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user100 [1]
3 years ago
12

Do you guys know this by any chance

Mathematics
2 answers:
Ne4ueva [31]3 years ago
8 0

Answer:

10^{17}

Step-by-step explanation:

Move the decimal so there is one non-zero digit to the left of the decimal point. The number of decimal places you move will be the exponent on the

10

10

. If the decimal is being moved to the right, the exponent will be negative. If the decimal is being moved to the left, the exponent will be positive.

sergeinik [125]3 years ago
8 0
10^17 is the answer

Hope this helped
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3 0
3 years ago
If f(x) = |(x2 − 9)(x2 + 1)|, how many numbers in the interval [−1, 1] satisfy the conclusion of the mean value theorem?
ivann1987 [24]

Answer:

only one number c=0 in the interval [-1,1]

Step-by-step explanation:

Given : Function f(x) = |(x^2-9)(x^2 + 1)|   in the interval [-1,1]

To find : How many numbers in the interval [−1, 1] satisfy the conclusion of the mean value theorem.

Mean value theorem : If f is a continuous function on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a point 'c' in (a,b) such that f'(c)=\frac{f(b)-f(a)}{b-a}

Solution : f(x) is a function that satisfies all of the following :

1) f(x)  is continuous on the closed interval [-1,1]  

\lim_{x\to a} f(x)=f(a)

2) f(x) is differentiable on the open interval  (-1,1)

Then there is a number  c such that  f'(c)=\frac{f(b)-f(a)}{b-a}

f(a)=f(-1) = |(-1^2-9)(-1^2 + 1)|=|(-8)(2)|=16

f(b)=f(1) = |(1^2-9)(1^2 + 1)|=|(8)(2)|=16

f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{16-16}{2}=0

f'(c)=0 ........[1]

Now, we find f'(x)

f(x) = |(x^2-9)(x^2 + 1)|

f(x) =x^4-8x^2-9

Differentiating w.r.t  x

f'(x) =4x^3-16x

In place of x we put x=c

f'(c) =4c^3-16c

f'(c) =4c^3-16c=0  (by [1], f'(c)=0)

4c(c^2-4)=0

4c=0,c^2-4=0

either c=0 or  c^2-4=0\rightarrow c=\pm2

we cannot take c=\pm2  because they don't lie in the interval [-1,1]

Therefore, there is only one number c=0 which lie in interval [-1,1] and satisfying the conclusion of the mean value theorem.



3 0
3 years ago
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