We have that
x²<span>-20x
</span><span>Group
terms that contain the same variable
</span>(x²-20x)
Complete
the square
(x²-20x+10²)------> (x-10)²
the answer is
the number must be 100
X - $42.15 > $20.00
<u> + $42.15 + $42.15</u>
x > $62.15
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
Find out more on equation at: brainly.com/question/2972832
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Step-by-step explanation:
I hope it's correct...
Answer:
The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.
Step-by-step explanation:
The volume of the sphere (
), measured in cubic feet, is represented by the following formula:
Where 
is the radius of the sphere, measured in feet.
The change in volume is obtained by means of definition of total difference:
The derivative of the volume as a function of radius is:
Then, the change in volume is expanded:
If 
and 
, the change in the volume of the sphere is approximately:
The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.
