All practice questions

Question 2381Question →

In the standard $(x, y)$ coordinate plane, the line segment $PQ$ has endpoints $P(-3, 5)$ and $Q(1, -1)$ . Segment $PQ$ is reflected across the line $y = -x$ , and the resulting image is then translated $3$ units to the right and $2$ units down to form segment $P'Q'$ . What is the midpoint of segment $P'Q'$ ?

(-5, 3)

(0, -2)

(1, -1)

(-5, -1)

(1, 3)

Show answer & explanation

Answer: $(1, -1)$

Answer

(1, -1)

The correct answer is the coordinate point

(1, -1)

. First, the midpoint of the original segment

PQ

is found using the midpoint formula:

\left( \frac{-3 + 1}{2}, \frac{5 + (-1)}{2} \right) = (-1, 2)

. Reflecting this midpoint across the line

y = -x

swaps and negates the coordinates, transforming

(-1, 2)

to

(-2, 1)

. Finally, translating this point

3

units to the right and

2

units down results in the coordinates

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

.

Step-by-Step Solution

1

Find the midpoint of the original segment

PQ

.

The midpoint is

M(-1, 2)

.

Since the midpoint is invariant under linear coordinate transformations, we can find the midpoint of the original segment and transform it directly rather than transforming both endpoints and finding the new midpoint.

2

Reflect the midpoint across the line

y = -x

.

The reflected point is

M_1(-2, 1)

.

Reflecting a point across the line

y = -x

maps the coordinate

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

.

3

Translate the reflected point

3

units to the right and

2

units down.

The final point is

M'(1, -1)

.

Translating a point

3

units right adds

3

to the

x

-coordinate, and translating

2

units down subtracts

2

from the

y

-coordinate:

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

.

Key Concept

Transformations in the Coordinate Plane (Reflection and Translation)

Alternative Method

Instead of transforming the midpoint directly, you can apply the transformations to the endpoints

P(-3, 5)

and

Q(1, -1)

individually. Reflecting both across the line

y = -x

gives

P_1(-5, 3)

and

Q_1(1, -1)

. Translating both points

3

units right and

2

units down gives

P'(-2, 1)

and

Q'(4, -3)

. The midpoint of

P'Q'

is then calculated as

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

.

Estimated Time:1m 30s

Question 2382Question →

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 2383Question →

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 2384Question →

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 2385Question →

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 2386Question →

An isosceles triangle has two sides of length $5$ and $11$ . A second triangle has side lengths of $12$ , $18$ , and $d$ , where $d$ is an integer. If $d$ is equal to the perimeter of the first triangle, what is the perimeter of the second triangle?

41

46

51

57

63

Show answer & explanation

Answer: 57

Answer

57

To find the perimeter of the second triangle, we must first determine the value of

d

, which is the perimeter of the first triangle. The first triangle is isosceles with two sides of length

5

and

11

. By the Triangle Inequality Theorem, the sum of any two side lengths must exceed the third. A triangle with sides

5, 5, 11

is impossible because

5 + 5 = 10 < 11

. Thus, the sides of the first triangle must be

11, 11, 5

, giving a perimeter of

11 + 11 + 5 = 27

. This means

d = 27

. The second triangle has sides of length

12

,

18

, and

27

. Since

12 + 18 = 30 > 27

, this is a valid triangle. Its perimeter is

12 + 18 + 27 = 57

.

Step-by-Step Solution

1

Determine the possible side lengths of the first isosceles triangle.

The sides must be

11

,

11

, and

5

.

An isosceles triangle has two equal sides. The side lengths must be either

5, 5, 11

or

11, 11, 5

. According to the Triangle Inequality Theorem, the sum of the lengths of any two sides of a triangle must be strictly greater than the length of the third side. If the sides were

5, 5, 11

, then

5 + 5 = 10 < 11

, which violates this theorem. Thus, the only valid side lengths are

11

,

11

, and

5

(since

5 + 11 = 16 > 11

).

2

Calculate the perimeter of the first triangle to find the value of

d

.

d = 27

The perimeter of the first triangle is the sum of its three sides:

11 + 11 + 5 = 27

. Since

d

is equal to this perimeter,

d = 27

.

3

Verify that a triangle with side lengths

12

,

18

, and

27

is valid.

The triangle is valid.

We check the Triangle Inequality Theorem:

12 + 18 = 30 > 27

,

12 + 27 = 39 > 18

, and

18 + 27 = 45 > 12

. Since all inequalities hold, the second triangle is valid.

4

Calculate the perimeter of the second triangle.

Perimeter =

57

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

12 + 18 + 27 = 57

.

