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laiz [17]
3 years ago
13

I need help with a math problem

Mathematics
1 answer:
Neko [114]3 years ago
7 0
Which one I can help with anyone

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10 1/2 I am pretty sure its right
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Solve the inequality and graph its solution: 2t+1>9or -18>5t-3​
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6 0
2 years ago
A student inscribes a triangle within a semicircle. What is the measure of ∠XYZ?
Dmitry [639]

Answer:

The correct answer is option A.  90°

Step-by-step explanation:

<u>Points to remember</u>

Measure of angles on the semicircle is 90°

From the figure we can see a semicircle and a triangle inscribes  on it.

<u>To find the measure of <XYZ</u>

Here <XYZ is on the semicircle that is angle XYZ is a right angle.

Therefore m<XYZ = 90°

The correct answer is option A. 90°

3 0
3 years ago
Read 2 more answers
What is the Greatest common factor of 18x^2 and 36x
adoni [48]

For this case we have that by definition, the Greatest Common Factor or GFC of two or more numbers, is the largest number that divides them without leaving residue.

So, we have to:

We look for the factors of 18 and 36:

18: 1,2,3,6,9,18

36: 1, 2,3,4, 6,9,18 ...

It is observed that the GFC of both numbers is 18.

Then, the GFC of 18x ^ 2 and 36x is:

18x

Answer:

18x

6 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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