Answer:
The quotient is 3x - 11 + 60/(x + 5) ⇒ 2nd answer
Step-by-step explanation:
* We will use the long division to solve the problem
- The dividend is 3x² + 4x + 5
- The divisor is x + 5
- The quotient is the answer of the division
- If the divisor not a factor of a dividend, the quotient has
a remainder
* Lets solve the problem
- At first divide the first term in the dividend by the first term in
the divisor
∵ 3x² ÷ x = 3x
- Multiply the divisor by 3x
∴ 3x (x + 5) = 3x² + 15x
-Subtract this expression from the dividend
∴ 3x² + 4x + 5 - (3x² + 15x) = 3x² + 4x + 5 - 3x² - 15x = -11x + 5
- Divide the first term -11x in the new dividend by the first
term x in the divisor
∴ -11x ÷ x = -11
- Multiply the divisor by -11
∴ -11(x + 5) = -11x - 55
-Subtract this expression from the new dividend
∴ -11x + 5 - (-11x - 55) = -11x + 5 + 11x + 55 = 60
∴ The quotient is 3x - 11 with remainder = 60
* The quotient is 3x - 11 + 60/(x + 5)
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)
|RZ|=0.5|SW| therefore 2|RZ|=|SW|.
5x - 20 = 2 · 30
5x - 20 = 60 |add 20 to both sides
5x = 80 |divide both sides by 5
x = 16