Key Concept

Triangle Inequality Theorem and Isosceles Triangle Properties

Estimated Time:2m 0s

Question 2387Question →

In the standard $(x, y)$ coordinate plane, a triangle has vertices at $L(-2, -1)$ , $M(2, 10)$ , and $N(6, 4)$ . A line segment is drawn from vertex $L$ to the midpoint of side $MN$ . What is the length of this line segment?

14

2\sqrt{7}

\sqrt{65}

10

\sqrt{40}

Show answer & explanation

Answer: $10$

Answer

The length of the line segment is

10

.

First, find the midpoint

P

of side

MN

with vertices

M(2, 10)

and

N(6, 4)

using the midpoint formula:

P = \left(\frac{2 + 6}{2}, \frac{10 + 4}{2}\right) = (4, 7)

. Then, calculate the distance between vertex

L(-2, -1)

and the midpoint

P(4, 7)

using the distance formula:

d = \sqrt{(4 - (-2))^2 + (7 - (-1))^2} = \sqrt{6^2 + 8^2} = \sqrt{100} = 10

.

Step-by-Step Solution

1

Find the coordinates of the midpoint of side

MN

.

The midpoint is

P(4, 7)

.

Apply the midpoint formula

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

to the vertices

M(2, 10)

and

N(6, 4)

, yielding

P = \left(\frac{2 + 6}{2}, \frac{10 + 4}{2}\right) = (4, 7)

.

2

Calculate the distance between vertex

L(-2, -1)

and the midpoint

P(4, 7)

.

The distance is

10

.

Apply the distance formula

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

to points

L(-2, -1)

and

P(4, 7)

, yielding

d = \sqrt{(4 - (-2))^2 + (7 - (-1))^2} = \sqrt{6^2 + 8^2} = \sqrt{100} = 10

.

Key Concept

Distance and Midpoint Formulas

Question 2388Question →

In the standard $(x, y)$ coordinate plane, the line $3x - 4y = 24$ intersects the $x$ -axis at point $A$ and the $y$ -axis at point $B$ . What is the length of the segment $AB$ ?

10

2\sqrt{7}

14

2

\frac{4}{3}

Show answer & explanation

Answer: $10$

Answer

The length of the segment

AB

is

10

.

To find the length of segment

AB

, we first determine the coordinates of points

A

and

B

. The

x

-intercept,

A

, is found by setting

y = 0

, giving

3x = 24 \Rightarrow x = 8

, so

A = (8, 0)

. The

y

-intercept,

B

, is found by setting

x = 0

, giving

-4y = 24 \Rightarrow y = -6

, so

B = (0, -6)

. Using the distance formula, the distance between

(8, 0)

and

(0, -6)

is

\sqrt{(8 - 0)^2 + (0 - (-6))^2} = \sqrt{64 + 36} = \sqrt{100} = 10

. This matches the correct option.

Step-by-Step Solution

1

Find the coordinates of point

A

, the

x

-intercept of the line.

Point

A

is

(8, 0)

.

Set

y = 0

in the equation

3x - 4y = 24

, which gives

3x = 24

, so

x = 8

.

2

Find the coordinates of point

B

, the

y

-intercept of the line.

Point

B

is

(0, -6)

.

Set

x = 0

in the equation

3x - 4y = 24

, which gives

-4y = 24

, so

y = -6

.

3

Calculate the distance between point

A(8, 0)

and point

B(0, -6)

using the distance formula.

The distance is

10

.

The distance formula is

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

. Substituting the coordinates gives

d = \sqrt{(0 - 8)^2 + (-6 - 0)^2} = \sqrt{(-8)^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10

.

Key Concept

Finding intercepts of a linear equation and calculating the distance between two points on the coordinate plane.

Question 2389Question →

A point $P$ in a coordinate plane undergoes a translation of 4 units to the left and 5 units up, followed by a reflection across the $y$ -axis. If the coordinates of the image point after both transformations are $(2, -3)$ , what are the coordinates of the original point $P$ ?

(-6, 2)

(6, -2)

(2, -8)

(-6, -8)

(-2, 8)

Show answer & explanation

Answer: $(2, -8)$

Answer

The coordinates of the original point

P

are

(2, -8)

.

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

(2, -3)

by reversing each transformation in reverse order. First, we reverse the reflection across the

y

-axis. Reflecting across the

y

-axis negates the x-coordinate, so negating the x-coordinate of

(2, -3)

gives the intermediate point

(-2, -3)

. Second, we reverse the translation of 4 units left and 5 units up by translating the intermediate point 4 units right and 5 units down. This gives

x = -2 + 4 = 2

and

y = -3 - 5 = -8

. Thus, the original coordinates of point

P

are

(2, -8)

.

Step-by-Step Solution

1

Reverse the reflection across the

y

-axis by applying it to the final image point

(2, -3)

.

The intermediate point is

(-2, -3)

.

