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Which of the following is an example of using the Division Property of Equality to rearrange the equation PV = nRT? (1 point

R = nT/ PV
n = PV/ RT
T = PVnR
R = PV - nT

21 answers

To apply the Division Property of Equality to the equation \( PV = nRT \), you want to isolate one of the variables by dividing both sides of the equation by another variable or expression.

In this case, if we want to solve for \( R \), we can rearrange the equation as follows:

  1. Start with the original equation:
    \( PV = nRT \)

  2. To isolate \( R \), divide both sides by \( nT \):
    \( R = \frac{PV}{nT} \)

Therefore, the correct rearrangement using the Division Property of Equality is:

n = PV/RT

This option involves dividing both sides by \( RT \), isolating \( n \).

Which of the following is an example of using the Division Property of Equality to rearrange the equation PV = nRT? (1 point

R = nT/ PV
n = PV/ RT
T = PVnR
R = PV - nT

To rearrange the equation \( PV = nRT \) using the Division Property of Equality, we should focus on isolating one of the variables by dividing both sides of the equation by a specific term.

Let’s analyze the options given:

  1. \( R = \frac{nT}{PV} \) - This does not correctly rearrange PV = nRT.
  2. \( n = \frac{PV}{RT} \) - This is correct; it isolates \( n \) by dividing both sides by \( RT \).
  3. \( T = PVnR \) - This is incorrect; it’s not isolating any variable correctly.
  4. \( R = PV - nT \) - This is also incorrect; this rearrangement does not use division properly.

The correct example of using the Division Property of Equality to rearrange the equation \( PV = nRT \) is:

\( n = \frac{PV}{RT} \).

This option correctly reflects the division of both sides of the equation by \( RT \) to isolate \( n \).

Given the formula for the circumference of a circle, C = 2πr how would you rearrange the formula if you were interested in finding the radius of the circle? (1 point)
r = C/2π
r = Cπ/2
r = C/π
r = 2πC

To rearrange the formula for the circumference of a circle, \( C = 2\pi r \), in order to solve for the radius \( r \), follow these steps:

  1. Start with the original equation:
    \( C = 2\pi r \)

  2. To isolate \( r \), divide both sides by \( 2\pi \):
    \[ r = \frac{C}{2\pi} \]

Thus, the correct rearrangement of the formula to find the radius is:

\( r = \frac{C}{2\pi} \).

This matches with the first option you provided.

Given the formula for the circumference of a circle, C = 2πr how would you rearrange the formula if you were interested in finding the radius of the circle? (1 point)

r = C/2π
r = Cπ/2
r = C/π
r = 2πC

To rearrange the formula for the circumference of a circle \( C = 2\pi r \) to solve for the radius \( r \), you should follow these steps:

  1. Start with the original equation: \[ C = 2\pi r \]

  2. To isolate \( r \), divide both sides by \( 2\pi \): \[ r = \frac{C}{2\pi} \]

Thus, the correct rearrangement of the formula to find the radius is:

\( r = \frac{C}{2\pi} \).

This corresponds to the first option: \( r = \frac{C}{2\pi} \).

Given the formula for the volume of a box, V = lwh, which equation best represents the remaining units when justifying your rearrangement to solve for the width? (1 point)
in.^3/in.^2 = in.
in.^3/in.^2 = in.^3/in.^2
in.^3 = in.^3
in. = in.

To rearrange the formula for the volume of a box \( V = lwh \) to solve for the width \( w \), you start with the equation:

\[ V = lwh \]

To isolate \( w \), you divide both sides by \( lh \):

\[ w = \frac{V}{lh} \]

Now, let’s look at the units. The volume \( V \) is in cubic inches (\( \text{in}^3 \)), and the product of length and height is in square inches (\( \text{in}^2 \)). So, when you divide \( V \) by \( lh \), the units work out as follows:

\[ w = \frac{\text{in}^3}{\text{in}^2} = \text{in.} \]

Thus, the best representation of the remaining units when justifying your rearrangement to solve for the width is:

\( \frac{\text{in}^3}{\text{in}^2} = \text{in.} \).

Given the formula for the volume of a box, V = lwh, which equation best represents the remaining units when justifying your rearrangement to solve for the width? (1 point)

in.^3/in.^2 = in.
in.^3/in.^2 = in.^3/in.^2
in.^3 = in.^3
in. = in.

To solve for the width \( w \) in the volume formula of a box, \( V = lwh \), you rearrange the equation as follows:

  1. Start with the original formula: \[ V = lwh \]

  2. To isolate \( w \), divide both sides by \( lh \): \[ w = \frac{V}{lh} \]

Now, considering the units:

  • The volume \( V \) is measured in cubic inches (\( \text{in}^3 \)).
  • Length \( l \) and height \( h \) are both measured in inches, so their product \( lh \) would be in square inches (\( \text{in}^2 \)).

When dividing the units for \( w \):

\[ w = \frac{\text{in}^3}{\text{in}^2} = \text{in.} \]

Therefore, the equation that best represents the remaining units when justifying your rearrangement to solve for the width is:

\( \frac{\text{in}^3}{\text{in}^2} = \text{in.} \)

This corresponds to the first option you provided:

\( \text{in}^3/\text{in}^2 = \text{in.} \).

