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8_murik_8 [283]
3 years ago
13

Which of the following compositions produces the same image as R 0,-180?

Mathematics
1 answer:
Damm [24]3 years ago
5 0
The transformation RO - 180° means rotation of - 180°.

That is half-turn clockwise.

When you apply that rotation, the coordinates of a point (x,y) transforms into coordinates (-x,-y).

So, any coomposition that is equivalent to half-turn will produce the same image as RO - 180°.

These are some examples:

1) RO 90° and RO 90° (because 90° + 90° = 180° and that is equivalent to  -180°).

2) RO - 90° and RO - 90° (because - 90° - 90° = - 180°).

3) RO 90° and RO - 270° (becasue 90° - 270° = - 180°).

At the end there are infinite combinations equivalent to Ro - 180°, so, given that you did not include the set of choices, I have provided the explanations and examples which teachs you how to test any combination of rotations.
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The sum of a number and twenty-six, divided by twelve is equal to seven! <br><br> Please!!
Travka [436]

Answer:

58

Step-by-step explanation:

1. let x be the unknown number

2. make an expression

x + 26/12 = 7/1

3. cross multiply

x +26 × 1 = x + 26

12 × 7  = 84

 x + 26 = 84

4. move variables on one side and numbers on other side of the equal sign

x + 26 = 84

   - 26    -26

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4 0
2 years ago
-12+8d &lt;52 <br> Please answer
Radda [10]

Answer:

d < 8

Step-by-step explanation:

Given

- 12 + 8d < 52 ( add 12 to both sides )

8d < 64 ( divide both sides by 8 )

d < 8

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3 years ago
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Simplify the expression:<br> 3(5s + 2) =
Licemer1 [7]

Answer:

The answer is =15s+6

Step-by-step explanation:

3(5s + 2)

=15s+6

5 0
3 years ago
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Please someone help me...​
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

<h3>-  \cot( - 4A)</h3>

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

That's

<h3>- ( -  \cot(4A) )</h3>

<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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Answer:

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Step-by-step explanation:

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