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Lets start by putting the known number on the other side.

Y3 -7500y=125000. There is your start. Take it from here.

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15y ago

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Related Questions

Iy y equals 3 then y3 y3-y equals?

72


Is the y3 equals x is linear function?

no


If y equals 3 then y3 y3-y equals?

y3 x y3 - y (3)3 x 3(3) - 3 9 x 9 - 3 = ? 9 x 9= 81 81 - 3 = 78 I hope that solves your problem


Whats y3+x3+z3 equals to?

3xyz


Can you solve y3 125?

No, because it is not an equation (or inequality), but simply two expressions separated by a space.


If y equals 3 then y3 pararenthesis y3 minus y equals?

y=3 y^3(y^3-y)= 3^3(3^3-3)= 27(27-3)= 27(24)=648


How do you solve 16y exponent 3 20y exponent 4?

16y3 / 20y4 = 16/20 * y3/y4 = 4/5 * y3-4 = 4/5 * y-1 or 4/5y


What is the solution to x3 plus 1331 equals y3?

One equation with two unknowns usually does not have a solution.


What is 6y-y3?

6y-y3 = 3


X y 3 equals x y3?

If one of them and the other One is exactly the same(capital,letters) then it is equal:D


How do you undo exponents when different exponents are on each side of the equation?

I'm assuming that you're asking how to solve equations like this "x2 = y3", right? If you solve for x in equations such as this, you should keep in mind that y will probably have a fractional power or will be the radicand in a radical whose index isn't 2 (in other words, it won't be in a square root). You should also be conscious of the rule for converting radicals to exponents: xa/b equals the b-root of x to the a power. So, to solve x2 = y3 for x, take the square root of both sides. This results in (x2)1/2 = (y3)1/2. You know that raising a quantity to a power when addition and substraction are absent means that you have to multiply the exponents, so you would then obtain x = y3/2. Using the conversion rule, x equals the square root of y cubed. It is a square root because the denominator of the exponent of y (or "b" if you're looking at the conversion rule) is 2, but things will not always work out so nicely. I hope this helped.


What is x3-y3-8?

You would need to know the value of either x or y, so that you could solve for the one independent variable.