(b+6)(b-2) = 0

. Since the y-intercept

b

must be negative, we have

b = -6

.

Step-by-Step Solution

1

Substitute the given point into the slope-intercept equation

m = \frac{3 - b}{4}

Since the line passes through

(4, 3)

, substituting these coordinates into

y = mx + b

allows us to express the slope

m

in terms of the y-intercept

b

.

2

Determine the intercepts and the dimensions of the bounded region

Base

= -\frac{b}{m}

and Height

= -b

The boundary of the region in the fourth quadrant is a right triangle formed by the origin, the x-intercept

(-\frac{b}{m}, 0)

, and the y-intercept

(0, b)

.

3

Set up the area of the triangle and equate it to 8

b^2 = 16m

The area of a right triangle is

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

, so

\frac{1}{2} \left(-\frac{b}{m}\right)(-b) = 8

simplifies to

b^2 = 16m

.

4

Substitute

m

into the area equation and solve the resulting quadratic equation

b = -6

(discarding

b = 2

)

Substituting

m = \frac{3 - b}{4}

yields

b^2 + 4b - 12 = 0

, which factors into

(b + 6)(b - 2) = 0

. Since the region is in the fourth quadrant, the y-intercept must be negative (

b < 0

).

Key Concept

Using linear equation intercepts to calculate bounded areas on the coordinate plane and relating variables using point substitution.

Question 2124Question →

In the standard $(x, y)$ coordinate plane, a triangle has vertices $A(1, 2)$ , $B(7, 2)$ , and $C$ . The midpoint of side $AC$ lies on the line $y = 3x - 3$ , and the midpoint of side $BC$ lies on the line $y = -2x + 13$ . What is the distance between point $C$ and the midpoint of side $AB$ ?

3

4

\sqrt{5}

2\sqrt{2}

2\sqrt{5}

Show answer & explanation

Answer: $\sqrt{5}$

Answer

The distance between point

C

and the midpoint of side

AB

is

\sqrt{5}

.

The coordinates of point

C(3, 4)

are determined by setting up the midpoint coordinates for sides

AC

and

BC

and substituting them into their respective line equations. The midpoint of

AB

is calculated to be

(4, 2)

. Using the distance formula between

C(3, 4)

and

(4, 2)

yields

\sqrt{(4-3)^2 + (2-4)^2} = \sqrt{5}

.

Step-by-Step Solution

1

Express the midpoint of side

AC

in terms of the unknown coordinates of point

C(x, y)

and substitute it into the given line equation.

The midpoint of

AC

is

M_{AC} = \left(\frac{x+1}{2}, \frac{y+2}{2}\right)

. Substituting this into

y = 3x - 3

yields:

\frac{y+2}{2} = 3\left(\frac{x+1}{2}\right) - 3

, which simplifies to

y = 3x - 5

.

Since the midpoint of

AC

lies on the line

y = 3x - 3

, its coordinates must satisfy the equation of the line.

2

Express the midpoint of side

BC

in terms of the unknown coordinates of point

C(x, y)

and substitute it into the second given line equation.

The midpoint of

BC

is

M_{BC} = \left(\frac{x+7}{2}, \frac{y+2}{2}\right)

. Substituting this into

y = -2x + 13

yields:

\frac{y+2}{2} = -2\left(\frac{x+7}{2}\right) + 13

, which simplifies to

y = -2x + 10

.

Since the midpoint of

BC

lies on the line

y = -2x + 13

, its coordinates must satisfy this equation.

3

Solve the system of two linear equations to find the coordinates of point

C(x, y)

.

Equating the two expressions for

y

:

3x - 5 = -2x + 10 \Rightarrow 5x = 15 \Rightarrow x = 3

. Substituting

x = 3

back into the first equation:

y = 3(3) - 5 = 4

. Thus,

C = (3, 4)

.

Point

C

must simultaneously satisfy the midpoint constraints on both sides

AC

and

BC

.

4

Find the coordinates of the midpoint of side

AB

.

The midpoint of segment

AB

with endpoints

A(1, 2)

and

B(7, 2)

is

M_{AB} = \left(\frac{1+7}{2}, \frac{2+2}{2}\right) = (4, 2)

.

The question asks for the distance between point

C

and the midpoint of

AB

, so we need to determine the coordinates of this midpoint first.

5

Calculate the distance between point

C(3, 4)

and the midpoint

M_{AB}(4, 2)

using the distance formula.

The distance

d

is:

d = \sqrt{(4-3)^2 + (2-4)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}

.

The distance formula

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

is used to find the straight-line distance between two points in a coordinate plane.

Key Concept

Applying the midpoint and distance formulas within coordinate geometry constraint systems.

