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kenny6666 [7]
3 years ago
13

PLZ HELP ME!!!!!!!!!!!

Mathematics
1 answer:
Fiesta28 [93]3 years ago
5 0
WAT SCHOOL U GO TO AND GRADE 

POLL 1. D
POLL 2. D
POLL 3. A
POLL 4. A
POLL 5. C
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HELLO CAN ANYONE ANSWER THIS PLEASE I WILL MARK YOU THE BRAINLIEST !!!
Arte-miy333 [17]
25 degrees is the answer. CBE is supplementary to ABC. That means those two angles add up to 180 degrees.
5 0
2 years ago
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REPLY ASAP PLEASE!!!
d1i1m1o1n [39]

Answer:

The answer is c.

Step-by-step explanation:

4 0
2 years ago
What is the value of m? (m is a whole number)
yuradex [85]

m is equal to 3. Need I say more?

6 0
2 years ago
(((QUICK QUESTION-WILL GIVE BRAINLYEST)))
svetoff [14.1K]

Given:

Point (7,12) is rotated 1260° counterclockwise about the origin.

To find:

The  x-coordinate of the point after this rotation.

Solution:

If a point is rotated 360 degrees then its coordinates remains unchanged.

If a point is rotated 180 counterclockwise about the origin degrees, then

(x,y)\to (-x,-y)

We know that,

1260^\circ=1080^\circ+180^\circ

1260^\circ=3\times 360^\circ+180^\circ

After 3\times 360^\circ rotation the coordinates of points remains same, i.e., (7,12). So, after that (7,12) is rotated 180° counterclockwise about the origin.

(7,12)\to (-7,-12)

The point (7,12) becomes (-7,-12) after rotation of  1260° counterclockwise about the origin.

Therefore, the x-coordinate of the required point is -7.

3 0
3 years ago
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

6 0
3 years ago
Read 2 more answers
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