3
x/y
Suppose the two numbers are X and Y and you are given XY = A and X/Y = B. Then AB = XY*X/Y = X2 so that X = sqrt(AB) and A/B = XY / (X/Y) = XY*Y/X = Y2 so that Y = sqrt(A/B) So take the product and quotient that are given. Find THEIR product and quotient. Your two numbers are the square roots of these numbers. Depending on the context (or requirements of the question) you can either use only the principal square roots or select the necassary signs for the square roots. For example, given the product and quotient are 6 and 2/3, AB = 6*2/3 = 4 sand A/B = 6 / (2/3) = 6*3/2 = 9 The possible solutions are X = 2 and Y = 3 OR X = -2 and Y = -3
(1) x + y = 15(2) x/y = 14Solving (1) and substituting in (2)...x = 15 - y15-y/y = 1414y = 15 - y15y = 15y = 1x = 14================================The sum of 1 + 14 = 15The quotient of 14/1 = 14
45/y
The Quotient of y and -2 is greater than 8
quotient means the answer after dividing, so your question can be expressed as: (x/y)-2
The quotient of 985/5 = 197
3
y/12 + 8 = 2
The quotient of two numbers, X and Y, is X/Y.
Y/3
x/y - 4 = 2
x/y
y/22
Suppose the two numbers are X and Y and you are given XY = A and X/Y = B. Then AB = XY*X/Y = X2 so that X = sqrt(AB) and A/B = XY / (X/Y) = XY*Y/X = Y2 so that Y = sqrt(A/B) So take the product and quotient that are given. Find THEIR product and quotient. Your two numbers are the square roots of these numbers. Depending on the context (or requirements of the question) you can either use only the principal square roots or select the necassary signs for the square roots. For example, given the product and quotient are 6 and 2/3, AB = 6*2/3 = 4 sand A/B = 6 / (2/3) = 6*3/2 = 9 The possible solutions are X = 2 and Y = 3 OR X = -2 and Y = -3
For a quotient x/y , then its log is logx - log y . NOT log(x/y)