Answer:
9x^2 - 10x -4
Step-by-step explanation:
It is set up like this
9x^2 - 8x
- (x-2)2
First solve (x-2)2
the answer is (2x-4)
so now you can do this instead
9x^2 - 8x
- 2x - 4
----------------------------------
9x^2 - 10x +4
9x^2 minus zero is 9x^2
-8x-2x is -10x
0- -4 is 4 because that is like saying 0 + 4 because double negatives equal a positive
A simpler way is to change all the numbers to the opposite sign and then change the minus to a plus
like this
9x^2 - 8x
+ -2x + 4
----------------------------------
9x^2 - 10x +4
Answer:
y=x-5
Step-by-step explanation:
y=mx+b
-5=0(3)+b
0(3)-5=b
-5=b
y=x-5
Correct me if im wrong
Answer: 2/3
Step-by-step explanation:
Answer:



Step-by-step explanation:
<u>Optimizing With Derivatives
</u>
The procedure to optimize a function (find its maximum or minimum) consists in
:
- Produce a function which depends on only one variable
- Compute the first derivative and set it equal to 0
- Find the values for the variable, called critical points
- Compute the second derivative
- Evaluate the second derivative in the critical points. If it results positive, the critical point is a minimum, if it's negative, the critical point is a maximum
We know a cylinder has a volume of 4
. The volume of a cylinder is given by

Equating it to 4

Let's solve for h

A cylinder with an open-top has only one circle as the shape of the lid and has a lateral area computed as a rectangle of height h and base equal to the length of a circle. Thus, the total area of the material to make the cylinder is

Replacing the formula of h

Simplifying

We have the function of the area in terms of one variable. Now we compute the first derivative and equal it to zero

Rearranging

Solving for r

![\displaystyle r=\sqrt[3]{\frac{4}{\pi }}\approx 1.084\ feet](https://tex.z-dn.net/?f=%5Cdisplaystyle%20r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B4%7D%7B%5Cpi%20%7D%7D%5Capprox%201.084%5C%20feet)
Computing h

We can see the height and the radius are of the same size. We check if the critical point is a maximum or a minimum by computing the second derivative

We can see it will be always positive regardless of the value of r (assumed positive too), so the critical point is a minimum.
The minimum area is

