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Ber [7]
3 years ago
9

Based on the similar triangles shown below, Theodore claims that ∆TUV is transformed to ∆WXY with a scale factor of 32. Is Theod

ore correct? A Yes, the triangles are similar with a scale factor of 32. B No, the triangles are similar with a scale factor of 21. C No, the triangles are similar with a scale factor of 23. D No, the triangles are similar with a scale factor of 43.

Mathematics
1 answer:
Anarel [89]3 years ago
8 0

*Correct Question:

Based on the similar triangles shown below, Theodore claims that ∆TUV is transformed to ∆WXY with a scale factor of 3/2. Is Theodore correct?

A. Yes, the triangles are similar with a scale factor of 3/2.

B. No, the triangles are similar with a scale factor of 2/1.

C. No, the triangles are similar with a scale factor of 2/3.

D. No, the triangles are similar with a scale factor of 4/3.

Answer:

C. No, the triangles are similar with a scale factor of 2/3.

Step-by-step explanation:

∆TUV is the original triangle. After transformation, the size reduced to give us ∆WXY. This means ∆TUV was reduced by a scale factor to give ∆WXY. The scale factor should be a fraction, suggesting, the original size of the ∆ was reduced upon transformation.

Thus, the ratio of their corresponding sides = the scale factor.

This is: \frac{8}{12} = \frac{16}{24} = \frac{12}{18} = \frac{2}{3}

If you multiply the side length of ∆TUV by ⅔, you'd get side length of ∆WXY.

So, Theodore is wrong.

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Answer:

p = \dfrac{I}{rt}

Step-by-step explanation:

I = prt

Switch sides.

prt = I

We are solving for p. We want p alone on the left side. p is being multiplied by r and t, so we divide both sides by r and t.

\dfrac{prt}{rt} = \dfrac{I}{rt}

p = \dfrac{I}{rt}

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3 years ago
Brad buys a jacket that is on sale for 30% off the original price. The expression p - 0.3p can be used to find the sale price of
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3 years ago
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Use the change-of-basis identity,

\log_x(y) = \dfrac{\ln(y)}{\ln(x)}

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xyz = \log_a(bc) \log_b(ac) \log_c(ab) = \dfrac{\ln(bc) \ln(ac) \ln(ab)}{\ln(a) \ln(b) \ln(c)}

Use the product-to-sum identity,

\log_x(yz) = \log_x(y) + \log_x(z)

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xyz = \dfrac{(\ln(b) + \ln(c)) (\ln(a) + \ln(c)) (\ln(a) + \ln(b))}{\ln(a) \ln(b) \ln(c)}

Redistribute the factors on the left side as

xyz = \dfrac{\ln(b) + \ln(c)}{\ln(b)} \times \dfrac{\ln(a) + \ln(c)}{\ln(c)} \times \dfrac{\ln(a) + \ln(b)}{\ln(a)}

and simplify to

xyz = \left(1 + \dfrac{\ln(c)}{\ln(b)}\right) \left(1 + \dfrac{\ln(a)}{\ln(c)}\right) \left(1 + \dfrac{\ln(b)}{\ln(a)}\right)

Now expand the right side:

xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} \\\\ ~~~~~~~~~~~~+ \dfrac{\ln(c)\ln(a)}{\ln(b)\ln(c)} + \dfrac{\ln(c)\ln(b)}{\ln(b)\ln(a)} + \dfrac{\ln(a)\ln(b)}{\ln(c)\ln(a)} \\\\ ~~~~~~~~~~~~ + \dfrac{\ln(c)\ln(a)\ln(b)}{\ln(b)\ln(c)\ln(a)}

Simplify and rewrite using the logarithm properties mentioned earlier.

xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} + \dfrac{\ln(a)}{\ln(b)} + \dfrac{\ln(c)}{\ln(a)} + \dfrac{\ln(b)}{\ln(c)} + 1

xyz = 2 + \dfrac{\ln(c)+\ln(a)}{\ln(b)} + \dfrac{\ln(a)+\ln(b)}{\ln(c)} + \dfrac{\ln(b)+\ln(c)}{\ln(a)}

xyz = 2 + \dfrac{\ln(ac)}{\ln(b)} + \dfrac{\ln(ab)}{\ln(c)} + \dfrac{\ln(bc)}{\ln(a)}

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6 0
2 years ago
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nexus9112 [7]

Answer:

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At 99%

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Thank you!

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