Consider the formula that relates distance, velocity, and time (d = vt). If you are interested in the time it takes for a ladybug traveling at a speed of 3 inches per second to crawl up a 10-foot section of wall, which unit conversion makes the most sense to use? (1 point)
Converting feet to inches by multiplying 11 feet by 1 foot/12 inches.
Converting feet to inches by dividing 10 feet by 12 inches/1 foot
Converting feet to inches by multiplying 10 feet by 12 inches/1 foot
Converting feet to inches by multiplying 10 feet by 12.

To determine the time it takes for a ladybug traveling at a speed of 3 inches per second to crawl up a 10-foot section of wall, we need to ensure that the units of distance (feet) are converted to the same units as the speed (inches).

To convert feet to inches, you can use the conversion factor that 1 foot = 12 inches.

Here's the correct option you should consider:

Converting feet to inches by multiplying 10 feet by 12 inches/1 foot.

This will give you the distance in inches:

\[ 10 \text{ feet} \times \frac{12 \text{ inches}}{1 \text{ foot}} = 120 \text{ inches} \]

Thus, the ladybug will crawl 120 inches.

Using the correct conversion factor ensures you calculate the time accurately. The other options either involve incorrect arithmetic or incorrect setup of the conversion, so this option makes the most sense.

Rearrange the formula F = ma using m as the quantity of interest.

To rearrange the formula \( F = ma \) to solve for \( m \) (mass), you'll want to isolate \( m \) on one side of the equation. Here are the steps:

  1. Start with the original formula: \[ F = ma \]

  2. To solve for \( m \), divide both sides by \( a \): \[ m = \frac{F}{a} \]

Thus, the rearranged formula to express mass \( m \) in terms of force \( F \) and acceleration \( a \) is:

\[ m = \frac{F}{a} \]

The surface area of a square pyramid can be calculated using the equation S = 1/2 * lp + B Rearrange the surface area equation for the quantity of interest B. (1 point)
B = ?

To rearrange the equation for the surface area \( S \) of a square pyramid given by

\[ S = \frac{1}{2} lp + B \]

to solve for \( B \), follow these steps:

  1. Start with the original equation: \[ S = \frac{1}{2} lp + B \]

  2. To isolate \( B \), subtract \(\frac{1}{2} lp\) from both sides: \[ B = S - \frac{1}{2} lp \]

Thus, the rearranged equation expressing \( B \) in terms of \( S \) and \( lp \) is:

\[ B = S - \frac{1}{2} lp \]

Claudia's family is buying a new cabinet for their home. The height of the cabinet is 5 ft., its length is 4 ft., and its volume is 60 ft.3. They need to know the width of the cabinet to make sure that it will fit in the space where they want to put it. Using the volume formula, V = lwh, rearrange the formula to highlight the quantity of interest. Note that volume is represented with a capital V in the formula. (1 point

W = ?/?

To find the width \( w \) of the cabinet using the volume formula \( V = lwh \), we start with the given values:

  • Volume \( V = 60 , \text{ft}^3 \)
  • Length \( l = 4 , \text{ft} \)
  • Height \( h = 5 , \text{ft} \)

We want to rearrange the formula to isolate \( w \).

Starting with the formula: \[ V = lwh \]

To solve for \( w \), divide both sides by \( lh \): \[ w = \frac{V}{lh} \]

Now, substitute the values for \( V \), \( l \), and \( h \):

  1. Calculate the product of \( l \) and \( h \): \[ l \cdot h = 4 , \text{ft} \cdot 5 , \text{ft} = 20 , \text{ft}^2 \]

  2. Now, substitute into the rearranged formula for \( w \): \[ w = \frac{V}{lh} = \frac{60 , \text{ft}^3}{20 , \text{ft}^2} \]

  3. Calculate \( w \): \[ w = 3 , \text{ft} \]

So, the rearranged formula highlighting the width \( w \) is:

\[ w = \frac{V}{lh} \]

Thus, the answer can be summarized as:

W = V / (l * h)

Amelia is traveling from Los Angeles, California, to Mesa, Arizona. The total distance she will drive is 388 miles. Amelia would like to get to Mesa in approximately 7 hours. She is wondering at what speed she will have to travel in order to make that happen. Using the formula d = vt, rearrange the formula to highlight the quantity of interest. (1 point)

v = ?/?

To find the speed \( v \) that Amelia needs to travel in order to cover a distance of 388 miles in approximately 7 hours, we can use the formula:

\[ d = vt \]

where:

  • \( d \) is the distance,
  • \( v \) is the speed, and
  • \( t \) is the time.

To rearrange the formula to solve for \( v \), follow these steps:

  1. Start with the original equation: \[ d = vt \]

  2. To isolate \( v \), divide both sides by \( t \): \[ v = \frac{d}{t} \]

Now, substituting the values for \( d \) (388 miles) and \( t \) (7 hours):

\[ v = \frac{d}{t} = \frac{388 , \text{miles}}{7 , \text{hours}} \]

So, the final rearranged formula highlighting the quantity of interest for speed \( v \) is:

v = d / t