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If \(y = 2 + 2k\) and \(y\ne 0,\) then \(\dfrac1{y} + \dfrac1{y} + \dfrac1{y} + \dfrac1{y} = ?\)

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If \(y = 2 + 2k\) and \(y\ne 0,\) then \(\dfrac1{y} + \dfrac1{y} + \dfrac1{y} + \dfrac1{y} = ?\)

by VJesus12 » Fri Jul 09, 2021 11:44 am

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If \(y = 2 + 2k\) and \(y\ne 0,\) then \(\dfrac1{y} + \dfrac1{y} + \dfrac1{y} + \dfrac1{y} = ?\)

A. \(\dfrac1{8+8k}\)

B. \(\dfrac2{1+k}\)

C. \(\dfrac1{8+k}\)

D. \(\dfrac4{8+k}\)

E. \(\dfrac4{1+k}\)

Answer: B

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Re: If \(y = 2 + 2k\) and \(y\ne 0,\) then \(\dfrac1{y} + \dfrac1{y} + \dfrac1{y} + \dfrac1{y} = ?\)

by regor60 » Sat Jul 10, 2021 12:29 pm
Set K=0, so Y=2.

Plug K=0 into the answer choices
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