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
.
a) To find the volume of a cube, you cube the side length.
So, that will be V = 9/2³ = 729/8 in³.
b) To find the volume of a rectangular prism, you multiply the length by the width by the height.
So, that is V = 18 * 9 * 9/2 = 729 in³.
c) To find how many times the smaller box can go into the larger one, you divide the larger one's volume over the smaller one's.
That is <em>729/(729/8) = 729*8/729 = </em><em>8 boxes</em><em>.</em>
Answer:
it is the first one
Step-by-step explanation:
I’m confused there are no numbers ?
There are 310 tens in 3100.
