Since both are variable, you need to plug in what the values are given in the problem, then subtract y from x.
If the answer is zero , then they are equal.
If the answer is a positive number then x is greater than y.
If the answer is a negative number then y is greater than x.
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When the second number is positive and has an absolute value that is bigger. Suppose x is the first number and y the second. Then x is smaller than y implies that if x is non-negative then y is larger, and if x is negative then y is less negative (nearer 0), 0 or positive In the first case, abs(x) < abs(y), that is 0 <= x < y In the second case, abs(x) < abs(y) if 0 < -x < y.
The number is 27let the number be xy (that is it is 10x + y). Then:Assume x is the bigger digit, thenx - y = 5 ⇒ x = y + 5x2 - y2 = 45Substituting for x in 2 using 1 above:(y + 5)2 - y2 = 45⇒ y2 + 10y + 25 - y2 = 45⇒ 10y = 20⇒ y = 2⇒ x = 7But y must be odd since xy is odd.If y was assumed to be the bigger digit, x and y would be swapped in the above calculation, thus the values of x and y can be swapped:x = 2, y = 7 ⇒ the number is 27
y = sin(x+y) cos( x + y )[(1 + y')] = y' cos(x + y ) + y'cos(x + y ) = y' y'-y'cos( x+ y) = cos( x + y ) y'[1-cos(x+y)]= cos(x+y) y'= [cos(x+y)]/ [1-cos(x+y)]
Given two numbers, x and y their difference is |x - y|. |x - y| = x - y if X ≥ y and |x - y| = y - x if x < y
X + Y (X + Y) ^2 = (X+Y)(X+Y) Factor = (X + Y)