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- anonymous

Evaluate 2x^2y for x = 2 1/2 and y = -3 3/5

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- anonymous

Evaluate 2x^2y for x = 2 1/2 and y = -3 3/5

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- anonymous

step by step please

- anonymous

i plugged in 2(2 1/2) ^2 (-3 3/5)

- anonymous

\[2(2\frac{ 1 }{ 2 })^2 (-3\frac{ 3 }{ 5})\]

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- Jhannybean

I think what will make it a little easier to evaluate would be if you changed the mixed integer into a fraction. \[2\frac{1}{2} \iff \frac{(2 \cdot 2)+1}{2}\]\[-3\frac{3}{5} \iff \frac{(-3 \cdot 5) +3}{5}\]

- anonymous

okay :)

- anonymous

and from there do i just multiply straight across?

- Jhannybean

Yup, you input the fraction in place of x and y and simplify! :)

- anonymous

okay thanks!

- anonymous

okay so after that i got 5/2 and then -12/5 is that right?

- Jhannybean

Lets check it out. I havent solved it yet! SO we'll see.

- anonymous

okay! im solving it right now

- anonymous

i figured it out! thanks

- Jhannybean

\[x = \frac{ 5}{2} ~,~ y = -\frac{12}{5}\]Now let's input these into our formula: \(2x^2y\) When I put these values in, I got:\[2\left(\frac{5}{2}\right)^2\left(\frac{-12}{5}\right)\]\[2\left(\frac{25}{4}\cdot \frac{-12}{5}\right)\]\[2(-15) = \boxed{-30}\]

- anonymous

Hmm. i got the answer as -45

- Jhannybean

can you write out your steps? Maybe we can compare and see which one of is wrong :o

- anonymous

I did 2(5^2/2^2) and then -18/5
because i multiped 2*2 then added 1 which gave me 5 so i then put 2(25/4) (-18/5)
and then put the 2 into a fraction so 2/1 (25/4) (-18/5) and then i crossed out numbers so i did (2*1/1) (25/2*2) and then it would be 25/2 (-18/5)
Cross out numbers again (5*5)/2) (-1*18/5*1)
(5/2) (-18/1)
= -90/2
= -45

- anonymous

sorry if you cant understand

- Jhannybean

Where did the \(-\frac{18}{5}\) come from?

- anonymous

i multiplied -3*5 and added 3

- anonymous

thank you for helping me! I have to go to school now, the answer sheet said it was -45 so maybe i did something wrong : )

- Jhannybean

Well... \((-3\cdot 5) +3 = -15 + 3 = -12...\)

- anonymous

yeah i moved the negative sign tho

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