Alternative Method

Instead of algebraically solving for the lines of midpoints, one can translate the lines using vectors. The set of possible points

C

when the midpoint of

AC

lies on line

L_1

is a line

L'_1

obtained by dilating

L_1

by a factor of 2 with respect to center

A

. Dilating

y = 3x - 3

from

A(1, 2)

gives the line

y = 3x - 5

. Similarly, dilating

y = -2x + 13

from

B(7, 2)

by a factor of 2 gives

y = -2x + 10

. The intersection of these two dilated lines is point

C(3, 4)

.

Estimated Time:2m 30s

Question 2125Question →

In the standard $(x,y)$ coordinate plane, a line with a negative slope passes through the point $(3, 2)$ and has positive integer intercepts $(a, 0)$ and $(0, b)$ . What is the sum of all possible values of $a$ ?

10

15

24

20

27

Show answer & explanation

Answer: 24

Answer

The sum of all possible values of

a

is 24.

The correct answer is the sum 24. Writing the line's equation in intercept form gives

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

. Substituting the point

(3, 2)

yields

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

. Solving for

b

results in

b = 2 + \frac{6}{a-3}

. Because both

a

and

b

must be positive integers, the expression

a-3

must be a positive divisor of

6

. The positive divisors of

6

are

1, 2, 3,

and

6

, which correspond to the

a

values of

4, 5, 6,

and

9

. Summing these values gives

4 + 5 + 6 + 9 = 24

.

Step-by-Step Solution

1

Set up the intercept form of the linear equation.

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

A line with

x

-intercept

(a,0)

and

y

-intercept

(0,b)

can be written in intercept form.

2

Substitute the given point

(3, 2)

into the equation.

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

Since the line passes through

(3,2)

, these coordinates must satisfy the equation.

3

Solve the equation for

b

in terms of

a

.

\frac{2}{b} = 1 - \frac{3}{a} \implies \frac{2}{b} = \frac{a-3}{a} \implies b = \frac{2a}{a-3}

Expressing

b

in terms of

a

helps analyze the integer constraints.

4

Rewrite the expression for

b

to isolate the fractional part.

b = \frac{2(a-3) + 6}{a-3} = 2 + \frac{6}{a-3}

This form allows us to see when

b

will be an integer based on the divisors of the numerator.

5

Determine the positive integer solutions for

a

and

b

.

a-3

must be a positive divisor of

6

. The positive divisors of

6

are

1, 2, 3,

and

6

. This yields:
- If

a-3=1 \implies a=4, b=8

- If

a-3=2 \implies a=5, b=5

- If

a-3=3 \implies a=6, b=4

- If

a-3=6 \implies a=9, b=3

Since

a

and

b

must be positive integers,

a-3

must be positive and divide 6 evenly. Negative divisors (like

-1

or

-2

) would make

b

negative or zero.

6

Sum the possible values of

a

.

4 + 5 + 6 + 9 = 24

To find the final answer, we sum all the valid

x

-intercept values.

Key Concept

Using the intercept form of a linear equation and applying integer constraints to find coordinates.

Question 2126Question →

If $x$ and $y$ are positive real numbers, what is the simplified form of the expression below?

\frac{4 x^{1/2} y^3 \cdot x^{1/3} y^{-1}}{2 x^{-1} y}

2x^{7/5}y

\frac{2y}{x^{1/6}}

2x^{11/6}y

2x^{7/6}y

8x^{11/6}y

Show answer & explanation

Answer: $2x^{11/6}y$

Answer

The simplified form of the expression is

2x^{11/6}y

.

The correct answer is

2x^{11/6}y

. First, simplify the constant coefficients to obtain

\frac{4}{2} = 2

. In the numerator, combine the bases by adding exponents:

x^{1/2} \cdot x^{1/3} = x^{1/2 + 1/3} = x^{5/6}

, and

y^3 \cdot y^{-1} = y^{3 - 1} = y^2

. Next, divide by the terms in the denominator using the quotient rule:

\frac{x^{5/6}}{x^{-1}} = x^{5/6 - (-1)} = x^{11/6}

, and

\frac{y^2}{y^1} = y^{2 - 1} = y

. This yields the fully simplified expression

2x^{11/6}y

.

Step-by-Step Solution

1

Simplify the coefficients of the fraction.

\frac{4}{2} = 2

Dividing the numerical constants in the numerator and denominator simplifies the constant multiplier of the expression.

2

Combine the

x

terms in the numerator using the product rule.

x^{1/2} \cdot x^{1/3} = x^{1/2 + 1/3} = x^{5/6}

When multiplying terms with the same base, add their exponents:

\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}

.

3

Combine the

y

terms in the numerator using the product rule.

y^3 \cdot y^{-1} = y^{3 + (-1)} = y^2

When multiplying terms with the same base, add their exponents:

3 - 1 = 2

.

4

Simplify the

x

terms in the fraction using the quotient rule.

\frac{x^{5/6}}{x^{-1}} = x^{5/6 - (-1)} = x^{5/6 + 1} = x^{11/6}

When dividing terms with the same base, subtract the denominator's exponent from the numerator's exponent.

