Step-by-step explanation:
could you make the question full please
Answer:
![\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx\approx 3.099558](https://tex.z-dn.net/?f=%5Cint_%7B0%7D%5E%7B%5Cfrac%7B3%20%5Cpi%7D%7B4%7D%7D3%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%5C%20dx%5Capprox%203.099558)
Step-by-step explanation:
We want to find the Riemann sum for
with n = 6, using left endpoints.
The Left Riemann Sum uses the left endpoints of a sub-interval:
![\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f(x_0)+f(x_1)+2f(x_2)+...+f(x_{n-2})+f(x_{n-1})\right)](https://tex.z-dn.net/?f=%5Cint_%7Ba%7D%5E%7Bb%7Df%28x%29dx%5Capprox%5CDelta%7Bx%7D%5Cleft%28f%28x_0%29%2Bf%28x_1%29%2B2f%28x_2%29%2B...%2Bf%28x_%7Bn-2%7D%29%2Bf%28x_%7Bn-1%7D%29%5Cright%29)
where
.
Step 1: Find ![\Delta{x}](https://tex.z-dn.net/?f=%5CDelta%7Bx%7D)
We have that ![a=0, b=\frac{3\pi }{4}, n=6](https://tex.z-dn.net/?f=a%3D0%2C%20b%3D%5Cfrac%7B3%5Cpi%20%7D%7B4%7D%2C%20n%3D6)
Therefore, ![\Delta{x}=\frac{\frac{3 \pi}{4}-0}{6}=\frac{\pi}{8}](https://tex.z-dn.net/?f=%5CDelta%7Bx%7D%3D%5Cfrac%7B%5Cfrac%7B3%20%5Cpi%7D%7B4%7D-0%7D%7B6%7D%3D%5Cfrac%7B%5Cpi%7D%7B8%7D)
Step 2: Divide the interval
into n = 6 sub-intervals of length ![\Delta{x}=\frac{\pi}{8}](https://tex.z-dn.net/?f=%5CDelta%7Bx%7D%3D%5Cfrac%7B%5Cpi%7D%7B8%7D)
![a=\left[0, \frac{\pi}{8}\right], \left[\frac{\pi}{8}, \frac{\pi}{4}\right], \left[\frac{\pi}{4}, \frac{3 \pi}{8}\right], \left[\frac{3 \pi}{8}, \frac{\pi}{2}\right], \left[\frac{\pi}{2}, \frac{5 \pi}{8}\right], \left[\frac{5 \pi}{8}, \frac{3 \pi}{4}\right]=b](https://tex.z-dn.net/?f=a%3D%5Cleft%5B0%2C%20%5Cfrac%7B%5Cpi%7D%7B8%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B%5Cpi%7D%7B8%7D%2C%20%5Cfrac%7B%5Cpi%7D%7B4%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B%5Cpi%7D%7B4%7D%2C%20%5Cfrac%7B3%20%5Cpi%7D%7B8%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B3%20%5Cpi%7D%7B8%7D%2C%20%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B%5Cpi%7D%7B2%7D%2C%20%5Cfrac%7B5%20%5Cpi%7D%7B8%7D%5Cright%5D%2C%20%5Cleft%5B%5Cfrac%7B5%20%5Cpi%7D%7B8%7D%2C%20%5Cfrac%7B3%20%5Cpi%7D%7B4%7D%5Cright%5D%3Db)
Step 3: Evaluate the function at the left endpoints
![f\left(x_{0}\right)=f(a)=f\left(0\right)=3=3](https://tex.z-dn.net/?f=f%5Cleft%28x_%7B0%7D%5Cright%29%3Df%28a%29%3Df%5Cleft%280%5Cright%29%3D3%3D3)
![f\left(x_{1}\right)=f\left(\frac{\pi}{8}\right)=3 \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}=2.77163859753386](https://tex.z-dn.net/?f=f%5Cleft%28x_%7B1%7D%5Cright%29%3Df%5Cleft%28%5Cfrac%7B%5Cpi%7D%7B8%7D%5Cright%29%3D3%20%5Csqrt%7B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B4%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7D%7D%3D2.77163859753386)