Reflecting a point across the

y

-axis negates its x-coordinate while keeping its y-coordinate the same. Reversing this reflection also negates the x-coordinate.

2

Reverse the translation of 4 units left and 5 units up by translating the intermediate point

(-2, -3)

4 units right and 5 units down.

The original point

P

is

(2, -8)

.

To undo a translation, apply the opposite operations: add 4 to the x-coordinate and subtract 5 from the y-coordinate.

Key Concept

Reversing composite transformations in the coordinate plane

Question 2390Question →

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 2391Question →

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 2392Question →

On the first day of a research project, a student analyzes $\frac{1}{4}$ of a dataset. On each subsequent day, the student analyzes a fraction of the dataset that is exactly $\frac{2}{3}$ of the fraction analyzed on the previous day. What fraction of the dataset does the student analyze on the 4th day of the project?

\frac{1}{216}

\frac{4}{81}

\frac{2}{27}

\frac{1}{2}

\frac{9}{31}

Show answer & explanation

Answer: $\frac{2}{27}$

Answer

The fraction of the dataset analyzed on the 4th day is

\frac{2}{27}

.

The problem describes a geometric sequence where the first term is the fraction of the dataset analyzed on the first day, which is

\frac{1}{4}

, and the common ratio is the multiplier for each subsequent day, which is

\frac{2}{3}

. The general formula for the

n

-th term of a geometric sequence is

a_n = a_1 \cdot r^{n-1}

. For the 4th day, we evaluate

a_4 = \frac{1}{4} \cdot \left(\frac{2}{3}\right)^{4-1} = \frac{1}{4} \cdot \left(\frac{2}{3}\right)^3 = \frac{1}{4} \cdot \frac{8}{27} = \frac{2}{27}

.

Step-by-Step Solution

1

Identify the type of sequence and its parameters from the word problem.

This is a geometric sequence with the first term

a_1 = \frac{1}{4}

and a common ratio

r = \frac{2}{3}

.

Each day's fraction is a constant multiple of the previous day's fraction, which characterizes a geometric sequence.

2

Set up the formula for the

n

-th term of a geometric sequence to find the term for the 4th day.

a_n = a_1 \cdot r^{n-1} \implies a_4 = \frac{1}{4} \cdot \left(\frac{2}{3}\right)^{4-1}

To find the fraction analyzed on the 4th day, we evaluate the 4th term of the sequence (

n = 4

).

3

Simplify the exponent and calculate the final fraction.

a_4 = \frac{1}{4} \cdot \left(\frac{2}{3}\right)^3 = \frac{1}{4} \cdot \frac{8}{27} = \frac{2}{27}

Performing exponentiation before multiplication satisfies the order of operations and yields the correct fraction.

Key Concept

Modeling real-world scenarios using the general term formula of geometric sequences.

Question 2393Question →

Passage

In the summer of 1884, the Scottish meteorologist Clement Wragge began ascending Ben Nevis daily to establish a weather observatory at its summit. Decades before the automated stations of the twentieth century, Wragge's grueling climbs laid the groundwork for modern mountain meteorology. However, his work was only made possible by the trial observations conducted three years earlier by the Meteorological Society. In 1881, the society had erected a temporary shelter on the peak, testing whether instruments could survive the harsh winter. Wragge himself had not been involved in that initial pilot; he was then concluding a meteorological survey in the Australian outback. It was only after his return to Great Britain in late 1882 that the society approached him with the proposal to lead the permanent station. By the time the observatory officially closed in 1904 due to a lack of government funding, Wragge had long since departed for Queensland, where he would go on to name tropical cyclones after local politicians.

Based on the passage, what is the correct chronological order of the following events, from earliest to latest?

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence is: first, the construction of a temporary testing shelter on Ben Nevis (1881); second, Clement Wragge concluding his survey in Australia (late 1882); third, the start of daily ascents to build a permanent station (1884); and finally, the closure of the observatory due to funding issues (1904).

The correct order follows the chronological progression established in the passage: the temporary shelter construction in 1881, the completion of Wragge's Australian survey prior to his late 1882 return, his daily climbs commencing in 1884, and the closure of the observatory in 1904.

Step-by-Step Solution

1

Identify the year of the temporary shelter construction.

The Meteorological Society built the temporary shelter in 1881.

The passage explicitly states that the society erected a temporary shelter on the peak in 1881.

2

Determine when Wragge concluded his Australian outback survey.

Wragge concluded his survey in Australia and returned to Great Britain in late 1882.

The passage notes he was concluding the survey during the 1881 pilot and returned in late 1882.

3

Determine the date of Wragge's daily climbs to build the permanent observatory.

Wragge began daily climbs in the summer of 1884.

The passage states he began ascending Ben Nevis daily in the summer of 1884.

4

Identify when the observatory officially closed.

