We are integrating f(x) = 9cos(9x) + 3x²: 
a) Apply the sum rule
b) Calculate each antiderivative
<u>First integral</u>
1. Take out the constant
2. Apply u-substitution, where u is 9x
3. Take out the constant (again)
4. Take the common integral of cos, which is sin
5. Substitute the original function back in for u and simplify
6. Always remember to add an arbitrary constant, C, at the end
<u>Second integral</u>
1. Take out the constant
2. Apply the power rule, 
, where <em>a</em> is your exponent
⇒ 
3. Add the arbitrary constant
c) Add the integrals
sin(9x) + C + x³ + C = sin(9x) + x³ + C
Notice the two arbitrary constants. Since we do not know what either constant is, we can combine them into one arbitrary constant.
<h3>
Answer:</h3>
F(x) = sin(9x) + x³ + C
Answer:
A8 = 66 [8th term]
Step-by-step explanation:
Given the explicit arithmetic sequence An = -14 + 10n → -4 + 10(n-1)
To find the 8th term. An nth term is where n is n in the sequence. So substitute 8 for n, and simplify.
So A8 = -4 + 10(8-1) → -4 + 10(7) → -4 + 70 → 70 - 4 → 66.
Answer:
c = $5,400
Step-by-step explanation:
Given:
c = 0.9s
where,
s = amount sold
c = commission on sales
How much commission will he or she earn if the amount sold is $6,000?
Find c when she = $6,000
c = 0.9s
= 0.9 × 6,000
= 5,400
c = $5,400
(c, s) (5400, 6000)
Answer:
![A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]](https://tex.z-dn.net/?f=A%5E%7B-1%7D%3D%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7D%20%5Cfrac%7B1%7D%7B9%7D%20%26%20%5Cfrac%7B4%7D%7B27%7D%20%26%20-%20%5Cfrac%7B2%7D%7B27%7D%20%5C%5C%5C%5C%20%5Cfrac%7B8%7D%7B9%7D%20%26%20%5Cfrac%7B5%7D%7B27%7D%20%26%20%5Cfrac%7B11%7D%7B27%7D%20%5C%5C%5C%5C%20-%20%5Cfrac%7B4%7D%7B9%7D%20%26%20%5Cfrac%7B2%7D%7B27%7D%20%26%20-%20%5Cfrac%7B1%7D%7B27%7D%20%5Cend%7Barray%7D%20%5Cright%5D)
Step-by-step explanation:
We want to find the inverse of ![A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7D%201%20%26%200%20%26%20-2%20%5C%5C%5C%5C%204%20%26%201%20%26%203%20%5C%5C%5C%5C%20-4%20%26%202%20%26%203%20%5Cend%7Barray%7D%20%5Cright%5D)
To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.
So, augment the matrix with identity matrix:
![\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 4&1&3&0&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%26-2%261%260%260%20%5C%5C%5C%5C%204%261%263%260%261%260%20%5C%5C%5C%5C%20-4%262%263%260%260%261%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 1 multiplied by 4 from row 2
![\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%26-2%261%260%260%20%5C%5C%5C%5C%200%261%2611%26-4%261%260%20%5C%5C%5C%5C%20-4%262%263%260%260%261%5Cend%7Barray%7D%5Cright%5D)
- Add row 1 multiplied by 4 to row 3
![\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%26-2%261%260%260%20%5C%5C%5C%5C%200%261%2611%26-4%261%260%20%5C%5C%5C%5C%200%262%26-5%264%260%261%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 2 multiplied by 2 from row 3
![\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%26-2%261%260%260%20%5C%5C%5C%5C%200%261%2611%26-4%261%260%20%5C%5C%5C%5C%200%260%26-27%2612%26-2%261%5Cend%7Barray%7D%5Cright%5D)
![\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%26-2%261%260%260%20%5C%5C%5C%5C%200%261%2611%26-4%261%260%20%5C%5C%5C%5C%200%260%261%26-%20%5Cfrac%7B4%7D%7B9%7D%26%5Cfrac%7B2%7D%7B27%7D%26-%20%5Cfrac%7B1%7D%7B27%7D%5Cend%7Barray%7D%5Cright%5D)
- Add row 3 multiplied by 2 to row 1
![\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%260%26%5Cfrac%7B1%7D%7B9%7D%26%5Cfrac%7B4%7D%7B27%7D%26-%20%5Cfrac%7B2%7D%7B27%7D%20%5C%5C%5C%5C%200%261%2611%26-4%261%260%20%5C%5C%5C%5C%200%260%261%26-%20%5Cfrac%7B4%7D%7B9%7D%26%5Cfrac%7B2%7D%7B27%7D%26-%20%5Cfrac%7B1%7D%7B27%7D%5Cend%7Barray%7D%5Cright%5D)
- Subtract row 3 multiplied by 11 from row 2
![\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7Cccc%7D1%260%260%26%5Cfrac%7B1%7D%7B9%7D%26%5Cfrac%7B4%7D%7B27%7D%26-%20%5Cfrac%7B2%7D%7B27%7D%20%5C%5C%5C%5C%200%261%260%26%5Cfrac%7B8%7D%7B9%7D%26%5Cfrac%7B5%7D%7B27%7D%26%5Cfrac%7B11%7D%7B27%7D%20%5C%5C%5C%5C%200%260%261%26-%20%5Cfrac%7B4%7D%7B9%7D%26%5Cfrac%7B2%7D%7B27%7D%26-%20%5Cfrac%7B1%7D%7B27%7D%5Cend%7Barray%7D%5Cright%5D)
As can be seen, we have obtained the identity matrix to the left. So, we are done.
180-(83+53)=44
That will be the answer