![f\left(x_{2}\right)=f\left(\frac{\pi}{4}\right)=\frac{3 \sqrt{2}}{2}=2.12132034355964](https://tex.z-dn.net/?f=f%5Cleft%28x_%7B2%7D%5Cright%29%3Df%5Cleft%28%5Cfrac%7B%5Cpi%7D%7B4%7D%5Cright%29%3D%5Cfrac%7B3%20%5Csqrt%7B2%7D%7D%7B2%7D%3D2.12132034355964)
![f\left(x_{3}\right)=f\left(\frac{3 \pi}{8}\right)=3 \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}=1.14805029709527](https://tex.z-dn.net/?f=f%5Cleft%28x_%7B3%7D%5Cright%29%3Df%5Cleft%28%5Cfrac%7B3%20%5Cpi%7D%7B8%7D%5Cright%29%3D3%20%5Csqrt%7B%5Cfrac%7B1%7D%7B2%7D%20-%20%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B4%7D%7D%3D1.14805029709527)
![f\left(x_{4}\right)=f\left(\frac{\pi}{2}\right)=0=0](https://tex.z-dn.net/?f=f%5Cleft%28x_%7B4%7D%5Cright%29%3Df%5Cleft%28%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cright%29%3D0%3D0)
![f\left(x_{5}\right)=f\left(\frac{5 \pi}{8}\right)=- 3 \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}=-1.14805029709527](https://tex.z-dn.net/?f=f%5Cleft%28x_%7B5%7D%5Cright%29%3Df%5Cleft%28%5Cfrac%7B5%20%5Cpi%7D%7B8%7D%5Cright%29%3D-%203%20%5Csqrt%7B%5Cfrac%7B1%7D%7B2%7D%20-%20%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B4%7D%7D%3D-1.14805029709527)
Step 4: Apply the Left Riemann Sum formula
![\frac{\pi}{8}(3+2.77163859753386+2.12132034355964+1.14805029709527+0-1.14805029709527)=3.09955772805315](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%7D%7B8%7D%283%2B2.77163859753386%2B2.12132034355964%2B1.14805029709527%2B0-1.14805029709527%29%3D3.09955772805315)
![\int_{0}^{\frac{3 \pi}{4}}3 \cos{\left(x \right)}\ dx\approx 3.099558](https://tex.z-dn.net/?f=%5Cint_%7B0%7D%5E%7B%5Cfrac%7B3%20%5Cpi%7D%7B4%7D%7D3%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%5C%20dx%5Capprox%203.099558)
<h2><u>Answers and Explanation</u></h2>
<u>(a) Betty’s expression:</u>
Cost of guest = 10 × x + 10 × x + 10 × x
= 10x + 10x + 10x
cost of the venue = 80
Total cost =<em> 10x + 10x + 10x + 80</em>
<u>(b) Dawnie’s expression:</u>
Cost of the location = $80
Number of guests = x + x + x
Cost of all guests = $10(x + x + x)
Total cost = <em> 10(x + x + x) + 80</em>
<u>(c) Joan’s expression:</u>
Cost of the location = 80
Cost of guest = 10 × x + 10 × x + 10 × x
= 10x + 10x + 10x
= x(10 + 10 + 10) ⇒ After factoring out x.
Total cost =<em> x(10 + 10 + 10) + 80</em>
Answer:
we sell orange at market is q
Parentheses indicate multiplication (no need to use the "*" sign).
So the question is basically asking for the usage of multiplication and addition only.
To get 13 through the given combination using only parentheses and addition, you can write it as follows:
3 + 2(4+1) = 13
Now, let's check this:
3 + 2(4+1)
= 3 + 2(5)
= 3 + 10
= 13 which is the required output.