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Mathematics

1 answer:

Mars2501 [29]3 years ago

7 0

Answer:

Scale Factor = 2

Step-by-step explanation:

ABC is your original triangle and A'B'C' is the dilation. You know these are similar triangles so only one side has to be used to find the scale factor. AB has a length of 3 and A'B' has a length of 6 so you know the side lengths of A'B'C' are going to be 2 times the size of triangle ABC side lengths. This gives you a scale factor of 2.

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Simplify expression cos^2(pi/2-x) / √1-sin^2(x) =

andrey2020 [161]

Answer:

$\frac{cos^2(\frac{\pi}{2}-x)}{\sqrt{1-sin^2(x)}}=\frac{sin^2(x)}{|cos(x)|}$

Step-by-step explanation:

To simplify this expression you must use the following trigonometric identities

$cos(\frac{\pi}{2}-x) = sinx$ I

$1-sin (x) ^ 2 = cos ^ 2(x)$ II

Remember that

$\sqrt{f(x)^2} =f(x)$

Only if $f(x)> 0$ for all x

If f(x) is not greater than 0 for all x then

$\sqrt{f(x)^2} =|f(x)|$

Now we have the expression:

$\frac{cos^2(\frac{\pi}{2}-x)}{\sqrt{1-sin^2(x)}}$

then using the trigonometric identities I and II we have to:

$\frac{cos^2(\frac{\pi}{2}-x)}{\sqrt{1-sin^2(x)}}=\frac{sin^2(x)}{\sqrt{1-sin^2(x)}}\\\\\\\frac{sin^2(x)}{\sqrt{1-sin^2(x)}}= \frac{sin^2(x)}{\sqrt{cos^2(x)}}$