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The answer to this math problem would equal to 20. This is taught in high school math.

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6q +1

Q: 6q - 1 -q plus 20?

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5p-7q

If -3q + 4 = 13 then 3q = -9 : q = -3 If 6q = 13 then q = 13/6 = 2 1/6 The question contains an anomaly and is invalid.

7p + 2q = 46 . . . . (A) 5p + 3q = 36 . . . . (B) 3*(A): 21p + 6q = 138 2*(B): 10p + 6q = 72 Subtracting gives 11p = 66 so that p = 6 Substitute for p in (A): 7*6 + 2q = 46 or 42 + 2q = 46 which gives 2q = 4 so that q = 2 Solution: (p, q) = (6,2)

It is not an equation, but q2 meaning q^2 represents q being multiplied by itself.

a/b = 1 so a = b. Then a b = q implies that a = b = q/2 So ab = (q/2)*(q/2) = q2/4

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5p-7q

If -3q + 4 = 13 then 3q = -9 : q = -3 If 6q = 13 then q = 13/6 = 2 1/6 The question contains an anomaly and is invalid.

6q

I'm not exactly sure what you mean by this, but if you'd like to know how to do this in C here: q ^ 2 + 20 q + c

7p + 2q = 46 . . . . (A) 5p + 3q = 36 . . . . (B) 3*(A): 21p + 6q = 138 2*(B): 10p + 6q = 72 Subtracting gives 11p = 66 so that p = 6 Substitute for p in (A): 7*6 + 2q = 46 or 42 + 2q = 46 which gives 2q = 4 so that q = 2 Solution: (p, q) = (6,2)

6

If: 1/2 = q+2/3 Then: q = -1/6

q3-q2+2q-2 = (q-1)(q2+2) = (q-1)(q+2.5i)(q-2.5i)

It is not an equation, but q2 meaning q^2 represents q being multiplied by itself.

p2 + 2pq + q2 = 1q2 + 2pq + (p2 - 1) = 0q = 1/2 [ -2p plus or minus sqrt( 4p2 - 4p2 + 4 ) ]q = -1 - pq = 1 - p

p=3 !

p + q + r = (2x - 9y) + (5y + 6 - 4x) + (3x + 3y - 5) = x - y + 1