To multiply two fractions, we multiply the top number (numerators) together and the bottom numbers together (denominators) and then simplify.
8 x 5 = 40
15 x 6 = 90
40/90 can be simplified (÷10) to 4/9
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
.
Translated means the points are moving across the plane without rotating or changing shape. In this case, the x-coordinate would be moving up 5 (x + 5) and the y-coordinate would be moving to the left 4 (y - 4).
A is (-8, 6). A' is the result of the translation from this point. The results of the solution above in A is the point (-3, 2) = A'.
Now you must find the distance between these two coordinates. To find the distance you must use the distance formula: √<span>(x2 - x1)^2 + (y2 - y1)^2. Since you now have two points, A and A', plug these into the distance formula.
</span>√(-3 - (-8))^2 + (2 - 6)^2
√5^2 + (-4)^2
√25 + 16
√41
The distance from A to A' is √41.
