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Karo-lina-s [1.5K]

3 years ago

8

Help. I got it wrong and have no clue how to get to the right answer!

1 answer:

natima [27]3 years ago

7 0

$\bf 2x+\sqrt{4x^2}+\sqrt[3]{8x^3}\qquad 
\begin{cases}
4\to 2^2\\
8\to 2^3
\end{cases}\qquad thus
\\\\\\
2x+\sqrt{2^2x^2}+\sqrt[3]{2^3x^3}\implies 2x+\sqrt{(2x)^2}+\sqrt[3]{(2x)^3}
\\\\\\
2x+2x+2x=\boxed{6x}$

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The pair of points is on the graph of an inverse variation. find the missing value. (5,4) and (x,7)

d1i1m1o1n [39]

One form of the relation of inverse variation is y = k/x.

first step: Find the value of k: Taking the values x=5 and y = 4, 4=k/5, or k=20. Then, y = 20/x.

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x = 20/7 (answer)

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Mandarinka [93]

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<h2>x = 3.1</h2>

Step-by-step explanation:

<h3> ${10}^{x} = 1200$ </h3>

<u>To solve first take logarithm to both sides</u>

That's

<h3> $log_{10}(10) ^{x} = log_{10}(1200)$ </h3><h3 /><h3> $log_{10}(10)^{x} = x log_{10}(10)$ </h3>

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So we have

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That's

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So we have

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<u>Using the rules of logarithms</u>

That's

<h3> $log_{a}(x \times y) = log_{a}(x) + log_{a}(y)$ </h3>

Rewrite the expression

That's

<h3> $x = log_{10}(100) + log_{10}(12)$ </h3><h3 /><h3> $x = log_{10}(10)^{2} + log_{10}(12)$ </h3><h3> $x = 2 log_{10}(10) + log_{10}(12)$ </h3><h3> $log_{10}(10) = 1$ </h3><h3 /><h3> $x = 2 + log_{10}(12)$ </h3>

x = 3.079

So we have the final answer as

<h3>x = 3.1 to one decimal place</h3>

Hope this helps you

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