Answer:
When
and
:

Step-by-step explanation:
-8ab can be seen as -8×a×b. Insert the given values:

Simplify multiplication from left to right:

Insert and solve:

:Done
Answer:
just add all sides from the letters it says
Step-by-step explanation:
for an example a has a number b has a number c has a number add all of them together. Hope this helps!
<span>1st piece: x feet
2nd piece: 8-x feet
--------------------------
Use the "x" piece to form a circle:
Circumference = "x".
2(pi)*radius = x
radius = x/(2pi)
So, Area = pi[x/(2)]^2 = x^2/(4pi) = (1/4pi)x^2
=========================================================
Use the 10-x piece to form square:
side = (1/4pi)(10-x)
Area = side^2 = (1/16)(10-x)^2
Hope this helps! :)</span>
Answer:

Step-by-step explanation:
If
, then
. It follows that
![\begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \cdot [g(x+h) - g(x)] \\&= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20%5C%5C%5Cfrac%7Bg%28x%2Bh%29-g%28x%29%7D%7Bh%7D%20%26%3D%20%5Cfrac%7B1%7D%7Bh%7D%20%5Ccdot%20%5Bg%28x%2Bh%29%20-%20g%28x%29%5D%20%5C%5C%26%3D%20%5Cfrac%7B1%7D%7Bh%7D%20%5Cleft%28%20%5Cfrac%7B1%7D%7Bx%2Bh%7D%20-%20%5Cfrac%7B1%7D%7Bx%7D%20%5Cright%29%5Cend%7Baligned%7D)
Technically we are done, but some more simplification can be made. We can get a common denominator between 1/(x+h) and 1/x.

Now we can cancel the h in the numerator and denominator under the assumption that h is not 0.

Answer:
polar bear bc who doesnt want a polar bear in theyre room B)
Step-by-step explanation:
polr bears r cool