Answer:
c=6
Step-by-step explanation:
c+25-19=100
isolate variable...
c+25-19+19=100+19
c+25=119
c+25-25=119-25
c=6
Comparing to the standard equation, the parabola with a vertex at (-2,0) is given by:
y = (x + 2)²
<h3>What is the equation of a parabola given it’s vertex?</h3>
The equation of a quadratic function, of vertex (h,k), is given by:
y = a(x - h)² + k
In which a is the leading coefficient.
In this problem, we have that h = -2 and k = 0, and all the options have leading coefficient a = 1, hence the equation is:
y = (x + 2)²
More can be learned about the equation of a parabola at brainly.com/question/17987697
Answer:
a) 31.4 cm b) 22.0 cm c) 9.4 mm d) 18.8 m e) 1.6 m f) 28.3 cm g) 78.5 cm h) 9.7 m
Step-by-step explanation:
All it really is multiply the diameter which is the straight line through the center of the circle
C = d × \pi
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)
