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Vitek1552 [10]
2 years ago
5

How would you write 12-^3 using a positive exponent? 12^3 12^0 12^3/1 1/12^3

Mathematics
1 answer:
Brums [2.3K]2 years ago
6 0

Answer:

\frac{1}{ {12}^{3} }

Step-by-step explanation:

{12}^{ - 3}  =  \frac{1}{ {12}^{3} }  \\

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3 years ago
Find the linear approximation of the function g(x) = 3 root 1 + x at a = 0. g(x). Use it to approximate the numbers 3 root 0.95
Virty [35]

Answer:

L(x)=1+\dfrac{1}{3}x

\sqrt[3]{0.95} \approx 0.9833

\sqrt[3]{1.1} \approx 1.0333

Step-by-step explanation:

Given the function: g(x)=\sqrt[3]{1+x}

We are to determine the linear approximation of the function g(x) at a = 0.

Linear Approximating Polynomial,L(x)=f(a)+f'(a)(x-a)

a=0

g(0)=\sqrt[3]{1+0}=1

g'(x)=\frac{1}{3}(1+x)^{-2/3} \\g'(0)=\frac{1}{3}(1+0)^{-2/3}=\frac{1}{3}

Therefore:

L(x)=1+\frac{1}{3}(x-0)\\\\$The linear approximating polynomial of g(x) is:$\\\\L(x)=1+\dfrac{1}{3}x

(b)\sqrt[3]{0.95}= \sqrt[3]{1-0.05}

When x = - 0.05

L(-0.05)=1+\dfrac{1}{3}(-0.05)=0.9833

\sqrt[3]{0.95} \approx 0.9833

(c)

(b)\sqrt[3]{1.1}= \sqrt[3]{1+0.1}

When x = 0.1

L(1.1)=1+\dfrac{1}{3}(0.1)=1.0333

\sqrt[3]{1.1} \approx 1.0333

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3 years ago
The slope of a line is 2. The y-intercept of the line is –6. Which statements accurately describe how to graph the function
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F(x) = 2x - 6
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2 is the slope
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