The volume of the trough is V(w) = w³ + 20w² - 429w and the rate of change of the volume over a width of 38 inches to 53 inches is 4695 in³/in
<h3>What is an
equation?</h3>
An equation is an expression that shows the relationship between two or more variables and numbers.
Let w represent the width, hence:
length = w + 33, height = w - 13
Volume (V) = w(w + 33)(w - 13) = w³ + 20w² - 429w
V(w) = w³ + 20w² - 429w
Rate of change = dV/dw = 3w² + 40w - 429
When w = 38, dV/dw = 3(38)² + 40(38) - 429 = 5423
When w = 53, dV/dw = 3(53)² + 40(53) - 429 = 10118
Rate = 10118 - 5423 = 4695 in³/in
The volume of the trough is V(w) = w³ + 20w² - 429w and the rate of change of the volume over a width of 38 inches to 53 inches is 4695 in³/in
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Use rates, so 8 garden hoses over 18 hours
then find the how many hours one garden house can fill up by doing this = 18/8 you divide the 8 for both sides which is about 2.25/1
so then you times 9 which is about 20/9 hoses
your answer is 20
Answer:
Step-by-step explanation:
10/25=20/50=40/100
6/8=3/4=12/16
3/5=6/10=60/100
1/10=10/100
For this case we have the following equation:
y = 150 * (1.06) ^ t
For the first month we have:
y = 150 * (1.06) ^ 1
y = 159 $
For the second month we have:
y = 150 * (1.06) ^ 2
y = 168.54 $
For the third month we have:
y = 150 * (1.06) ^ 1
y = 178.65 $
Answer:
d. $ 159.00 + $ 168.54 + $ 178.65
