How would i solve for Y:
3.2*10^-10= (y)(2y)^2/ (0.140-2y)^2
3.2*10^-10= (y)(2y)^2/ (0.140-2y)^2 or
3.2*10^-10= 2y^3/ (0.140-2y)^2
for all practical purposes, the left side of your equation is "zero"
(it has value 0.00000000032)
so we end up with 2y^3 = appr. 0
so y = appr. 0
(0.140-2y)^2 * (3.2*10^-10) = 4 y^5
(0.0196 -0.280y + 4y^2)*(3.2*10^-10)
= 4 y^5
6.282*10^-12 -8.96*10^-11 y +1.28*10^-9 y^2
= 4 y^5
This is a messy fifth order polynomial that is best solved by iteration. As a first approximation, I would assume 4 y^5 = 6.282*10^-12, since the y and Y^2 terms will be negligible compared to the constant.
y^5 = 1.57*10^-12
y = 4.36*10^-3
Solve 7/18 using bar notation
To solve for Y in the equation 3.2 * 10^(-10) = (y)(2y)^2 / (0.140 - 2y)^2, you can follow these steps:
Step 1: Simplify the expression on the right side of the equation.
- Square the term (2y)^2, which gives you 4y^2 (since (a^2)^2 = a^4).
- Square the denominator (0.140 - 2y)^2, which gives you (0.140 - 2y)(0.140 - 2y) = (0.140 - 2y)^2.
Your equation now becomes 3.2 * 10^(-10) = (y)(4y^2) / (0.140 - 2y)^2.
Step 2: Multiply both sides of the equation by (0.140 - 2y)^2.
- This will remove the denominator.
You now have: (0.140 - 2y)^2 * 3.2 * 10^(-10) = (y)(4y^2).
Step 3: Multiply out both sides of the equation to distribute the terms.
- Apply the distributive property to both sides.
This gives you: (0.140 - 2y)(0.140 - 2y)(3.2 * 10^(-10)) = (y)(4y^2).
Step 4: Simplify the expression on both sides.
- Expand the left side of the equation by multiplying (0.140 - 2y)(0.140 - 2y) using the FOIL method.
You will have two terms in the expansion: (0.140 * 0.140) - (2y * 0.140) - (2y * 0.140) + (2y * 2y).
Simplify each term: 0.0196 - 0.28y - 0.28y + 4y^2 = (y)(4y^2).
Combine like terms: 0.0196 - 0.56y + 4y^2 = 4y^3.
Step 5: Rearrange the equation and set it equal to zero.
- Move the terms to one side of the equation to create a quadratic equation.
- Set the equation equal to zero to solve for Y.
You will have: 4y^3 - 4y^2 + 0.56y - 0.0196 = 0.
Step 6: Use numerical methods or factorization to solve for Y.
- The resulting equation is a cubic equation, which may require numerical methods or factoring to find the solutions.
- One approach is to use numerical methods such as the Newton-Raphson method or the bisection method to approximate the roots of the equation.
- Another approach is to factor if possible. However, cubic equations do not always have rational roots that can be easily factored.
Unfortunately, solving the cubic equation exactly may be difficult, and it may require advanced mathematical techniques or the use of software or calculators specifically designed to solve these types of equations.