Answer:
4.21399176955
Step-by-step explanation:
4.21399176955
I'm positive the property is commutative property because you only switched the position of the numbers and got the same outcome.
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)
For more information on inequality click on the link below:
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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
.
Hello!
We can solve this system using substitution.
Since y = -3x + 5, we can substitute this into the following equation:
5x - 4(-3x + 5) = -3
Simplify:
5x - 4(-3x) - 4(5) = -3
5x + 12x - 20 = -3
Combine like terms:
17x - 20 = -3
Add 20 to both sides:
17x = 17
x = 1
Substitute to find y:
y = -3(1) + 5
y = -3 + 5
y = 2
Solution: (1, 2)
6,824 rounded to the nearest ten is 6,820.
That is because 6,824 is closer to 6,820 than it is to 6,830.
