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Mathematics 16 Online
OpenStudy (anonymous):

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

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

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

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

OpenStudy (shinalcantara):

is that your answer? well think about it, is 4 having a perfect "4th root"? think it again

OpenStudy (shinalcantara):

try to think of another factor

OpenStudy (anonymous):

16 and 3

OpenStudy (shinalcantara):

|dw:1410687399287:dw|

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