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gladu [14]
3 years ago
11

Perimeter is a square clock is 16 6/9 inches. Perimeter of a rectangular clock is 16 16/18 inches. Write an inequality comparing

the two perimeters.
Mathematics
1 answer:
worty [1.4K]3 years ago
4 0
<span>16 6/9 inches < 16 16/18 inches or Perimeter of square clock < Perimeter of rectangular clock First we would put convert the perimeter fractions into equivalent terms. So for the square clock, 16 6/9 inches becomes 16 12/18 inches (multiplying the fraction by 2/2). Now it is obvious that that the square clock at 16 12/18 inches has a smaller perimeter than the rectangular clock with a perimeter of 16 16/18 inches.</span>
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Trigonometric Formula's:

\boxed{\sf \ \sf \sin^2 \theta + \cos^2 \theta = 1}

\boxed{ \sf tan\theta = \frac{sin\theta}{cos\theta} }

Given to verify the following:

\bf (cos^2a) (2 + tan^2 a) = 2 - sin^2 a

\texttt{\underline{rewrite the equation}:}

\rightarrow \sf (cos^2a) (2 + \dfrac{sin^2 a}{cos^2 a} )

\texttt{\underline{apply distributive method}:}

\rightarrow \sf 2 (cos^2a) + (\dfrac{sin^2 a}{cos^2 a} ) (cos^2a)

\texttt{\underline{simplify the following}:}

\rightarrow \sf 2cos^2 a  + sin^2 a

\texttt{\underline{rewrite the equation}:}

\rightarrow \sf 2(1 - sin^2a )  + sin^2 a

\texttt{\underline{distribute inside the parenthesis}:}

\rightarrow \sf 2 - 2sin^2a   + sin^2 a

\texttt{\underline{simplify the following}} :

\rightarrow \sf 2 - sin^2a

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Consider the following figure:
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Answer:

x=90

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Step-by-step explanation:

Since we know a line is 180 degrees, and we know that angle Q is 90, x must be a 90-degree angle as well.

With our given information we can add up the two given angles in the triangle which are 90 and 58

90+58=148

To find what R is, we must subtract 148 from 180 because all the angles in a triangle sum up to 180.

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Now that we know that R is 32, because we know that a line is 180 degrees, we can subtract 32 from 180 to get our final answer for y as 148.

180-32=148

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