Answer:
<h2><u><em>
2 × [(3x² + x + 2) + (-x² – 5x + 1)] =</em></u></h2>
Step-by-step explanation:
Write an expression for the perimeter of a rectangle with a length of
(3x² + x + 2) and a width of (-x² – 5x + 1).
P = 2 × (L + W)
<u><em>2 × [(3x² + x + 2) + (-x² – 5x + 1)] =</em></u>
if you want the result:
4x²−8x+6
Answer:
The measure of angle LMW is 
Step-by-step explanation:
see the attached figure to better understand the problem
step 1
Find the measure of arc MW
we know that
The inscribed angle measures half that of the arc comprising
so
![m\angle MLK=\frac{1}{2}[arc\ MW+arc\ WK]](https://tex.z-dn.net/?f=m%5Cangle%20MLK%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20MW%2Barc%5C%20WK%5D)
substitute the given values
![65\°=\frac{1}{2}[arc\ MW+68\°]](https://tex.z-dn.net/?f=65%5C%C2%B0%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20MW%2B68%5C%C2%B0%5D)
![130\°=[arc\ MW+68\°]](https://tex.z-dn.net/?f=130%5C%C2%B0%3D%5Barc%5C%20MW%2B68%5C%C2%B0%5D)

step 2
Find the measure of arc LK
we know that
-----> by complete circle
substitute the given values
step 3
Find the measure of angle LMW
we know that
The inscribed angle measures half that of the arc comprising
so
![m\angle LMW=\frac{1}{2}[arc\ LK+arc\ WK]](https://tex.z-dn.net/?f=m%5Cangle%20LMW%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20LK%2Barc%5C%20WK%5D)
substitute the given values
![m\angle LMW=\frac{1}{2}[66\°+68\°]=67\°](https://tex.z-dn.net/?f=m%5Cangle%20LMW%3D%5Cfrac%7B1%7D%7B2%7D%5B66%5C%C2%B0%2B68%5C%C2%B0%5D%3D67%5C%C2%B0)
Answer:
<r=29
Step-by-step explanation:
Answer:
Answer is 88
Step-by-step explanation:
We have Given function

we have to find f'(x) at x=3 Where f'(x) shows derivative of the given function.
This means we need
f'(3)=?
So By definition of the Derivative that is

so this our definition of derivative of the function
Now We have to find out at x=3, So By putting x =3 in definition ,We get
f'(3)=lim(h---->0)(f(3+h)-f(3))/h
Here lim(h--->0) means limit h approaches to zero(right arrow 0limh→0)
=lim(h---->0)((88(3+h)-77))-(88(3)-77))/h
=lim(h---->0)((264+88h-77)-264+77)/h
=lim(h----->0)(264+88h-77-264+77)/h
now by performing simple arithematic we get result
f'(3) = lim(h---->0)(88h/h)
f('3) = lim(h---->0)(88)
here we use law of the limit we limit of the constant is that constant
lim(h----->0)c=c
so
f'(3)=88
So this our answer
Answer:
35
Step-by-step explanation:
<u>7x2</u>+21
<u>14+21</u>
35
*The underlined problems is what you do first*