Answer:
The volume is 
cubic units.
Step-by-step explanation:
The given curve is
The given line is
Equate both the functions to find the intersection point of both line and curve.
According to washer method:
![V=\pi \int_{a}^{b}[f(x)^2-g(x)^2]dx](https://tex.z-dn.net/?f=V%3D%5Cpi%20%5Cint_%7Ba%7D%5E%7Bb%7D%5Bf%28x%29%5E2-g%28x%29%5E2%5Ddx)
Using washer method, where a=0 and b=1, we get
![V=\pi \int_{0}^{1}[(7x)^2-(7x^6)^2]dx](https://tex.z-dn.net/?f=V%3D%5Cpi%20%5Cint_%7B0%7D%5E%7B1%7D%5B%287x%29%5E2-%287x%5E6%29%5E2%5Ddx)
![V=\pi \int_{0}^{1}[49x^2-49x^{12}]dx](https://tex.z-dn.net/?f=V%3D%5Cpi%20%5Cint_%7B0%7D%5E%7B1%7D%5B49x%5E2-49x%5E%7B12%7D%5Ddx)
![V=49\pi \int_{0}^{1}[x^2-x^{12}]dx](https://tex.z-dn.net/?f=V%3D49%5Cpi%20%5Cint_%7B0%7D%5E%7B1%7D%5Bx%5E2-x%5E%7B12%7D%5Ddx)
![V=49\pi [\frac{x^3}{3}-\frac{x^{13}}{13}]_0^1](https://tex.z-dn.net/?f=V%3D49%5Cpi%20%5B%5Cfrac%7Bx%5E3%7D%7B3%7D-%5Cfrac%7Bx%5E%7B13%7D%7D%7B13%7D%5D_0%5E1)
![V=49\pi [\frac{1^3}{3}-\frac{1^{13}}{13}-(0-0)]](https://tex.z-dn.net/?f=V%3D49%5Cpi%20%5B%5Cfrac%7B1%5E3%7D%7B3%7D-%5Cfrac%7B1%5E%7B13%7D%7D%7B13%7D-%280-0%29%5D)
![V=49\pi [\frac{1}{3}-\frac{1}{13}]](https://tex.z-dn.net/?f=V%3D49%5Cpi%20%5B%5Cfrac%7B1%7D%7B3%7D-%5Cfrac%7B1%7D%7B13%7D%5D)
Therefore the volume is cubic units.
Answer: (0, 0)
Step-by-step explanation:
start at (-3, 3)
go 3 units right
(0, 3)
go 3 units down
(0, 0)
The absolute function will be f(x) = | x - 5 | - 4.
<h3 /><h3>What is an absolute function?</h3>
The absolute function is also known as the mode function. The value of the absolute function is always positive.
The absolute function is given as
f(x) = | x - h | + k
Where h,k is the vertex of the absolute function.
f(x) = | x - 5 | - 4
Therefore the absolute function will be f(x) = | x - 5 | - 4.
More about the absolute function link is given below.
brainly.com/question/10664936
#SPJ1
Answer: -136
Step-by-step explanation:
123+13 is 136 you then change 136 into a negative since you were subtracting 123 from -13
Answer:
Step-by-step explanation:
substitute x = r*cos(θ), y = r*sin(θ) ==> r²(cos²(θ) + sin²(θ)) = 2r²cos(θ)sin(θ). Cancel the r² on both sides. On the left, use pythagorean identity cos²(θ) + sin²(θ) = 1. On the right apply double angle identity sin(2θ) = 2cos(θ)sin(θ).
This yields 1=sin(2θ). (I assume you meant to type sin(2θ) on the right hand side of the equation).
