Step-by-step explanation:
I think it's right answer is c
We can create two equations here:
(1) Volume = area of square * height of box
85.75 = s^2 h
(2) Cost = 3 * area of square + 1.5 * area of side box
C = 3 s^2 + 1.5 s h
From (1), we get:
h = 85.75 / s^2
Combining this with (2):
C = 3 s^2 + 1.5 s (85.75 / s^2)
C = 3 s^2 + 128.625 s-
Taking the 1st derivative and equating dC/ds =
0:
dC/ds = 6s – 128.625 / s^2 = 0
Multiply all by s^2:
6s^3 – 128.625 = 0
6s^3 = 128.625
s = 2.78 cm
So h is:
h = 85.75 / s^2 = 85.75 / (2.78)^2
h = 11.10 cm
So the dimensions are 2.78 cm x 2.78 cm x 11.10 cm
The total cost now is:
C = 3 (2.78)^2 + 1.5 (2.78) (11.10)
C = $69.47
The answer is 5:9 or 5 to 9.
Please mark me as the brainiest answer if there's a chance :)
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Answer:
(x + 4)(2x - 3)
Step-by-step explanation:
Given
f(x) = 2x² + 5x - 12
Consider the factors of the product of the x² term and the constant term which sum to give the coefficient of the x- term.
product = 2 × -12 = - 24 and sum = + 5
The factors are + 8 and - 3
Use these factors to split the x- term
2x² + 8x - 3x - 12 ( factor the first/second and third/fourth terms )
2x(x + 4) - 3(x + 4) ← factor out (x + 4) from each term
(x + 4)(2x - 3)
Thus
f(x) = 2x² + 5x - 12 = (x + 4)(2x - 3) ← in factored form
Answer:
The volume of the solid is 
Step-by-step explanation:
In this case, the washer method seems to be easier and thus, it is the one I will use.
Since the rotation is around the y-axis we need to change de dependency of our variables to have 
. Thus, our functions with 
as independent variable are:
For the washer method, we need to find the area function, which is given by:
![A=\pi\cdot [(\rm{outer\ radius)^2 -(\rm{inner\ radius)^2 ]](https://tex.z-dn.net/?f=A%3D%5Cpi%5Ccdot%20%5B%28%5Crm%7Bouter%5C%20radius%29%5E2%20-%28%5Crm%7Binner%5C%20radius%29%5E2%20%5D)
By taking a look at the plot I attached, one can easily see that for a rotation around the y-axis the outer radius is given by the function 
and the inner one by 
. Thus, the area function is:
![A(y)=\pi\cdot [(\sqrt{y} )^2-(y^2)^2]\\A(y)=\pi\cdot (y-y^4)](https://tex.z-dn.net/?f=A%28y%29%3D%5Cpi%5Ccdot%20%5B%28%5Csqrt%7By%7D%20%29%5E2-%28y%5E2%29%5E2%5D%5C%5CA%28y%29%3D%5Cpi%5Ccdot%20%28y-y%5E4%29)
Now we just need to integrate. The integration limits are easy to find by just solving the equation 
, which has two solutions 
and 
. These are then, our integration limits.
