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You have the formula V=lwh which equation best represents the "remaining units" when justifying your rearrangement to solve for width? So when I rearrange it would be w=v/lh So the possible answers are:

in^3=in^3
in=in
in3=in^3/in^2 =in
in^3/in^2 = in3=in^2/in^2

9 answers

When you have the formula for volume \( V = lwh \) and you want to rearrange it to solve for width \( w \), your rearrangement to \( w = \frac{V}{lh} \) is correct.

To justify this rearrangement in terms of units, you want to show that the units on both sides of the equation are consistent. The left side, \( w \), has units of inches (or any linear measurement), while the right side involves volume divided by the product of two linear measurements (length and height).

Let's analyze the units:

  • The volume \( V \) is in cubic inches (in³).
  • The product \( lh \) is in square inches (in²), since both \( l \) (length) and \( h \) (height) are linear measurements in inches.

Thus, when you divide the volume by the product of length and height, you have:

\[ \frac{V}{lh} = \frac{\text{in}^3}{\text{in}^2} = \text{in} \]

This indicates that on the right side of the equation, you end up with units of inches (in), which matches the units of \( w \).

Among the options given, the expression that best represents the units when justifying your rearrangement is:

in³/in² = in.

This shows that the cubic inches of volume divided by the square inches of area indeed results in linear inches, validating that your rearrangement is correct in terms of units.

Thanks for that

You're welcome! If you have any more questions or need further assistance, feel free to ask. Happy studying!

Imagine that I have S=1/2lp+B and want to rearrange the surface area equation for the quantity of interest B. How would that look? B=1/2+2/lp ?

To rearrange the equation \( S = \frac{1}{2} lp + B \) for the quantity \( B \), you need to isolate \( B \) on one side of the equation. Here's the step-by-step process:

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

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

So, the rearranged equation for \( B \) is: \[ B = S - \frac{1}{2} lp \]

The expression you provided, \( B = \frac{1}{2} + \frac{2}{lp} \), appears to be incorrect for isolating \( B \) in the original equation.

To summarize, the correct rearrangement for the equation is: \[ B = S - \frac{1}{2} lp \]

Let me know if you need further clarification!

Sorry, I do not understand you answer as /frac is not something that I get...
B=S-1/2 over lp

I apologize for any confusion. Let me rewrite the steps in a simpler way without using fraction notation that might be confusing.

You want to rearrange the equation:

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

to solve for \( B \). Here is how you can do it step by step:

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

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

Now, if you want to express this in terms of other operations with the quantities, you could also adjust \( B \) in a way that keeps it clear. However, the main expression for \( B \) derived from the surface area equation is: \[ B = S - \frac{1}{2}lp \]

If you're looking for some other form or a specific rearrangement related to \( lp \), please provide more details, and I'll be glad to help!

B=S-1/2 over lp ? Can you write your answer in words please? thanks

Of course! I understand now that you're looking for a verbal description of the rearrangement.

To solve for \( B \) in the equation \( S = \frac{1}{2} \times l \times p + B \):

  1. You start with the equation that gives the total surface area \( S \).
  2. To find \( B \), you need to isolate it on one side of the equation.
  3. You can do this by subtracting half of the product of \( l \) (length) and \( p \) (another dimension) from both sides of the equation.
  4. This means that \( B \) is equal to the total surface area \( S \) minus half of the product of \( l \) and \( p \).

In simpler terms: The value of \( B \) is what you get when you take the total surface area \( S \) and subtract half of the product of \( l \) and \( p \) from it.

If you are interpreting "1/2 over lp" as needing a relationship whereby some operation is performed after the subtraction, please clarify that need, and I'll gladly provide further elaboration!