We have been given the expression

We have the exponent rule

Using this rule, we have

Now, using the fact that
, we get
![x^{\frac{9}{7}}= \sqrt[7]{x^9}\\ \\ x^{\frac{9}{7}}=\sqrt[7]{x^7\times x^2}\\ \\ x^{\frac{9}{7}}=x\sqrt[7]{x^2}](https://tex.z-dn.net/?f=x%5E%7B%5Cfrac%7B9%7D%7B7%7D%7D%3D%20%5Csqrt%5B7%5D%7Bx%5E9%7D%5C%5C%0A%5C%5C%0Ax%5E%7B%5Cfrac%7B9%7D%7B7%7D%7D%3D%5Csqrt%5B7%5D%7Bx%5E7%5Ctimes%20x%5E2%7D%5C%5C%0A%5C%5C%0Ax%5E%7B%5Cfrac%7B9%7D%7B7%7D%7D%3Dx%5Csqrt%5B7%5D%7Bx%5E2%7D)
D is the correct option.
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:
b
Step-by-step explanation:
The median of a triangle is a line that from the vertex touches the middle lets eliminate options
a=ab starts at a vertex but doensn't touch a the middle of a line X
b=cd starts at a vertex and touches the middle of a line Ye
C= ce stars at a vertex but doesn't end at the middle of a line X
d=mb starts in the middle of the triangle and and ends a vertex X
<u>SO THE CORRECT ANSWER IS B</u>
<em>-hope this helps :)</em>
Answer:
Your answer: 87
Step-by-step explanation:
Since you are given what both variables are in the equation, this problem requires you to plug it in.
12(6)+3(5)= ?
You can either do this in a calculator, or do this one step at a time.
12(6)=72
3(5)=15
72+15=87