5

Simplify the

y

terms in the fraction using the quotient rule.

\frac{y^2}{y^1} = y^{2 - 1} = y

When dividing terms with the same base, subtract the denominator's exponent from the numerator's exponent.

6

Combine the simplified components.

2x^{11/6}y

Multiply the simplified coefficient,

x

term, and

y

term together to get the final simplified expression.

Key Concept

Properties of Exponents in Algebraic Expressions

Estimated Time:1m 30s

Question 2127Question →

A parabola in the standard $(x, y)$ coordinate plane is defined by the equation $x^2 - 6x - 8y + 25 = 0$ . What is the distance, in coordinate units, between the focus and the directrix of this parabola?

Show answer & explanation

Answer: 4

Answer

The distance between the focus and the directrix of the parabola is 4.

By completing the square on the equation

x^2 - 6x - 8y + 25 = 0

, we get

(x-3)^2 = 8(y-2)

. Since the coefficient of the linear factor is

8

, we set

4p = 8

, which yields

p = 2

. The distance from the focus to the directrix is

2p = 2(2) = 4

.

Step-by-Step Solution

1

Isolate the terms containing

x

on one side of the equation.

x^2 - 6x = 8y - 25

To set up the equation for completing the square on the

x

terms.

2

Complete the square for the quadratic expression in

x

by adding

9

to both sides.

x^2 - 6x + 9 = 8y - 16 \implies (x-3)^2 = 8y - 16

Adding

( -6/2 )^2 = 9

creates a perfect square trinomial on the left side.

3

Factor out the coefficient of

y

on the right side to write the equation in standard form.

(x-3)^2 = 8(y-2)

This matches the standard form equation

(x-h)^2 = 4p(y-k)

for a vertical parabola.

4

Determine the value of the focal parameter

p

from the standard form.

4p = 8 \implies p = 2

Comparing the standard form coefficient

4p

with the value

8

gives

p = 2

.

5

Calculate the total distance between the focus and the directrix.

2p = 2(2) = 4

The vertex is situated halfway between the focus and the directrix, making the distance between them

2p

.

Key Concept

Finding the geometric properties of a parabola by completing the square to convert its general equation to standard form.

Question 2128Question →

In the standard $(x,y)$ coordinate plane, a line is defined by the equation $3x - 2y = 12$ . What is the sum of the $x$ -intercept and the $y$ -intercept of this line?

-2

10

2

16

1

Show answer & explanation

Answer: -2

Answer

The sum of the

x

-intercept and the

y

-intercept of the line is

-2

.

To find the

x

-intercept, set

y = 0

in the equation

3x - 2y = 12

, which gives

3x = 12

, so

x = 4

. To find the

y

-intercept, set

x = 0

, which gives

-2y = 12

, so

y = -6

. Adding these two values together gives

4 + (-6) = -2

.

Step-by-Step Solution

1

Find the

x

-intercept by setting

y = 0

in the equation.

Substitute

y = 0

into

3x - 2y = 12

to get

3x - 2(0) = 12

, which simplifies to

3x = 12

, yielding

x = 4

.

The

x

-intercept of a line is the point where the line crosses the

x

-axis, which occurs when the

y

-coordinate is

0

.

2

Find the

y

-intercept by setting

x = 0

in the equation.

Substitute

x = 0

into

3x - 2y = 12

to get

3(0) - 2y = 12

, which simplifies to

-2y = 12

, yielding

y = -6

.

The

y

-intercept of a line is the point where the line crosses the

y

-axis, which occurs when the

x

-coordinate is

0

.

3

Add the

x

-intercept and

y

-intercept together.

Calculate

4 + (-6) = -2

.

The question asks for the sum of the

x

-intercept and the

y

-intercept.

Key Concept

Finding the

x

- and

y

-intercepts of a linear equation in standard form.

Estimated Time:1m 0s

Question 2129Question →

If $8^{2x - 1} = \left(\frac{1}{4}\right)^{x - 3}$ , what is the value of $x$ ?

-\frac{3}{8}

\frac{1}{2}

\frac{9}{8}

1

-\frac{3}{4}

Show answer & explanation

Answer: $\frac{9}{8}$

Answer

The correct answer is

\frac{9}{8}

.

By writing both sides of the equation with a common base of

2

, we get

(2^3)^{2x-1} = (2^{-2})^{x-3}

. Applying the power-to-a-power exponent rule, this simplifies to

2^{6x-3} = 2^{-2x+6}

. Since the bases are now identical, their exponents must be equal:

6x-3 = -2x+6

. Adding

2x

and

3

to both sides results in

8x = 9

, which gives

x = \frac{9}{8}

.

Step-by-Step Solution

1

Express both bases as powers of

2

.

8 = 2^3

and

\frac{1}{4} = 2^{-2}

, so the equation becomes

(2^3)^{2x-1} = (2^{-2})^{x-3}

.

Finding a common base allows us to equate the exponents directly.

2

Apply the power-to-a-power rule

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

to simplify the exponents.

2^{3(2x-1)} = 2^{-2(x-3)} \Rightarrow 2^{6x-3} = 2^{-2x+6}

.

This simplifies the exponential expressions on both sides of the equation.

3

Equate the exponents and solve for

x

.

6x - 3 = -2x + 6 \Rightarrow 8x = 9 \Rightarrow x = \frac{9}{8}

.

Since the bases are equal, their exponents must be equal.

Key Concept

Solving exponential equations by finding a common base and applying exponent properties.

Estimated Time:1m 30s

Question 2130Question →

An ellipse in the standard $(x, y)$ coordinate plane is defined by the equation $x^2 + 4y^2 + 6x - 8y + 9 = 0$ . What is the length of the major axis of this ellipse?

1

2

4

8

16

Show answer & explanation

Answer: 4

Answer

4

To find the length of the major axis, rewrite the general equation of the ellipse in standard form by completing the square. Grouping the terms yields

(x^2 + 6x) + 4(y^2 - 2y) = -9

. Completing the square for both variables gives

(x+3)^2 - 9 + 4[(y-1)^2 - 1] = -9

, which simplifies to

(x+3)^2 + 4(y-1)^2 = 4

. Dividing both sides by 4 gives the standard form

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

. In this form, the horizontal axis is the major axis because the denominator under the

x

-term (

a^2 = 4

) is larger than the denominator under the

y

-term (

b^2 = 1

). Since

a^2 = 4

, the semi-major axis is

a = 2

. Therefore, the total length of the major axis is

2a = 2(2) = 4

.

Step-by-Step Solution

1

Group the

x

-terms and

y

-terms and move the constant to the right side of the equation.

(x^2 + 6x) + (4y^2 - 8y) = -9

Grouping like terms allows completing the square for each variable independently.

2

Factor out the coefficient of

y^2

from the

y

-terms.

(x^2 + 6x) + 4(y^2 - 2y) = -9

Before completing the square, the leading coefficient of the squared terms inside the parentheses must be 1.

3

Complete the square for both the

x

and

y

expressions by adding and subtracting the square of half of the linear coefficients.

((x+3)^2 - 9) + 4((y-1)^2 - 1) = -9

This rewrites the quadratic expressions into perfect square trinomial form.

4

Distribute the coefficients and simplify the constant terms.

(x+3)^2 - 9 + 4(y-1)^2 - 4 = -9 \implies (x+3)^2 + 4(y-1)^2 - 13 = -9 \implies (x+3)^2 + 4(y-1)^2 = 4

Isolating the squared terms on one side helps convert the equation to the standard form of an ellipse.

5

Divide both sides of the equation by 4 to set the right side equal to 1.

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

The standard form of a horizontal ellipse is

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

.

6

Identify the values of

a^2

and

b^2

and calculate the length of the major axis.

a^2 = 4 \implies a = 2

. The major axis length is

2a = 2(2) = 4

.

The length of the major axis is twice the length of the semi-major axis (

a

).

Key Concept

Rewriting the general equation of an ellipse into standard form by completing the square to find its key features, such as the length of the major axis.

Alternative Method

Another way to find the length of the major axis is to find the vertices of the ellipse by finding the maximum and minimum x-values where the equation has real solutions for y, though completing the square is the standard and most direct method.

Estimated Time:1m 30s

Question 2131Question →

In the standard $(x, y)$ coordinate plane, a parabola is defined by the equation $y = x^2 + 2x + 7$ and a line is defined by the equation $y = mx + 3$ , where $m$ is a constant. If the system of equations consisting of this parabola and line has exactly one real solution, and this solution lies in the first quadrant, what is the value of $m$ ?

-2

-6

6

2\sqrt{3}

2 + 2\sqrt{7}

Show answer & explanation

Answer: 6

Answer

6

The correct answer is

6

. When setting the equations of the parabola and line equal, we get

x^2 + (2-m)x + 4 = 0

. Setting the discriminant to zero yields

(2-m)^2 - 16 = 0

, which gives

m = 6

or

m = -2

. Substituting

m = 6

back gives a single intersection point of

(2, 15)

, which is in the first quadrant since both coordinates are positive. The other value,

m = -2

, gives an intersection point of

(-2, 7)

, which is in the second quadrant.

Step-by-Step Solution

1

Set the equations of the parabola and the line equal to each other to find their intersection points.

x^2 + 2x + 7 = mx + 3

To find the coordinates where the two graphs intersect.

2

Rearrange the equation into standard quadratic form:

ax^2 + bx + c = 0

.

x^2 + (2 - m)x + 4 = 0

To express the intersection condition as a single quadratic equation where we can analyze the number of solutions.

3

Set the discriminant of the quadratic equation to zero.

\Delta = (2 - m)^2 - 4(1)(4) = 0 \implies (2 - m)^2 - 16 = 0

For the system to have exactly one real solution, the quadratic equation must have a discriminant of zero (tangency).

4

Solve for the possible values of

m

.

(2 - m)^2 = 16 \implies 2 - m = \pm 4

, giving

m = -2

or

m = 6

.

To find all values of

m

that result in exactly one intersection point.

5

Find the intersection point for each value of

m

and determine which lies in the first quadrant.

For

m = -2

,

(x+2)^2 = 0 \implies x = -2

and

y = -2(-2)+3 = 7

, yielding

(-2, 7)

(Quadrant II). For

m = 6

,

(x-2)^2 = 0 \implies x = 2

and

y = 6(2)+3 = 15

, yielding

(2, 15)

(Quadrant I).

To satisfy the condition that the single solution must lie in the first quadrant (where both

x > 0

and

y > 0

).

Key Concept

Solving systems of linear and quadratic equations and applying the discriminant to find conditions for tangency.

Estimated Time:2m 0s

Question 2132Question →

A thermometer is considered accurate if its temperature reading, $T$ degrees Fahrenheit, differs from the actual temperature by less than $1.5^\circ\text{F}$ . If the actual temperature is $72.0^\circ\text{F}$ , which of the following inequalities represents the range of reading temperatures, $T$ , that are NOT considered accurate?

|T - 1.5| \ge 72.0

T - 72.0 \ge 1.5

|T - 72.0| \ge 1.5

|T - 72.0| < 1.5

|T - 72.0| > 1.5

Show answer & explanation

Answer: $|T - 72.0| \ge 1.5$

Answer

|T - 72.0| \ge 1.5

The difference between the temperature reading,

T

, and the actual temperature of

72.0^\circ\text{F}

is represented by

|T - 72.0|

. A thermometer is accurate when this difference is less than

1.5^\circ\text{F}

(

|T - 72.0| < 1.5

). Therefore, the thermometer is not accurate when the difference is greater than or equal to

1.5^\circ\text{F}

, which is written as

|T - 72.0| \ge 1.5

.

Step-by-Step Solution

1

Represent the difference between the reading and the actual temperature.

|T - 72.0|

The difference between the reading temperature and the actual temperature of

72.0^\circ\text{F}

is given by the absolute value expression, representing the distance between the two temperatures on a thermometer.

2

Formulate the condition for an accurate reading.

|T - 72.0| < 1.5

An accurate reading differs from the actual temperature by less than

1.5^\circ\text{F}

.

3

Determine the complement condition for a reading that is NOT accurate.

|T - 72.0| \ge 1.5

To find the range of readings that are NOT accurate, we take the complement of the accurate condition. The opposite of 'less than

1.5

' is 'greater than or equal to

1.5

'.

Key Concept

Absolute Value Inequalities in Real-World Contexts

Estimated Time:1m 0s

Question 2133Question →

In the standard $(x, y)$ coordinate plane, a line $L$ passes through the point $(2, -3)$ and has a $y$ -intercept of $(0, b)$ , where $b > 0$ . If the area of the triangular region bounded by the line $L$ , the $x$ -axis, and the $y$ -axis is $4$ square units, what is the value of $b$ ?

2

4

6

8

12

Show answer & explanation

Answer: 6

Answer

6

The correct answer is the option containing 6. Using the two points

(2, -3)

and

(0, b)

, we find the slope of the line is

m = -\frac{b+3}{2}

, which gives the equation

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

. Setting

y=0

shows that the

x

-intercept is at

x = \frac{2b}{b+3}

. The area of the right triangle formed by the intercepts and the origin is

\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{2b}{b+3} \times b = \frac{b^2}{b+3}

. Setting this area equal to

4

yields

b^2 - 4b - 12 = 0

. Factoring gives

(b - 6)(b + 2) = 0

. Since we are given

b > 0

,

b

must be

6

.

Step-by-Step Solution

1

Find the slope of the line

L

in terms of

b

.

The slope

m

is given by

\frac{-3 - b}{2 - 0} = -\frac{b+3}{2}

.

The line passes through

(2, -3)

and its

y

-intercept

(0, b)

.

2

Write the equation of the line

L

and determine its

x

-intercept.

The equation is

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

. Setting

y = 0

gives the

x

-intercept

x = \frac{2b}{b+3}

.

The

x

-intercept is the point where the line crosses the

x

-axis (

y = 0

).

3

Set up the area equation for the triangle formed by the axes and the line.

The area is

\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{2b}{b+3} \times b = \frac{b^2}{b+3} = 4

.

The base of the triangle is the

x

-intercept, and the height is the

y

-intercept

b

(since both are positive for

b > 0

).

4

Solve the quadratic equation for

b

.

b^2 = 4(b + 3) \Rightarrow b^2 - 4b - 12 = 0 \Rightarrow (b - 6)(b + 2) = 0

. Since

b > 0

,

b = 6

.

