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Nesterboy [21]
4 years ago
13

What is the value of y^-5 and x^-3 for x = 2 and y = -4?

Mathematics
1 answer:
Ilya [14]4 years ago
7 0
a^{-n}=\dfrac{1}{a^n}\\\\x^{-3}=\dfrac{1}{x^3};\ y^{-5}=\dfrac{1}{y^5}\\\\\dfrac{y^{-5}}{x^{-3}}=\dfrac{1}{y^5}:\dfrac{1}{x^3}=\dfrac{1}{y^5}\cdot\dfrac{x^3}{1}=\dfrac{x^3}{y^5}\\\\x=2;\ y=-4=-2^2\\\\\dfrac{x^3}{y^5}=\dfrac{2^3}{(-2^2)^5}=\dfrac{2^3}{-2^{2\cdot5}}=\dfrac{2^3}{-2^{10}}=-\dfrac{1}{2^7}=\boxed{-\dfrac{1}{128}}\to\fbox{C.}
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Let f(x)= 100 / −10+ e^ −0.1x . What is f(−6) ?
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The value of f(-6) is -12.2

Explanation:

Given that the function f(x)=\frac{100}{-10+e^{-0.1\left(x\right)}}

We need to determine the value of f(-6)

The value of f(-6) can be determined by substituting the value for x=-6 in the function and simplify the function.

Hence, let us substitute x=-6 in the function, we get,

f(-6)=\frac{100}{-10+e^{-0.1\left(-6\right)}}

Let us apply the rule -\left(-a\right)=a , we get,

f(-6)=\frac{100}{-10+e^{0.1\left(6\right)}}

Multiplying the numbers, we get,

f(-6)=\frac{100}{-10+e^{0.6}}

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Subtracting the denominator, we have,

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Rounding off to the nearest tenth, we have,

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