The volume of a rectangular box is given by the formula V = lwh, where l is length, w is width, and h is height. What happens to the volume if the length and height are doubled and the width is tripled?
I don't know how to solve this problem at all.
Thank you for your help!
I really don't understand this honors algebra
new volume = (2h)(2l)(3w) = 12lwh
so the volume is 12 times as big
12 times
vol = l*w*h
the question is that
l is doubled to give 2l
h is doubled to give 2h and
w is tripled to give 3h.
so the new vol = (2l)*(2h)*(3w) = 12*l*w*h
so 'new vol' = 12*old vol
Hope that helps
To solve this problem, we need to understand how changes in the dimensions of a rectangular box affect its volume. Let's break it down step by step.
The formula for calculating the volume of a rectangular box is V = lwh, where l is the length, w is the width, and h is the height.
In this case, we are given that the length and height are doubled, and the width is tripled.
Let's assume the original length, width, and height are denoted as l1, w1, and h1, respectively.
According to the given information:
- The new length is 2 times the original length: l2 = 2 * l1.
- The new height is 2 times the original height: h2 = 2 * h1.
- The new width is 3 times the original width: w2 = 3 * w1.
To find the new volume (V2) after these changes, we substitute these new values into the volume formula:
V2 = l2 * w2 * h2
= (2 * l1) * (3 * w1) * (2 * h1)
= 4 * 3 * 2 * l1 * w1 * h1
= 24 * l1 * w1 * h1
From the above equation, we can see that the new volume (V2) is 24 times the original volume (V1 = l1 * w1 * h1).
Therefore, if the length and height are doubled and the width is tripled, the volume will increase by a factor of 24.
Hope this explanation helps!