Answer:
y = -1x + 6 or y = -x + 6
Step-by-step explanation:
First, let's identify what slope-intercept form is.
y = mx + b
m is the slope. b is the y-intercept.
We know the slope is -1, so m = -1. Plug this into our standard equation.
y = -1x + b
To find b, we want to plug in a value that we know is on this line: (2, 4). Plug in the x and y values into the x and y of the standard equation.
4 = -1(2) + b
To find b, multiply the slope and the input of x(2)
4 = -2 + b
Now, add 2 from both sides to isolate b.
6 = b
Plug this into your standard equation.
y = -1x + 6
This is your equation.
Check this by plugging in the point again.
y = -1x + 6
4 = -1(2) + 6
4 = -2 + 6
4 = 4
Your equation is correct.
Hope this helps!
You find a unit rate by multiplying by 100<span />
Answer:
30 degrees
Step-by-step explanation:
180-56-90=34
34 rounded off to nearest tenth is 30
Answer:
![\textsf{Midpoint rule}: \quad \dfrac{2\pi}{\sqrt[3]{2}}](https://tex.z-dn.net/?f=%5Ctextsf%7BMidpoint%20rule%7D%3A%20%5Cquad%20%5Cdfrac%7B2%5Cpi%7D%7B%5Csqrt%5B3%5D%7B2%7D%7D)
Step-by-step explanation:
<u>Midpoint rule</u>
![\displaystyle \int_{a}^{b} f(x) \:\:\text{d}x \approx h\left[f(x_{\frac{1}{2}})+f(x_{\frac{3}{2}})+...+f(x_{n-\frac{3}{2}})+f(x_{n-\frac{1}{2}})\right]\\\\ \quad \textsf{where }h=\dfrac{b-a}{n}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7Ba%7D%5E%7Bb%7D%20f%28x%29%20%5C%3A%5C%3A%5Ctext%7Bd%7Dx%20%5Capprox%20h%5Cleft%5Bf%28x_%7B%5Cfrac%7B1%7D%7B2%7D%7D%29%2Bf%28x_%7B%5Cfrac%7B3%7D%7B2%7D%7D%29%2B...%2Bf%28x_%7Bn-%5Cfrac%7B3%7D%7B2%7D%7D%29%2Bf%28x_%7Bn-%5Cfrac%7B1%7D%7B2%7D%7D%29%5Cright%5D%5C%5C%5C%5C%20%5Cquad%20%5Ctextsf%7Bwhere%20%7Dh%3D%5Cdfrac%7Bb-a%7D%7Bn%7D)
<u>Trapezium rule</u>
![\displaystyle \int_{a}^{b} y\: \:\text{d}x \approx \dfrac{1}{2}h\left[(y_0+y_n)+2(y_1+y_2+...+y_{n-1})\right] \quad \textsf{where }h=\dfrac{b-a}{n}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7Ba%7D%5E%7Bb%7D%20y%5C%3A%20%5C%3A%5Ctext%7Bd%7Dx%20%5Capprox%20%5Cdfrac%7B1%7D%7B2%7Dh%5Cleft%5B%28y_0%2By_n%29%2B2%28y_1%2By_2%2B...%2By_%7Bn-1%7D%29%5Cright%5D%20%5Cquad%20%5Ctextsf%7Bwhere%20%7Dh%3D%5Cdfrac%7Bb-a%7D%7Bn%7D)
<u>Simpson's rule</u>
<u>Given definite integral</u>:
![\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%5E%7B2%20%5Cpi%7D_0%20%5Csqrt%5B3%5D%7B%5Csin%5E2%20%28x%29%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx)
Therefore:
Calculate the subdivisions:
<u>Midpoint rule</u>
Sub-intervals are:
![\left[0, \dfrac{1}{2}\pi \right], \left[\dfrac{1}{2}\pi, \pi \right], \left[\pi , \dfrac{3}{2}\pi \right], \left[\dfrac{3}{2}\pi, 2 \pi \right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%20%5Cdfrac%7B1%7D%7B2%7D%5Cpi%20%5Cright%5D%2C%20%5Cleft%5B%5Cdfrac%7B1%7D%7B2%7D%5Cpi%2C%20%5Cpi%20%5Cright%5D%2C%20%5Cleft%5B%5Cpi%20%2C%20%5Cdfrac%7B3%7D%7B2%7D%5Cpi%20%5Cright%5D%2C%20%5Cleft%5B%5Cdfrac%7B3%7D%7B2%7D%5Cpi%2C%202%20%5Cpi%20%5Cright%5D)
