To determine the value of y when x is 100, we need more information about the relationship between x and y. If we have an equation or function relating x and y, we can substitute x=100 into the equation to find the corresponding value of y. Without this information, it is impossible to determine the value of y when x is 100.
You are basically looking for an equation with an x and y value where when 4 is set as x, y would equal 7. An equation of this is: 3 + x = y when x = 4, then y = 3 + 4 = 7
If you mean: x+y = 30 and the value of y is 15 then the value of x is also 15
There is no one solution to this problem. For example: y = x2 y = 5x y = x + 20 y = 2x / 5 + 23 These are just four of the infinite set functions that would give "y" a value of 25 when x = 5.
Y = X - 16 X= Y + 16 X is the original value. Y is the new value of the original value - 16.
If y varies directly as x and y is 36 when x is 9, then y is always four times the value of x. So if y is 12, then x is 3.
You don't.You could find percent of change but first you would have to learn to check your spelling!Suppose the value changes from X to Ythe change is (Y - X)the proportional change is (Y - X)/X = (Y/X - 1)the percentage change is 100*(Y/X - 1)that is,100*(New Value/Old Value - 1)
X as a percentage of Y is 100*X/Y.
y = 100 - 4x ( a ) Find the value of y when x = 20.
By formula, x/y*100 = Percentage. 6/8*100 = 75% ie) The formula for percentile value is x/y *100 where X is the part of y for which percentage needs to be known, and Y is our standard. eg) What percent is is 3 of 8. 3 is our 'x' of which value needs to be known, 8, or 'Y' is our standard. 3/8*100 = 37.5%
At first glance, this would appear to be a multivariable problem. But in actuality, with the two equations we have to work with: f(x,y) = xy x - y = 100 We can rearrange the second one to find what y is equal to in terms of x: y = 100 - x (We can also solve x in terms of y, but solving y in terms of x is conventional). then we can plug this in for y into f(x,y) to make it simply a function of x: f(x) = x(100 - x) = 100x - x2 We then take the derivative of this, setting f'(x) equal to 0: f'(x) = 0 = 100 - 2x 2x = 100 x = 100/2 = 50 This is the value of x that minimizes the product of x and y when x - y = 100. We then solve for y by plugging in the x-value into this equation: 50 - y = 100 -y = 100 - 50 y = -50 By plugging in values for which x > 50 and x < 50 into the equation f(x) = 100x - x2, we find that f(x) is indeed greater on both the intervals where x < 50 and where x > 50, proving that x = 50 is indeed an absolute minimum.
In expressions such as "x-y", both "x" and "y" can have any value. The value of "x-y" will depend on what the value of "x" and the value of "y" are.
Let [x percent of y] = (x/100)*y. And [y percent of x] = (y/100)*x.Use the ruSo the first one (x/100)*y = y*(x /100) = y*x/100.The second one: (y/100)*x = x*(y/100) = x * y / 100, which equals y*x/100.So yes they equal the same thing, so they are equal.
The formula to find the value of X would be Y-2X. This would equal to y-9 times 2 X.
If you start with a value x and end with a value y thenPercentage change = 100*(y/x - 1)If y > x then the above is positive and is a percentage increase andif y < x then the above is negative and is a percentage decrease.
It depends on the relationship between x and y.
There is no way to know unless you had an equation that gives the relationship between x and y. For example, if y=2x, then when x equals 99 or x equals 100, then the value of y has a difference of 2.
Y would decrease in value as X increases in value.