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f′(x)=2∗(8x(2−1))+1∗11x(1−1)
which is
f′(x)=16x+11
then let
x = 7 gives us
f′(7)=123
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<span>Hope my answer would be a great help for you. </span> </span>
<span> </span>
<h2>Here we go ~ </h2>
According to given figure,


[ By linear pair ]

now, we can see that :

[ By Exterior angle property of Triangle ]


Answer:
The answer will be A. 27f
Answer:
m=31/2, n=-24. (31/2, -24).
Step-by-step explanation:
2m+n=7
4m+3n=-10
-------------------
n=7-2m
4m+3(7-2m)=-10
4m+21-6m=-10
-2m=-10-21
-2m=-31
2m=31
m=31/2
2(31/2)+n=7
31+n=7
n=7-31
n=-24
Answer:
9/5 (K-273.15) + 32=F
Step-by-step explanation:
K=5 /9(F−32)+273.15
Subtract 273.15 from each side
K-273.15=5/9(F−32)+273.15-273.15
K-273.15=5/9(F−32)
Multiply by 9/5 on each side
9/5 (K-273.15)= 9/5 *5/9(F−32)
9/5 (K-273.15)=(F−32)
Add 32 to each side
9/5 (K-273.15) + 32=F−32 +32
9/5 (K-273.15) + 32=F