We must solve the equation and select the positive solution because the problem specifies

b > 0

.

Key Concept

Using the coordinates of a point and intercepts to write a linear equation, finding intercepts, and calculating the area of a coordinate triangle.

Alternative Method

Instead of setting up the area algebraically first, you can test the answer choices. For example, testing the correct value 6: the y-intercept is

(0, 6)

. The slope of the line passing through

(0, 6)

and

(2, -3)

is

m = \frac{-3 - 6}{2 - 0} = -4.5

. The equation of the line is

y = -4.5x + 6

. The x-intercept is found by setting

y = 0

, giving

x = \frac{6}{4.5} = \frac{4}{3}

. The area of the triangle is

\frac{1}{2} \times \frac{4}{3} \times 6 = 4

, which matches the given area of 4 square units.

Estimated Time:1m 30s

Question 2134Question →

The equation of a parabola is given by $(x - 4)^2 = 12(y + 1)$ . What is the $y$ -coordinate of the focus of this parabola?

Show answer & explanation

Answer: 2

Answer

The correct answer is 2.

The equation

(x - 4)^2 = 12(y + 1)

is a parabola with a vertical axis of symmetry, vertex at

(4, -1)

, and focal length

p = 3

. The focus is located

p

units above the vertex, yielding a

y

-coordinate of

-1 + 3 = 2

.

Step-by-Step Solution

1

Identify the standard form of the parabola's equation.

The equation

(x - 4)^2 = 12(y + 1)

matches

(x - h)^2 = 4p(y - k)

.

This form allows us to find the vertex and the focal distance

p

directly.

2

Determine the vertex and focal distance

p

.

The vertex is

(4, -1)

and

p = 3

since

4p = 12

.

Matching the given equation terms to the standard form reveals these properties.

3

Find the coordinates of the focus.

The focus is at

(4, 2)

.

The focus is located

p

units vertically above the vertex for a parabola opening upward.

Key Concept

Focus of a Parabola

Question 2135Question →

The measures of the three interior angles of a triangle are in the ratio $2:3:5$ . What is the measure of the largest angle in the triangle?

36^\circ

54^\circ

90^\circ

100^\circ

180^\circ

Show answer & explanation

Answer: $90^\circ$

Answer

The correct answer is

90^\circ

.

Since the interior angles of a triangle sum to

180^\circ

and their ratio is

2:3:5

, the sum of the ratio parts is

2 + 3 + 5 = 10

. Dividing the total

180^\circ

by 10 parts yields

18^\circ

per ratio unit. The largest angle corresponds to the largest part of the ratio, which is 5. Therefore, the measure of the largest angle is

5 \times 18^\circ = 90^\circ

.

Step-by-Step Solution

1

Find the total number of parts in the ratio by adding the terms together.

The total number of parts is

2 + 3 + 5 = 10

parts.

This determines how many equal units the total angle measure is divided into.

2

Divide the total sum of the interior angles of a triangle by the total number of parts to find the degree measure of one part.

The sum of the interior angles of a triangle is

180^\circ

. Dividing by the total parts gives

\frac{180^\circ}{10} = 18^\circ

per part.

The Triangle Angle Sum Theorem states that the interior angles of a triangle always sum to

180^\circ

.

3

Multiply the value of one part by the ratio term representing the largest angle.

The largest angle is represented by the term 5, so its measure is

5 \times 18^\circ = 90^\circ

.

This gives the measure of the largest angle in the triangle.

Key Concept

The interior angles of a triangle always sum to

180^\circ

, and individual angle measures can be found from a given ratio by dividing the total degrees by the sum of the ratio parts.

Question 2136Question →

What is the product of all real values of $x$ that satisfy the equation $\log_3(x) - 6\log_x(3) = 1$ ?

Show answer & explanation

Answer: 3

Answer

The product of all real values of

x

that satisfy the equation is 3.

By applying the change-of-base formula, the equation becomes

\log_3(x) - \frac{6}{\log_3(x)} = 1

. Substituting

y = \log_3(x)

leads to

y^2 - y - 6 = 0

, which has roots

y = 3

and

y = -2

. These roots correspond to

x = 3^3 = 27

and

x = 3^{-2} = \frac{1}{9}

. Both solutions are valid because they are positive and do not equal 1. The product of these solutions is

27 \times \frac{1}{9} = 3

. Alternatively, using Vieta's formulas, the sum of the roots of the quadratic equation is

y_1 + y_2 = 1

. The product of the solutions is

x_1 x_2 = 3^{y_1} \cdot 3^{y_2} = 3^{y_1 + y_2} = 3^1 = 3

.

Step-by-Step Solution

1

Apply the change-of-base formula to rewrite the variable base term.

\log_3(x) - \frac{6}{\log_3(x)} = 1

This expresses the equation in terms of logarithms with the same base.

