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4 years ago

(x^-7)^6 without using negative exponents

1 answer:

4 0

Answer: To eliminate negative exponents, you move the whole factor to the denominator, if possible. Repeated exponentiation multiplies exponents.

Step-by-step explanation:

$(x^{-7})^6 = \left({\frac{1}{x^7}}\right)^6 = \frac{1}{x^7} \frac{1}{x^7} \frac{1}{x^7}\frac{1}{x^7} \frac{1}{x^7} \frac{1}{x^7}=\frac{1}{x^{42}}$

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Please help and thank you.

Dimas [21]

Answer:

B the parent function is periodic

Step-by-step explanation:

We can see from the graph that

A it is a polynomial

C that is has 3 real root ( where it crosses the x axis)

D that is contains the point (0,-20)

What is not true is that it is periodic. That would mean it repeats after a certain time.

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4 years ago

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What is 16 times 4? help

Hatshy [7]

Answer:

Step-by-step explanation:

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3 years ago

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Pls help! URGENT ONLY DO QUESTION 21 PLS

Alekssandra [29.7K]

Answer:

y = x - 8

Step-by-step explanation:

There's no magic method to doing these, and problem 21 makes it harder to see a pattern by using inputs that are not consecutive integers (like problem 20 does)!

Looking at inputs 20 and 21, the outputs are 8 less than the inputs. Check the other inputs and notice that the same pattern holds up.

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3 years ago

What is f(8)=0.8(8)^2

Montano1993 [528]

Answer:

f = 5.12

Step-by-step explanation:

f(8)=0.8(8)^2

f(8)=6.4^2

f(8)=40.96

f(8)/8 = 40.96/8

f = 5.12

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3 years ago

F(x) = 3x + x3

____ [38]

Answer:

Please check the explanation.

Step-by-step explanation:

Given

f(x) = 3x + x³

Taking differentiate

$\frac{d}{dx}\left(3x+x^3\right)$

$\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'$

$=\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(x^3\right)$

solving

$\frac{d}{dx}\left(3x\right)$

$\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'$

$=3\frac{d}{dx}\left(x\right)$

$\mathrm{Apply\:the\:common\:derivative}:\quad \frac{d}{dx}\left(x\right)=1$

$=3\cdot \:1$

$=3$

now solving

$\frac{d}{dx}\left(x^3\right)$

$\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}$

$=3x^{3-1}$

$=3x^2$

Thus, the expression becomes

$\frac{d}{dx}\left(3x+x^3\right)=\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(x^3\right)$

$=3+3x^2$

Thus,

f'(x) = 3 + 3x²

Given that f'(x) = 15

substituting the value f'(x) = 15 in f'(x) = 3 + 3x²

f'(x) = 3 + 3x²

15 = 3 + 3x²

switch sides

3 + 3x² = 15

3x² = 15-3

3x² = 12

Divide both sides by 3

x² = 4

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{4},\:x=-\sqrt{4}$

$x=2,\:x=-2$

Thus, the value of x will be:

$x=2,\:x=-2$

5 0

3 years ago