Carlos rode an average of 9 miles every hour on his bicycle. It is because 3*9=27 and the distance is rate multiplied by time. use the formula and check it!
Hope I helped.
Good Luck!
Answer:
<h2>0.00390625 /
![4^{-4}](https://tex.z-dn.net/?f=4%5E%7B-4%7D)
</h2>
Step-by-step explanation:
so you are saying do:
which is 16 /4096
16/4096=0.00390625
<u>another way you can do this is</u>
just subtract the powers
2-6=-4
so the answer is
Split up the integration interval into 4 subintervals:
![\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]](https://tex.z-dn.net/?f=%5Cleft%5B0%2C%5Cdfrac%5Cpi8%5Cright%5D%2C%5Cleft%5B%5Cdfrac%5Cpi8%2C%5Cdfrac%5Cpi4%5Cright%5D%2C%5Cleft%5B%5Cdfrac%5Cpi4%2C%5Cdfrac%7B3%5Cpi%7D8%5Cright%5D%2C%5Cleft%5B%5Cdfrac%7B3%5Cpi%7D8%2C%5Cdfrac%5Cpi2%5Cright%5D)
The left and right endpoints of the
-th subinterval, respectively, are
![\ell_i=\dfrac{i-1}4\left(\dfrac\pi2-0\right)=\dfrac{(i-1)\pi}8](https://tex.z-dn.net/?f=%5Cell_i%3D%5Cdfrac%7Bi-1%7D4%5Cleft%28%5Cdfrac%5Cpi2-0%5Cright%29%3D%5Cdfrac%7B%28i-1%29%5Cpi%7D8)
![r_i=\dfrac i4\left(\dfrac\pi2-0\right)=\dfrac{i\pi}8](https://tex.z-dn.net/?f=r_i%3D%5Cdfrac%20i4%5Cleft%28%5Cdfrac%5Cpi2-0%5Cright%29%3D%5Cdfrac%7Bi%5Cpi%7D8)
for
, and the respective midpoints are
![m_i=\dfrac{\ell_i+r_i}2=\dfrac{(2i-1)\pi}8](https://tex.z-dn.net/?f=m_i%3D%5Cdfrac%7B%5Cell_i%2Br_i%7D2%3D%5Cdfrac%7B%282i-1%29%5Cpi%7D8)
We approximate the (signed) area under the curve over each subinterval by
![T_i=\dfrac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)](https://tex.z-dn.net/?f=T_i%3D%5Cdfrac%7Bf%28%5Cell_i%29%2Bf%28r_i%29%7D2%28%5Cell_i-r_i%29)
so that
![\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4T_i\approx\boxed{3.038078}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E%7B%5Cpi%2F2%7D%5Cfrac3%7B1%2B%5Ccos%20x%7D%5C%2C%5Cmathrm%20dx%5Capprox%5Csum_%7Bi%3D1%7D%5E4T_i%5Capprox%5Cboxed%7B3.038078%7D)
We approximate the area for each subinterval by
![M_i=f(m_i)(\ell_i-r_i)](https://tex.z-dn.net/?f=M_i%3Df%28m_i%29%28%5Cell_i-r_i%29)
so that
![\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4M_i\approx\boxed{2.981137}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E%7B%5Cpi%2F2%7D%5Cfrac3%7B1%2B%5Ccos%20x%7D%5C%2C%5Cmathrm%20dx%5Capprox%5Csum_%7Bi%3D1%7D%5E4M_i%5Capprox%5Cboxed%7B2.981137%7D)
We first interpolate the integrand over each subinterval by a quadratic polynomial
, where
![p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}](https://tex.z-dn.net/?f=p_i%28x%29%3Df%28%5Cell_i%29%5Cdfrac%7B%28x-m_i%29%28x-r_i%29%7D%7B%28%5Cell_i-m_i%29%28%5Cell_i-r_i%29%7D%2Bf%28m%29%5Cdfrac%7B%28x-%5Cell_i%29%28x-r_i%29%7D%7B%28m_i-%5Cell_i%29%28m_i-r_i%29%7D%2Bf%28r_i%29%5Cdfrac%7B%28x-%5Cell_i%29%28x-m_i%29%7D%7B%28r_i-%5Cell_i%29%28r_i-m_i%29%7D)
so that
![\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E%7B%5Cpi%2F2%7D%5Cfrac3%7B1%2B%5Ccos%20x%7D%5C%2C%5Cmathrm%20dx%5Capprox%5Csum_%7Bi%3D1%7D%5E4%5Cint_%7B%5Cell_i%7D%5E%7Br_i%7Dp_i%28x%29%5C%2C%5Cmathrm%20dx)
It so happens that the integral of
reduces nicely to the form you're probably more familiar with,
![S_i=\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))](https://tex.z-dn.net/?f=S_i%3D%5Cdisplaystyle%5Cint_%7B%5Cell_i%7D%5E%7Br_i%7Dp_i%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac%7Br_i-%5Cell_i%7D6%28f%28%5Cell_i%29%2B4f%28m_i%29%2Bf%28r_i%29%29)
Then the integral is approximately
![\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E%7B%5Cpi%2F2%7D%5Cfrac3%7B1%2B%5Ccos%20x%7D%5C%2C%5Cmathrm%20dx%5Capprox%5Csum_%7Bi%3D1%7D%5E4S_i%5Capprox%5Cboxed%7B3.000117%7D)
Compare these to the actual value of the integral, 3. I've included plots of the approximations below.
Answer: second option.
Step-by-step explanation:
You must apply ![sin\alpha=\frac{opposite}{hypotenuse}](https://tex.z-dn.net/?f=sin%5Calpha%3D%5Cfrac%7Bopposite%7D%7Bhypotenuse%7D)
Then, given the right triangle and the angle D, the opposite side and the hypotenuse of the triangle are the following:
![opposite=20\\hypotenuse=25](https://tex.z-dn.net/?f=opposite%3D20%5C%5Chypotenuse%3D25)
Therefore, when you substitute values, you have:
![sinD=\frac{20}{25}](https://tex.z-dn.net/?f=sinD%3D%5Cfrac%7B20%7D%7B25%7D)
Reduce the fraction, then you obtain:
![sinD=\frac{4}{5}](https://tex.z-dn.net/?f=sinD%3D%5Cfrac%7B4%7D%7B5%7D)