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likoan [24]
2 years ago
11

Ok so x=40+50+90? Am I right

Mathematics
2 answers:
posledela2 years ago
5 0

hi <3

no, that wouldn't be correct.

angles on a straight line sum to 180 degrees so you have to do 180 - 90 (the right angle) to get your answer of 90 degrees

hope this helps :)

Oksi-84 [34.3K]2 years ago
3 0
X = 90° since it’s a right angle.
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Read 2 more answers
If <img src="https://tex.z-dn.net/?f=%5Crm%20%5C%3A%20x%20%3D%20log_%7Ba%7D%28bc%29" id="TexFormula1" title="\rm \: x = log_{a}(
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Use the change-of-basis identity,

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

to write

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)

to write

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)}

xyz = 2 + \log_b(ac) + \log_c(ab) + \log_a(bc)

\implies \boxed{xyz = x + y + z + 2}

(C)

6 0
2 years ago
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