3 years ago

Question 10 of 11 Find the measure of the exterior 1 60 252 O A. 95 OB. 85 OC. 35 O D. 120

Mathematics

1 answer:

kramer3 years ago

3 0

Answer:

would help if u had a picture im assuming those numbers are a with shape

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If x-1/x=5 find x^3+1/x^3=?

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$\dfrac{x-1}{x}=5\ \ \ \ /x\neq0\\\\\dfrac{x-1}{x}=\dfrac{5}{1}\ \ \ \ |\text{cross multiply}\\\\5x=x-1\ \ \ \ |-x\\\\5x=1\ \ \ \ |:5\\\\x=0.2\\\\\dfrac{x^3+1}{x^3}=\dfrac{x^3}{x^3}+\dfrac{1}{x^3}=1+\dfrac{1}{x^3}\to1+\dfrac{1}{0.2^3}=1+\dfrac{1}{0.008}\\\\=1+\dfrac{1}{\frac{8}{1,000}}=1+\dfrac{1,000}{8}=1+125=126\\\\\text{Answer:}\ \dfrac{x^3+1}{x^3}=126$

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Question 10 of 11 Find the measure of the exterior 1 60 252 O A. 95 OB. 85 OC. 35 O D. 120​

Question 10 of 11 Find the measure of the exterior 1 60 252 O A. 95 OB. 85 OC. 35 O D. 120