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What is the value of the expression above when x = 2 and y = 3? Please show your work! :)

 

4x + 1/3y to the power of 2.    Oops!!

 May 24, 2017
edited by Sunrisesk8r  May 24, 2017
 #1
avatar+51 
0

Can someone please help me?

 May 24, 2017
 #2
avatar+8 
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What expression??

 May 24, 2017
 #3
avatar+7220 
+1

\((4x +\dfrac{1}{3}y)^2\)This?

\((4x + \dfrac{1}{3y})^2\)This?

\(4x + \dfrac{1}{3y^2}\)Or this?

\(4x + \dfrac{1}{(3y)^2}\)Or this?

\(4x + (\dfrac{1}{3y})^2\)This?

\(4x + \dfrac{1}{3}y^2\)Or even this?

Welp. Confused. Parentheses are important.

Nvm. I will do all of them

When x = 2, y = 3,

\((4x +\dfrac{1}{3}y)^2\\ =(4\times 2+\dfrac{1}{3}\times 3)^2\\ =(8 + 1)^2\\ =9^2\\ =81\)

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\((4x + \dfrac{1}{3y})^2\\ =(4\times 2 + \dfrac{1}{3\times 3})^2\\ =(8 + \dfrac{1}{9})^2\\ =(\dfrac{73}{9})^2\\ =\dfrac{73^2}{9^2}\\ =\dfrac{5329}{81}\)

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\(4x + \dfrac{1}{3y^2}\\ =4\times 2 + \dfrac{1}{3\times 3^2}\\ =8 + \dfrac{1}{27}\\ =\dfrac{217}{27}\)

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\(4x + \dfrac{1}{(3y)^2}\\ =4\times 2 + \dfrac{1}{(3\times 3)^2}\\ = 8 + \dfrac{1}{81}\\ =\dfrac{649}{81}\)

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\(4x + (\dfrac{1}{3y})^2\\ =4\times 2 + (\dfrac{1}{3\times 3})^2\\ =8 + \dfrac{1}{81}\\ =\dfrac{649}{81}\)

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\(4x + \dfrac{1}{3}y^2\\ =4\times 2 + \dfrac{1}{3}\times 3^2\\ =8 + 3\\ =11\)

.
 May 25, 2017
 #4
avatar+51 
0

Oh. Thanks.

Sunrisesk8r  Jun 1, 2017

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