Answer:
x = 9
Step-by-step explanation:
Well, let's solve it!
x - 9 = 0
Let's add 9 to both sides to isolate the variable and maintain equality.
We end up with :
x = 9 since -9 and 9 cancel out and 0 + 9 = 9.
I'd suggest you begin by subtracting 4 / (x-4) from both sides. Doing that would leave you with 1 / (x-4) - 2 / (x+2) = 0.
LCD is (x-4)(x+2). Mult. all three terms by (x-4)(x+2). The resulting equation is
x+2 - 2(x-4) = 0. Then x+2 = 2x - 8 => x = 10
Subst. 10 for x in the original equation to verify that 10 is indeed a solution.
The area of D is given by:
The average value of f over D is given by:
![\frac{1}{ \frac{343}{3} } \int\limits^7_0 \int\limits^{x^2}_0 {4x\sin(y)} \, dydx = -\frac{3}{343} \int\limits^7_0 {4x\cos(x^2)} \, dx \\ \\ =-\frac{3}{343} \int\limits^{49}_0 {2\cos(t)} \, dt=-\frac{6}{343} \left[\sin(t)\right]_0^{49} \, dt=-\frac{6}{343}\sin49](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%20%5Cfrac%7B343%7D%7B3%7D%20%7D%20%20%5Cint%5Climits%5E7_0%20%20%5Cint%5Climits%5E%7Bx%5E2%7D_0%20%7B4x%5Csin%28y%29%7D%20%5C%2C%20dydx%20%20%3D%20-%5Cfrac%7B3%7D%7B343%7D%20%20%5Cint%5Climits%5E7_0%20%7B4x%5Ccos%28x%5E2%29%7D%20%5C%2C%20dx%20%20%5C%5C%20%20%5C%5C%20%3D-%5Cfrac%7B3%7D%7B343%7D%20%5Cint%5Climits%5E%7B49%7D_0%20%7B2%5Ccos%28t%29%7D%20%5C%2C%20dt%3D-%5Cfrac%7B6%7D%7B343%7D%20%5Cleft%5B%5Csin%28t%29%5Cright%5D_0%5E%7B49%7D%20%5C%2C%20dt%3D-%5Cfrac%7B6%7D%7B343%7D%5Csin49)
For this case we have the following polynomials:
(2n ^ 2-7n + 2)
(3n + 1)
Multiplying we have:
(2n ^ 2-7n + 2) * (3n + 1)
(6n ^ 3-21n ^ 2 + 6n) + (2n ^ 2-7n + 2)
Rewriting we have:
6n ^ 3 + n ^ 2 (-21 + 2) + n (6-7) + 2
6n ^ 3 - 19n ^ 2 - n + 2
Answer:
6n ^ 3 - 19n ^ 2 - n + 2