2

Use substitution to convert the equation into a quadratic form.

y - \frac{6}{y} = 1 \Rightarrow y^2 - y - 6 = 0

where

y = \log_3(x)

Substitution simplifies the logarithmic equation into a polynomial equation.

3

Solve the quadratic equation by factoring.

(y-3)(y+2) = 0 \Rightarrow y = 3

or

y = -2

Finding the roots for

y

is the intermediate step to solving for

x

.

4

Back-substitute to find the values of

x

.

x = 3^3 = 27

and

x = 3^{-2} = \frac{1}{9}

Converting from logarithmic form back to exponential form yields the values of

x

.

5

Multiply the solutions together.

27 \times \frac{1}{9} = 3

The question asks for the product of all real solutions.

Key Concept

Solving logarithmic equations using the change-of-base formula and quadratic substitution.

Alternative Method

Instead of solving for individual values of

x

, note that if

y_1

and

y_2

are the roots of the quadratic equation

y^2 - y - 6 = 0

, then

y_1 + y_2 = 1

by Vieta's formulas. Since

x_1 = 3^{y_1}

and

x_2 = 3^{y_2}

, the product of the solutions is

x_1 x_2 = 3^{y_1} \cdot 3^{y_2} = 3^{y_1 + y_2} = 3^1 = 3

.

Estimated Time:2m 0s

Question 2137Question →

In $\triangle ABC$ , the measure of exterior angle $\angle BCD$ is $115^\circ$ . If the measure of interior angle $\angle A$ is $45^\circ$ , what is the measure, in degrees, of interior angle $\angle B$ ?

Show answer & explanation

Answer: 70

Answer

The measure of interior angle

\angle B

is

70

degrees.

By the Exterior Angle Theorem, the measure of exterior angle

\angle BCD

is equal to the sum of the two remote interior angles,

\angle A

and

\angle B

. We can write this relationship as

m\angle BCD = m\angle A + m\angle B

. Substituting

115^\circ

for

m\angle BCD

and

45^\circ

for

m\angle A

gives

115 = 45 + m\angle B

. Solving for

m\angle B

yields

70^\circ

.

Step-by-Step Solution

1

Set up the equation using the Exterior Angle Theorem.

m\angle BCD = m\angle A + m\angle B

The measure of an exterior angle of a triangle is equal to the sum of the measures of its two remote interior angles.

2

Substitute the given measurements into the equation.

115 = 45 + m\angle B

The exterior angle

\angle BCD

measures

115^\circ

and the remote interior angle

\angle A

measures

45^\circ

.

3

Solve for the unknown angle measure by subtraction.

m\angle B = 70

Subtracting

45

from both sides isolates

m\angle B

.

Key Concept

Exterior Angle Theorem

Question 2138Question →

In the standard $(x, y)$ coordinate plane, a line passes through the point $(3, 2)$ and has a $y$ -intercept of $-4$ . If the point $(k, 5k + 2)$ also lies on this line, what is the value of $k$ ?

Show answer & explanation

Answer: -2

Answer

The value of

k

is

-2

.

The line has a slope of

2

and a y-intercept of

-4

, giving the equation

y = 2x - 4

. Substituting the coordinates of

(k, 5k + 2)

results in

5k + 2 = 2k - 4

, which simplifies to

3k = -6

, yielding

k = -2

.

Step-by-Step Solution

1

Identify the coordinates of the y-intercept.

The y-intercept of

-4

corresponds to the point

(0, -4)

.

The y-intercept is the point where the line crosses the y-axis, meaning the x-coordinate is 0.

2

Calculate the slope of the line.

The slope

m

is

2

.

Using the slope formula

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

with the points

(3, 2)

and

(0, -4)

gives

m = \frac{2 - (-4)}{3 - 0} = \frac{6}{3} = 2

.

3

Write the equation of the line.

The equation of the line is

y = 2x - 4

.

Using the slope-intercept form

y = mx + b

, where the slope

m = 2

and the y-intercept

b = -4

.

4

Substitute the point

(k, 5k + 2)

into the line's equation.

The equation becomes

5k + 2 = 2k - 4

.

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

5

Solve the linear equation for

k

.

k = -2

.

Subtracting

2k

from both sides gives

3k + 2 = -4

. Subtracting

2

from both sides gives

3k = -6

. Dividing by

3

gives

k = -2

.

Key Concept

Finding the equation of a line from a point and an intercept, and solving for parameters of points on that line.

Question 2139Question →

Two cyclists start at opposite ends of a $90$ -mile trail at the same time and ride toward each other. One cyclist rides at a constant speed that is $3$ miles per hour faster than the other cyclist. If the two cyclists meet after exactly $3$ hours, what is the constant speed, in miles per hour, of the faster cyclist?

Show answer & explanation

Answer: 16.5

Answer

The speed of the faster cyclist is

16.5

miles per hour.

