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Iteru [2.4K]
3 years ago
12

Can someone help me on this

Mathematics
1 answer:
barxatty [35]3 years ago
5 0

A) as the exponent decreases by one, the number is divided by 5

B) as the exponent decreases by one, the number is divided by 4

C) as the exponent decreases by one, the number is divided by 3

D) 4⁰ = 4 ÷ 4 = 1

    4⁻¹ = 1 ÷ 4 = \frac{1}{4}

    4⁻² = \frac{1}{4} ÷ 4 = \frac{1}{16}

E) 3⁰ = 3 ÷ 3 = 1

    3⁻¹ = 1 ÷ 3 = \frac{1}{3}

    3⁻² = \frac{1}{3} ÷ 3 = \frac{1}{9}


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3 years ago
F(x) = 3x + x3
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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

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19=p-12 does that equal 29,30,31?​
Likurg_2 [28]
19+12=p-12+12
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