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# Help!

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

May 24, 2017
#3
+7220
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$$(4x +\dfrac{1}{3}y)^2$$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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$$4x + \dfrac{1}{3}y^2\\ =4\times 2 + \dfrac{1}{3}\times 3^2\\ =8 + 3\\ =11$$

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May 25, 2017
#4
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Oh. Thanks.

Sunrisesk8r  Jun 1, 2017