The correct answer of

16.5

is found by setting the speed of the slower cyclist to

s

and the faster cyclist to

s + 3

. Since both cyclists ride toward each other for

3

hours, their combined distance is

3s + 3(s + 3) = 90

. Solving for

s

yields

6s + 9 = 90

, which simplifies to

6s = 81

, or

s = 13.5

. Adding

3

to

13.5

gives the speed of the faster cyclist, which is

16.5

miles per hour.

Step-by-Step Solution

1

Define variables for the speeds of both cyclists in terms of a single variable.

Let

s

be the speed of the slower cyclist. The speed of the faster cyclist is

s + 3

.

Using a single variable simplifies the setup of a solvable linear equation.

2

Write a linear equation using the relationship

\text{Distance} = \text{Speed} \times \text{Time}

.

The equation is

3s + 3(s + 3) = 90

.

The sum of the distances traveled by both cyclists when they meet must equal the total length of the trail, which is

90

miles.

3

Solve the equation for

s

.

6s + 9 = 90 \implies 6s = 81 \implies s = 13.5

.

This yields the speed of the slower cyclist.

4

Calculate the speed of the faster cyclist.

13.5 + 3 = 16.5

.

The question specifically asks for the speed of the faster cyclist, which is represented by

s + 3

.

Key Concept

Translating distance-rate-time relationships from word problems into solvable linear equations.

Alternative Method

An alternative approach is to use the concept of relative speed. Since the two cyclists are moving directly toward each other, their relative speed of approach is the sum of their individual speeds. They cover a total of

90

miles in

3

hours, which means their combined speed is

\frac{90}{3} = 30

miles per hour. If the speed of the faster cyclist is

f

and the slower is

s

, then

f + s = 30

and

f - s = 3

. Adding these two equations gives

2f = 33

, which yields

f = 16.5

miles per hour.

Estimated Time:1m 15s

Question 2140Question →

In the standard $(x, y)$ coordinate plane, an ellipse is defined by the equation $7x^2 + 16y^2 - 42x - 32y - 33 = 0$ . A parabola has its vertex at the focus of the ellipse with the smaller $x$ -coordinate, and its focus at the focus of the ellipse with the larger $x$ -coordinate. What is the larger of the two $y$ -coordinates of the points on the parabola that have an $x$ -coordinate of 6?

Show answer & explanation

Answer: 13

Answer

The larger of the two

y

-coordinates of the points on the parabola is 13.

By completing the square on the general ellipse equation, we get

\frac{(x-3)^2}{16} + \frac{(y-1)^2}{7} = 1

. The center is

(3, 1)

and the focal distance is

c = \sqrt{16-7} = 3

, meaning the foci are at

(0, 1)

and

(6, 1)

. The parabola has its vertex at

(0, 1)

and focus at

(6, 1)

, which means it opens to the right with

p = 6

. Its equation is

(y - 1)^2 = 24x

. Substituting

x = 6

yields

(y - 1)^2 = 144

, so

y - 1 = \pm 12

. The two possible

y

-coordinates are

13

and

-11

, of which

13

is the larger value.

Step-by-Step Solution

1

Complete the square for the given ellipse equation to rewrite it in standard form.

\frac{(x-3)^2}{16} + \frac{(y-1)^2}{7} = 1

Converting the equation to standard form is necessary to determine the center and semi-axis lengths of the ellipse.

2

Find the focal distance

c

and calculate the coordinates of the foci.

Focal distance

c = 3

; Foci at

(0, 1)

and

(6, 1)

For an ellipse, the distance

c

from the center

(h, k)

to the foci is

\sqrt{a^2 - b^2}

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

(h \pm c, k)

.

3

Use the foci coordinates to identify the vertex and focus of the parabola.

Vertex:

(0, 1)

; Focus:

(6, 1)

The problem defines the parabola's vertex as the ellipse focus with the smaller

x

-coordinate, and the parabola's focus as the ellipse focus with the larger

x

-coordinate.

4

Determine the equation of the parabola using its vertex and focus.

(y - 1)^2 = 24x

The parabola is horizontal and opens to the right with focal distance

p = 6

. The standard form is

(y - k)^2 = 4p(x - h)

.

5

Substitute

x = 6

into the parabola equation and solve for the larger

y

-value.

y = 13

Substituting

x = 6

yields

(y - 1)^2 = 144

, which gives

y = 1 + 12 = 13

or

y = 1 - 12 = -11

. The larger value is 13.

Key Concept

Determining the equations and key features (foci, vertices, focal parameters) of ellipses and parabolas by rewriting equations into standard forms.

Alternative Method

Once the equation

(y - 1)^2 = 24x

is established, recognize that at

x = 6

(which is the

x

-coordinate of the focus), the points on the parabola form the endpoints of the latus rectum. The length of the latus rectum is

4p = 24

, so the points lie at distance

2p = 12

vertically above and below the focus

(6, 1)

. Thus, the

y

-coordinates are

1 \pm 12

, immediately yielding the larger coordinate as 13.

Estimated Time:3m 0s

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