The given inequality holds for the open interval (2.97,3.03)
It is given that
f(x)=6x+7
cL=25
c=3
ε=0.18
We have,
|f(x)−L| = |6x+7−25|
= |6x−18|
= |6(x−3)|
= 6|x−3|
Now,
6|x−3| <0.18 then |x−3|<0.03 ----->−0.03<x-3<0.03---->2.97<x<3.03
the given inequality holds for the open interval (2.97,3.03)
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Although part of your question is missing, you might be referring to this full question: For the given function f(x) and values of L,c, and ϵ0, find the largest open interval about c on which the inequality |f(x)−L|<ϵ holds. Then determine the largest value for δ>0 such that 0<|x−c|<δ→|f(x)−|<ϵ.
f(x)=6x+7,L=25,c=3,ϵ=0.18
.
L=2W+4
L+L+W+W=74
substitute L with 2W+4: 2W+4+2W+4+W+W=74
6W=74-8=66
W=11
L=2W+4=22+4=26
Answer:
B C and E
Step-by-step explanation:
Answer:
2
Step-by-step explanation:
x=4 and y=2