Answer:
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Step-by-step explanation:




Answer:
8y^2 (y+2)
Step-by-step explanation:
8y^3 + 16y^2
both terms contain an 8 and a y^2 so we can factor them out
8y^2 (y+2)
Answer:
Step-by-step explanation:

<h2 /><h2>
<u>Consider</u></h2>

<h2>
<u>W</u><u>e</u><u> </u><u>K</u><u>n</u><u>o</u><u>w</u><u>,</u></h2>




So, on substituting all these values, we get




<h2>Hence,</h2>

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<h2>ADDITIONAL INFORMATION :-</h2>
Sign of Trigonometric ratios in Quadrants
- sin (90°-θ) = cos θ
- cos (90°-θ) = sin θ
- tan (90°-θ) = cot θ
- csc (90°-θ) = sec θ
- sec (90°-θ) = csc θ
- cot (90°-θ) = tan θ
- sin (90°+θ) = cos θ
- cos (90°+θ) = -sin θ
- tan (90°+θ) = -cot θ
- csc (90°+θ) = sec θ
- sec (90°+θ) = -csc θ
- cot (90°+θ) = -tan θ
- sin (180°-θ) = sin θ
- cos (180°-θ) = -cos θ
- tan (180°-θ) = -tan θ
- csc (180°-θ) = csc θ
- sec (180°-θ) = -sec θ
- cot (180°-θ) = -cot θ
- sin (180°+θ) = -sin θ
- cos (180°+θ) = -cos θ
- tan (180°+θ) = tan θ
- csc (180°+θ) = -csc θ
- sec (180°+θ) = -sec θ
- cot (180°+θ) = cot θ
- sin (270°-θ) = -cos θ
- cos (270°-θ) = -sin θ
- tan (270°-θ) = cot θ
- csc (270°-θ) = -sec θ
- sec (270°-θ) = -csc θ
- cot (270°-θ) = tan θ
- sin (270°+θ) = -cos θ
- cos (270°+θ) = sin θ
- tan (270°+θ) = -cot θ
- csc (270°+θ) = -sec θ
- sec (270°+θ) = cos θ
- cot (270°+θ) = -tan θ
Answer:
The scale factor is 1/2
Step-by-step explanation:
Given
and 
Required
The scale factor from
to 
Take a corresponding side in both triangles.
We have:

The scale factor from
to
is:



Answer:
theres no picture.
Step-by-step explanation: