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Mathematics 19 Online
OpenStudy (goldphenoix):

Assume that the notation (o, p, q, r) means " Multiply o and p. Subtract q from the product, and then raise to the power of r, " What is the value of (8, 2, 16, 3) + (7, 7, 39, 2)?

OpenStudy (calculusxy):

The equation would be [(8x2)-16]^3 for the first one.

OpenStudy (goldphenoix):

Explain?

jimthompson5910 (jim_thompson5910):

(o, p, q, r) turns into the expression (o*p-q)^r based on the instructions given

OpenStudy (anonymous):

This is similar to the questions that use a made up operation. They're making up a notation and telling you what that notation means. The answers above pretty much explained it. It's saying, if you have four values in parenthesis like that, you multiply the first two, subtract the third, and then raise to the power of the fourth.

OpenStudy (jhannybean):

((o * p)-q)^r Jim.... right?

jimthompson5910 (jim_thompson5910):

correct Jhannybean

OpenStudy (goldphenoix):

So we substitute the values of o, p, q, and r into the expression.

jimthompson5910 (jim_thompson5910):

yes

OpenStudy (calculusxy):

Sorry for the delay. Yes, we substitute the numbers by he rules of the way that the alphabets are lined up.

OpenStudy (goldphenoix):

I see... So the equation would be: [(8 * 2)-16]^3 + [(7 * 7) - 16 39]^2 ?

OpenStudy (calculusxy):

For the second one it would be [(7x7)-39]^2

OpenStudy (calculusxy):

Sorry there was supposed to be a parenthesis after the 39.

OpenStudy (calculusxy):

Yes you are right. Then you use PEMDAS to evaluate the expression.

OpenStudy (goldphenoix):

[(7 * 7) - 39]^2 *

OpenStudy (calculusxy):

That would equal to 100.

OpenStudy (calculusxy):

49-39= 10. Then 10x10=100

OpenStudy (goldphenoix):

[(8 * 2)-16]^3 + [(7 * 7) - 39]^2 (16 - 16)^3 + [(49)-39]^2 (0)^3 + (10)^2 0 + 100 = 100

OpenStudy (calculusxy):

So the answer is 100

OpenStudy (calculusxy):

Good job @GoldPhenoix

OpenStudy (goldphenoix):

Ah, I see. This take some time to answer this question.

OpenStudy (goldphenoix):

Toughest part, who do I give the medal to...

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