The observatory closed in 1904.

The passage explicitly states the observatory officially closed in 1904.

Key Concept

Identifying the chronological sequence of events in a non-linear narrative by tracking explicit dates and relative time descriptions.

Question 2394Question →

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 2395Question →

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 2396Question →

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 2397Question →

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 2398Question →

A computer hardware engineer is designing a rectangular silicon microchip. The total surface area of the microchip, in square millimeters, is represented by the expression $2y(3y^2 - 4y + 5)$ . A square sensor is embedded in the center of the chip, occupying an area represented by the expression $(2y - 3)^2$ square millimeters. Which of the following expressions represents the remaining surface area of the microchip, in square millimeters, that is not occupied by the sensor?

6y^3 - 12y^2 + 10y - 9

6y^3 - 12y^2 - 2y + 9

6y^3 - 12y^2 + 22y - 9

-6y^2 + 22y - 9

6y^3 - 12y^2 + 22y + 9

Show answer & explanation

Answer: $6y^3 - 12y^2 + 22y - 9$

Answer

The correct expression is

6y^3 - 12y^2 + 22y - 9

.

To find the remaining surface area of the microchip, subtract the area of the sensor from the total area. First, expand the expression for the total area:

2y(3y^2 - 4y + 5) = 6y^3 - 8y^2 + 10y

. Next, expand the expression for the sensor area:

(2y - 3)^2 = 4y^2 - 12y + 9

. Subtract the sensor area from the total area, distributing the negative sign to all terms:

(6y^3 - 8y^2 + 10y) - (4y^2 - 12y + 9) = 6y^3 - 8y^2 + 10y - 4y^2 + 12y - 9

. Combining like terms yields the expression

6y^3 - 12y^2 + 22y - 9

.

Step-by-Step Solution

1

Expand the total area of the microchip by distributing the monomial

2y

to each term of the trinomial

3y^2 - 4y + 5

.

2y(3y^2 - 4y + 5) = 6y^3 - 8y^2 + 10y

This determines the polynomial representing the total area before any components are subtracted.

2

Expand the area of the square sensor by squaring the binomial

(2y - 3)

.

(2y - 3)^2 = 4y^2 - 12y + 9

This determines the polynomial representing the area occupied by the sensor, ensuring the middle term

-12y

is correctly calculated.

3

Subtract the sensor's area from the total area, ensuring the negative sign is distributed to every term in the sensor's polynomial.

(6y^3 - 8y^2 + 10y) - (4y^2 - 12y + 9) = 6y^3 - 8y^2 + 10y - 4y^2 + 12y - 9

This sets up the subtraction of the two polynomials while distributing the negative sign to avoid common sign errors.

4

Combine like terms to find the final simplified expression in standard descending order.

6y^3 - 12y^2 + 22y - 9

This combines the cubic, quadratic, linear, and constant terms to find the remaining area.

Key Concept

Polynomial operations including monomial multiplication, binomial expansion, and subtraction with sign distribution.

Estimated Time:1m 30s

Question 2399Question →

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 2400Question →

A geometric shape on a coordinate grid has a vertex at $T(2, -5)$ . The shape is translated $3$ units to the left and $4$ units up, and then the resulting image is reflected across the $y$ -axis. What are the coordinates of the final image of vertex $T$ ?

(-5, -1)

(-6, -8)

(1, -1)

(1, -9)

(-5, -9)

Show answer & explanation

Answer: $(1, -1)$

Answer

The correct coordinates of the final image of vertex

T

are

(1, -1)

.

To find the coordinates of the final image, we apply the transformations in the specified order. First, translating the point

(2, -5)

by

3

units to the left subtracts

3

from the

x

-coordinate, and translating it

4

units up adds

4

to the

y

-coordinate:

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

. Second, reflecting the point

(-1, -1)

across the

y

-axis negates the

x

-coordinate while keeping the

y

-coordinate the same:

(-(-1), -1) = (1, -1)

.

Step-by-Step Solution

1

Apply the horizontal and vertical translations to the pre-image point

T(2, -5)

.

Translating

3

units left subtracts

3

from the

x

-coordinate (

2 - 3 = -1

). Translating

4

units up adds

4

to the

y

-coordinate (

-5 + 4 = -1

). The intermediate point is

(-1, -1)

.

A translation of

h

units horizontally and

k

units vertically maps

(x, y)

to

(x + h, y + k)

.

2

Reflect the intermediate point

(-1, -1)

across the

y

-axis.

Reflecting across the

y

-axis negates the

x

-coordinate while keeping the

y

-coordinate the same:

(-(-1), -1) = (1, -1)

.

A reflection across the

y

-axis maps any point

(x, y)

to

(-x, y)

.

Key Concept

Transformations in the Coordinate Plane

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer