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yawa3891 [41]
3 years ago
14

Question is attached​

Mathematics
1 answer:
77julia77 [94]3 years ago
7 0

Answer:

Please check explanations

Step-by-step explanation:

Mathematically;

csc 2 theta = 1 / sin 2 theta

Sin 2 theta = sin (theta + theta)

= 2sin theta cos theta

So therefore;

csc 2 theta = 1/2(sin theta cos theta)

= 1/2 * 1/sin theta * 1/ cos theta

1/sin theta = csc theta

1/cos theta = sec theta

Thus;

csc 2 theta = 1/2 * csc theta * sec theta

csc 2 theta = 1/2 sec theta csc theta

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Find the following:
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Step-by-step explanation:

Limit refers to the value that the function approaches as the input approaches some value.

We say \displaystyle \lim_{x\rightarrow a}f(x)=L, if f(x) approaches L as x approaches 'a'.

(a)

\displaystyle \lim_{x\rightarrow 5}\left ( \frac{f(x)-8}{x-5} \right )=4\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-\displaystyle \lim_{x\rightarrow 5}8}{\displaystyle \lim_{x\rightarrow 5}x-\displaystyle \lim_{x\rightarrow 5}5}=4\\

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\displaystyle \lim_{x\rightarrow 5}f(x)-8=20-20=0\\\displaystyle \lim_{x\rightarrow 5}f(x)=8

(b)

\displaystyle \lim_{x\rightarrow 5}\left ( \frac{f(x)-8}{x-5} \right )=7\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-\displaystyle \lim_{x\rightarrow 5}8}{\displaystyle \lim_{x\rightarrow 5}x-\displaystyle \lim_{x\rightarrow 5}5}=7\\\frac{\displaystyle \lim_{x\rightarrow 5}f(x)-8}{\displaystyle \lim_{x\rightarrow 5}x-5}=7\\

\displaystyle \lim_{x\rightarrow 5}f(x)-8=7\left ( \displaystyle \lim_{x\rightarrow 5}x-5 \right )\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=7\displaystyle \lim_{x\rightarrow 5}x-7(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=7(5)-7(5)\\\displaystyle \lim_{x\rightarrow 5}f(x)-8=35-35=0\\\displaystyle \lim_{x\rightarrow 5}f(x)=8

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4 years ago
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Answer:

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Consider the provided information.

The following property can be used to rewrite each radical as an exponent.

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x^{\frac{m}{n}}=(\sqrt[n]{x})^m

For example:

(27)^{\frac{2}{3}}=(\sqrt[3]{27})^2

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I think this question is technically asking
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