When solving for a variable, you get the variable you're trying to solve for on one side and everything to the opposite of that variable.
We have the equation <span>5w + 9z = 2z + 3w.
Usually the variable we're solving for we want on the left. But it's fine to have it on the right side, too.
Let's subtract 9z from the left-hand side. That way, the 5w will be alone on the left-hand side.
And remember, anything we do on one side we do to the other side.
</span><span>5w + 9z - 9z = 2z + 3w - 9z
</span><span>5w = -7z + 3w
The 3w term on the right-hand side needs to be removed. So, subtract each side by 3w.
5w - 3w = -7z + 3w - 3w
2w = -7z
Now, we need to divide each side by 2 to see what the w variable is equal to.
2w / 2 = -7z / 2
w = -7z / 2 or w = -3.5z
So, w is equal to -3.5z.
</span>
Use the pythagorean theorem. You should come out with C. Hope this helps!
Answer with explanation:
--------------------------------------------------------Dividing both sides by 8 x
This Integration is of the form ⇒y'+p y=q,which is Linear differential equation.
Integrating Factor
Multiplying both sides by Integrating Factor
![x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\times [y'+y\times\frac{1+4x^2}{8x}]=\frac{1}{8}\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\\\\ \text{Integrating both sides}\\\\y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\frac{1}{8}\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx \\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=-[x^{\frac{9}{8}}]\times\frac{ \Gamma(0.5625, -x^2)}{(-x^2)^{\frac{9}{16}}}\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=(-1)^{\frac{-1}{8}}[ \Gamma(0.5625, -x^2)]+C-----(1)](https://tex.z-dn.net/?f=x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%5Ctimes%20%5By%27%2By%5Ctimes%5Cfrac%7B1%2B4x%5E2%7D%7B8x%7D%5D%3D%5Cfrac%7B1%7D%7B8%7D%5Ctimes%20x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%5C%5C%5C%5C%20%5Ctext%7BIntegrating%20both%20sides%7D%5C%5C%5C%5Cy%5Ctimes%20x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%3D%5Cfrac%7B1%7D%7B8%7D%5Cint%20%7Bx%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%7D%20%5C%2C%20dx%20%5C%5C%5C%5C8y%5Ctimes%20x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%3D%5Cint%20%7Bx%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%7D%20%5C%2C%20dx%5C%5C%5C%5C8y%5Ctimes%20x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%3D-%5Bx%5E%7B%5Cfrac%7B9%7D%7B8%7D%7D%5D%5Ctimes%5Cfrac%7B%20%5CGamma%280.5625%2C%20-x%5E2%29%7D%7B%28-x%5E2%29%5E%7B%5Cfrac%7B9%7D%7B16%7D%7D%7D%5C%5C%5C%5C8y%5Ctimes%20x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%3D%28-1%29%5E%7B%5Cfrac%7B-1%7D%7B8%7D%7D%5B%20%5CGamma%280.5625%2C%20-x%5E2%29%5D%2BC-----%281%29)
When , x=1, gives , y=9.
Evaluate the value of C and substitute in the equation 1.
Answer:1
Step-by-step explanation:
Here is the equation:
Since both decimals are negative, the answer will be positive.
- Negative times Negative = Positive
Divide both sides by -3.9 to leave the variable alone:
r is equal to 4 → r = 4
