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LiRa [457]
3 years ago
14

I need help plz help me

Mathematics
2 answers:
irinina [24]3 years ago
7 0
The answer is A
i think so but i might be wrong 
sorry if i am.
DanielleElmas [232]3 years ago
4 0
The answer is 35 cats are siamese
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PLEASE HELP <br> WILL GIVE BRAINLIEST AND 5.0 RATING
jasenka [17]

Answer:

<h2>(-4,-2)</h2>

Step-by-step explanation:

(-4,-5)

move up 3 units

change in y-axis

-5+3=-2

(-4,-2)

4 0
3 years ago
Jason has applied for two summer jobs, one as a lifeguard and the other as a camp counselor. The lifeguard job is three days a w
Anastasy [175]

Answer:

Lifegard

Step-by-step explanation:

3 0
3 years ago
Find the image of the given point
saul85 [17]

Answer:

Step-by-step explanation:

p(-3,-3)

here x is -3 and y is also -3 so

p(-3,-3)=p'(x-8,y-3

=p'(-3-8,-3-3)

=p'(-11,0)

4 0
3 years ago
2 tan 30°<br>II<br>1 + tan- 300​
shusha [124]

Question:

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}

Answer:

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}= sin(60^{\circ})

Step-by-step explanation:

Given

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}

Required

Simplify

In trigonometry:

tan(30^{\circ}) = \frac{1}{\sqrt{3}}

So, the expression becomes:

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}

Simplify the denominator

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}

Express the fraction as:

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}= \frac{2}{\sqrt 3} / \frac{4}{3}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{2}{\sqrt 3} * \frac{3}{4}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{1}{\sqrt 3} * \frac{3}{2}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{3}{2\sqrt 3}

Rationalize

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{3\sqrt{3}}{2* 3}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{\sqrt{3}}{2}

In trigonometry:

sin(60^{\circ}) =  \frac{\sqrt{3}}{2}

Hence:

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}= sin(60^{\circ})

3 0
3 years ago
Please help me solve for13 and 14
deff fn [24]

Answer:

13. A

14. True

Step-by-step explanation:

The months cancel out because they are the same unit in the right positions, and all you have to do for the second one is multiply

4 0
3 years ago
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