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s2008m [1.1K]
3 years ago
12

Please help me the photo is my queston

Mathematics
1 answer:
Studentka2010 [4]3 years ago
3 0

Answer:

i think it would be 30

Step-by-step explanation:

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6
GalinKa [24]

Answer:

I don't know sorry

sorry

Step-by-step explanation:

I don't know sorry

sorry

3 0
2 years ago
Arthur purchased a 2010 model sedan for $20,000. The dealership offered
JulsSmile [24]

Answer: the answer is $ 17,349.21 APEX

3 0
3 years ago
I try to look this up and people said yes or no I just need a straight answer
Anna11 [10]

start inside the square root, replace X with -7.

-7 + 11 = 4

The number 4 comes out of the square root as 2.

Now there's a minus factor out there. Let's multiply this negative factor by 2 and get -2.

Now let's add -2 and -3.

Our output is -5.

3 0
2 years ago
PLEASE HELP W/ MATH PROBLEM
Ilia_Sergeevich [38]

Answer:

11. 736080 cars //   12. 5397920 cars

Step-by-step explanation:

given total number of cars produced : 6,134,000

100% -->6,134,000

1%--> 6,134,000 /100

1% --> 61340

given big cars of America is 12%

so to find this 12% = 12 * 61340 = 736080 cars produced by America

by other manufactures =  total cars - big cars of America

                                       = 6,134,000 - 7,36,080

                                       = 5397920 cars

3 0
3 years ago
Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer
agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming:  the function is f(x)=x^{2} in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if  both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

g'(0)=g'(1)

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0

g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]

This is what the Bilateral Theorem says:

\lim_{x\rightarrow c^{-}}f(x)=L\Leftrightarrow \lim_{x\rightarrow c^{+}}f(x)=L\:and\:\lim_{x\rightarrow c^{-}}f(x)=L

4 0
3 years ago
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