Medal awarded can someone please help me on 20., 22, and 26. For 20 I got 3 as my answer but i'm not 100% sure and on 22 and 26 I have no clue how to solve
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OpenStudy (anonymous):
OpenStudy (shinalcantara):
for #20, you substitute the values of x,y&z to the expression
OpenStudy (anonymous):
i got 4 square root of 27+56-2 I used pemdas and got 4 square root of 81 which is 3
OpenStudy (shinalcantara):
that's correct
OpenStudy (shinalcantara):
now proceed to #22. substitute the values
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OpenStudy (anonymous):
(3*4)^2*-1 I get 12^-2
OpenStudy (shinalcantara):
yes. now what happens to a number raised to a power of a negative number?
OpenStudy (anonymous):
1/12^2
OpenStudy (anonymous):
1/144
OpenStudy (shinalcantara):
yep
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OpenStudy (anonymous):
for 26 I looked at the previous problem we solved, I subtracted 48-3 which is 45 the only thing was when I was trying to find a number raised to the 4th power it didn't turn out to be 45
OpenStudy (shinalcantara):
the previous one involves multiplication that's why it's okay but since this involves subtraction, your answer is invalid
OpenStudy (shinalcantara):
|dw:1410687110518:dw|
OpenStudy (anonymous):
12*4
OpenStudy (anonymous):
the square root of 4 is 2
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OpenStudy (shinalcantara):
is that your answer? well think about it, is 4 having a perfect "4th root"?
think it again