The midpoints of these sub-intervals are:
Therefore:
![\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{2}\pi \left[f \left(\dfrac{1}{4} \pi \right)+f \left(\dfrac{3}{4} \pi \right)+f \left(\dfrac{5}{4} \pi \right)+f \left(\dfrac{7}{4} \pi \right)\right]\\\\& = \dfrac{1}{2}\pi \left[\sqrt[3]{\dfrac{1}{2}} +\sqrt[3]{\dfrac{1}{2}}+\sqrt[3]{\dfrac{1}{2}}+\sqrt[3]{\dfrac{1}{2}}\right]\\\\ & = \dfrac{2\pi}{\sqrt[3]{2}}\\\\& = 4.986967483...\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cdisplaystyle%20%5Cint%5E%7B2%20%5Cpi%7D_0%20%5Csqrt%5B3%5D%7B%5Csin%5E2%20%28x%29%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%20%26%20%5Capprox%20%5Cdfrac%7B1%7D%7B2%7D%5Cpi%20%5Cleft%5Bf%20%5Cleft%28%5Cdfrac%7B1%7D%7B4%7D%20%5Cpi%20%5Cright%29%2Bf%20%5Cleft%28%5Cdfrac%7B3%7D%7B4%7D%20%5Cpi%20%5Cright%29%2Bf%20%5Cleft%28%5Cdfrac%7B5%7D%7B4%7D%20%5Cpi%20%5Cright%29%2Bf%20%5Cleft%28%5Cdfrac%7B7%7D%7B4%7D%20%5Cpi%20%5Cright%29%5Cright%5D%5C%5C%5C%5C%26%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%5Cpi%20%5Cleft%5B%5Csqrt%5B3%5D%7B%5Cdfrac%7B1%7D%7B2%7D%7D%20%2B%5Csqrt%5B3%5D%7B%5Cdfrac%7B1%7D%7B2%7D%7D%2B%5Csqrt%5B3%5D%7B%5Cdfrac%7B1%7D%7B2%7D%7D%2B%5Csqrt%5B3%5D%7B%5Cdfrac%7B1%7D%7B2%7D%7D%5Cright%5D%5C%5C%5C%5C%20%26%20%3D%20%5Cdfrac%7B2%5Cpi%7D%7B%5Csqrt%5B3%5D%7B2%7D%7D%5C%5C%5C%5C%26%20%3D%204.986967483...%5Cend%7Baligned%7D)
<u>Trapezium rule</u>
![\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{2} \cdot \dfrac{1}{2} \pi \left[(0+0)+2(1+0+1)\right]\\\\& = \dfrac{1}{4} \pi \left[4\right]\\\\& = \pi\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cdisplaystyle%20%5Cint%5E%7B2%20%5Cpi%7D_0%20%5Csqrt%5B3%5D%7B%5Csin%5E2%20%28x%29%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%20%26%20%20%5Capprox%20%5Cdfrac%7B1%7D%7B2%7D%20%5Ccdot%20%5Cdfrac%7B1%7D%7B2%7D%20%5Cpi%20%5Cleft%5B%280%2B0%29%2B2%281%2B0%2B1%29%5Cright%5D%5C%5C%5C%5C%26%20%3D%20%5Cdfrac%7B1%7D%7B4%7D%20%5Cpi%20%5Cleft%5B4%5Cright%5D%5C%5C%5C%5C%26%20%3D%20%5Cpi%5Cend%7Baligned%7D)
<u>Simpson's rule</u>
<u />
<u />![\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{3}\cdot \dfrac{1}{2} \pi \left(0+4(1)+2(0)+4(1)+0\right)\\\\& = \dfrac{1}{3}\cdot \dfrac{1}{2} \pi \left(8\right)\\\\& = \dfrac{4}{3} \pi\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cdisplaystyle%20%5Cint%5E%7B2%20%5Cpi%7D_0%20%5Csqrt%5B3%5D%7B%5Csin%5E2%20%28x%29%7D%5C%3A%5C%3A%5Ctext%7Bd%7Dx%20%26%20%5Capprox%20%5Cdfrac%7B1%7D%7B3%7D%5Ccdot%20%5Cdfrac%7B1%7D%7B2%7D%20%5Cpi%20%5Cleft%280%2B4%281%29%2B2%280%29%2B4%281%29%2B0%5Cright%29%5C%5C%5C%5C%26%20%3D%20%5Cdfrac%7B1%7D%7B3%7D%5Ccdot%20%5Cdfrac%7B1%7D%7B2%7D%20%5Cpi%20%5Cleft%288%5Cright%29%5C%5C%5C%5C%26%20%3D%20%5Cdfrac%7B4%7D%7B3%7D%20%5Cpi%5Cend%7Baligned%7D)
Answer:
uhh you didnt give the options
Step-by-step explanation:
10<x<25 mabye
