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Alex73 [517]
2 years ago
9

Consider the enlargement of the complex figure. 2 complex figures. a corresponds to 18 centimeters, b corresponds to 24 centimet

ers, and a side with length 14 meters corresponds to a side with length 42 feet. what can be concluded about the scale factor and missing measures? check all that apply. the scale factor is 3. the scale factor is one-third. add the scale factor to 18 and 24 to find a and b’s lengths. subtract the scale factor from 18 and 24 to find a and b’s lengths. multiply 18 and the scale factor to find a’s length. divide 24 by the scale factor to find b’s length.
Mathematics
2 answers:
maw [93]2 years ago
8 0

Answer:

The scale factor is 3 and Divide by 24 to find b's length.

Step-by-step explanation:

I think this is the answer

(Im just a bored high schooler searching this up) :P

MatroZZZ [7]2 years ago
4 0

Answer:

The scale factor is 3 and Divide by 24 to find b's length.

Step-by-step explanation:

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Help me in this algebra1 problem
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laiz [17]

Step-by-step explanation:

First factor out the negative sign from the expression and reorder the terms

That's

\frac{1}{ - (( \tan(2A) -  \tan(6A)  )}  -  \frac{1}{ \cot(6A)  -  \cot(2A) }

<u>Using trigonometric </u><u>identities</u>

That's

<h3>\cot(x)  =  \frac{1}{ \tan(x) }</h3>

<u>Rewrite the expression</u>

That's

\frac{1}{ - (( \tan(2A) -  \tan(6A)  )} -    \frac{1}{ \frac{1}{ \tan(6A) } }  -  \frac{1}{ \frac{1}{ \tan(2A) } }

We have

<h3>-  \frac{1}{  \tan(2A) -  \tan(6A)  } -   \frac{1}{ \frac{ \tan(2A) -  \tan(6A)  }{ \tan(6A) \tan(2A)  } }</h3>

<u>Rewrite the second fraction</u>

That's

<h3>-  \frac{1}{  \tan(2A) -  \tan(6A)  } -   \frac{ \tan(6A)  \tan(2A) }{ \tan(2A) -  \tan(6A)  }</h3>

Since they have the same denominator we can write the fraction as

-  \frac{1 +  \tan(6A) \tan(2A)  }{ \tan(2A) -  \tan(6A)  }

Using the identity

<h3>\frac{x}{y}  =  \frac{1}{ \frac{y}{x} }</h3>

<u>Rewrite the expression</u>

We have

<h3>-  \frac{1}{ \frac{ \tan(2A)  -  \tan(6A) }{1 +  \tan(6A) \tan(2A)  } }</h3>

<u>Using the trigonometric identity</u>

<h3>\frac{ \tan(x) -  \tan(y)  }{1 +  \tan(x)  \tan(y) }  =  \tan(x - y)</h3>

<u>Rewrite the expression</u>

That's

<h3>- \frac{1}{ \tan(2A -6A) }</h3>

Which is

<h3>-  \frac{1}{ \tan( - 4A) }</h3>

<u>Using the trigonometric identity</u>

<h3>\frac{1}{ \tan(x) }  =  \cot(x)</h3>

Rewrite the expression

That's

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<u>Simplify the expression using symmetry of trigonometric functions</u>

That's

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<u>Remove the parenthesis </u>

We have the final answer as

<h2>\cot(4A)</h2>

As proven

Hope this helps you

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