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
.
10. After all the Pythagorean theorem the answer is 51.7215 cm
Answer:
if it's not multiple choice, I'd say 36 + 4x
Step-by-step explanation:
Commutative property
Hi!
We need to find the value of x by isolating it on one side of one of the equations.
5y = x - 7
Add 7 to both sides.
5y + 7 = x - 7 + 7
5y + 7 = x
Now put the value of x in the other equation.
2y = 5y + 7 + 3
Simplify
2y = 5y + 10
Subtract 5y from both sides.
2y - 5y = 5y - 5y + 10
-3y = 10
Divide by -3 on both sides.
y = -10/3
Now put the value of y in the other original equation to solve for x!
5 * -10/3 = x - 7
-50/3 = x - 7
Add 7 to both sides.
-50/3 + 7 = x - 7 + 7
-29/3 = x
The answer is C.){(-29/3, -10/3)}
Hope this helps! :)
-Peredhel
Yes the treasury can determine whether the gold bar is
really weighing 500 ounces.
We can use the equation:
7 x + 11 y = 500
where x is the amount of the 7 ounce weights while y is
the amount of 11 ounce weights.
We know that x and y should be whole numbers therefore
this is solved through trial and error.
By trial and error I got:
x = 7 and y = 41
<span>
So that;</span>
7 * 7 + 11 * 41 = 500
49 + 451 = 500
500 = 500
<span>Hence they can use 41 11-ounce weights and 7 7-ounce
weights to test</span>
