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Kazeer [188]
4 years ago
10

Please help me and show your work!!

Mathematics
1 answer:
11Alexandr11 [23.1K]4 years ago
4 0
\bf \cfrac{\sqrt{250x^{16}}}{\sqrt{2x}}\qquad 
\begin{cases}
250=2\cdot 5\cdot 5\cdot 5\\
\qquad 2\cdot 5\cdot 5^2\\
x^{16}=x^{8\cdot 2}\\
\qquad (x^8)^2
\end{cases}\implies \cfrac{\sqrt{2\cdot 5\cdot 5^2\cdot (x^8)^2}}{\sqrt{2\cdot  x}}

\bf \cfrac{5x^8\sqrt{2\cdot 5}}{\sqrt{2}\cdot \sqrt{x}}\implies \cfrac{5x^8\underline{\sqrt{2}}\cdot \sqrt{5}}{\underline{\sqrt{2}}\cdot \sqrt{x}}\implies \cfrac{5x^8\sqrt{5}}{\sqrt{x}}\impliedby 
\begin{array}{llll}
\textit{and rationalizing the}\\
\textit{denominator}
\end{array}

\bf \cfrac{5x^8\sqrt{5}}{\sqrt{x}}\cdot \cfrac{\sqrt{x}}{\sqrt{x}}\implies \cfrac{5x^8\sqrt{5}\cdot \sqrt{x}}{(\sqrt{x})^2}\implies \cfrac{5x^8\sqrt{5x}}{x}\implies 5x^8x^{-1}\sqrt{5x}
\\\\\\
5x^{8-1}\sqrt{5x}\implies 5x^7\sqrt{5x}
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Answer: 150mm

Step-by-step explanation:

Take the two adjecent sides of the triangle to be x and y and X is the angle opposite the side, x.

From the trigonometric identities for a right-angled triangle, it is known that

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Squaring both sides hence,

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Recall that sin^2(X) + cos^2(X) = 1 and 2sin X.cos X = sin 2X.

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2X = 56.14° or (180 - 56.14)° = 56.14° or 123.86° for X existing within a triangle.

X = 56.14/2 or 123.86/2

X = 28.07° or 61.93°

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x = 170 sin (28.07) = 80mm

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Using the value of 61.93° as X would give the same values but interchanged for x and y.

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