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sashaice [31]
3 years ago
11

Value of this problem?

Mathematics
1 answer:
Rudik [331]3 years ago
5 0
\frac{4^3\times4^{-1}\times5^{-2}}{4^4\times5^{-3}\times5^0}=\frac{4^{3-1}\times5^{-2}}{4^4\times5^{-3+0}}\\=\frac{4^{2}\times5^{-2}}{4^4\times5^{-3}}=4^{2-4}\times5^{-2-(-3)}\\=4^{-2}\times5^{1}=\frac{5}{4^2}\\=\frac{5}{16}
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If we substitute in the known values 16^2 + y^2 = 20^2 and solve for y we get that y = sqrt(20^2 - 16^2), this then simplifies to y = 12

Finding x is much more annoying, the easiest way I can immediately see is to find the upper angles by doing sin(16/20) and then 90 - sin(16/20) since the complementary angle is the one you want. I don't have a calculator or a trig table with me right now but I will tell you that x will be equal to 12 ÷ the inverse cosine of the angle (90degrees - sin(16/20)).

I am pretty sure the answer is D though because we know for sure y = 12 and x has to be greater than y because the hypotenuse must be larger than both legs. It could be E but you won't know unless you do the math for x. So it is either D or E but I would be surprised if a Professor made you do all of the work just to say it doesn't work...
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