Answer:
-6
Step-by-step explanation:
Given that :
we are to evaluate the Riemann sum for
from 2 ≤ x ≤ 14
where the endpoints are included with six subintervals, taking the sample points to be the left endpoints.
The Riemann sum can be computed as follows:
![L_6 = \int ^{14}_{2}3- \dfrac{1}{2}x \dx = \lim_{n \to \infty} \sum \limits ^6 _{i=1} \ f (x_i -1) \Delta x](https://tex.z-dn.net/?f=L_6%20%3D%20%5Cint%20%5E%7B14%7D_%7B2%7D3-%20%5Cdfrac%7B1%7D%7B2%7Dx%20%5Cdx%20%3D%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Csum%20%5Climits%20%5E6%20_%7Bi%3D1%7D%20%5C%20f%20%28x_i%20-1%29%20%5CDelta%20x)
where:
![\Delta x = \dfrac{b-a}{a}](https://tex.z-dn.net/?f=%5CDelta%20x%20%3D%20%5Cdfrac%7Bb-a%7D%7Ba%7D)
a = 2
b =14
n = 6
∴
![\Delta x = \dfrac{14-2}{6}](https://tex.z-dn.net/?f=%5CDelta%20x%20%3D%20%5Cdfrac%7B14-2%7D%7B6%7D)
![\Delta x = \dfrac{12}{6}](https://tex.z-dn.net/?f=%5CDelta%20x%20%3D%20%5Cdfrac%7B12%7D%7B6%7D)
![\Delta x =2](https://tex.z-dn.net/?f=%5CDelta%20x%20%3D2)
Hence;
![x_0 = 2 \\ \\ x_1 = 2+2 =4\\ \\ x_2 = 2 + 2(2) \\ \\ x_i = 2 + 2i](https://tex.z-dn.net/?f=x_0%20%3D%202%20%5C%5C%20%5C%5C%20%20x_1%20%3D%202%2B2%20%3D4%5C%5C%20%5C%5C%20%20x_2%20%3D%202%20%2B%202%282%29%20%5C%5C%20%5C%5C%20%20x_i%20%3D%202%20%2B%202i)
Here, we are using left end-points, then:
![x_i-1 = 2+ 2(i-1)](https://tex.z-dn.net/?f=x_i-1%20%3D%202%2B%202%28i-1%29)
Replacing it into Riemann equation;
![L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} \begin {pmatrix}3 - \dfrac{1}{2} \begin {pmatrix} 2+2 (i-1) \end {pmatrix} \end {pmatrix}2](https://tex.z-dn.net/?f=L_6%20%3D%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%20%5Csum%20%5Cimits%20%5E%7B6%7D_%7Bi%3D1%7D%20%5Cbegin%20%7Bpmatrix%7D3%20-%20%5Cdfrac%7B1%7D%7B2%7D%20%5Cbegin%20%7Bpmatrix%7D%20%202%2B2%20%28i-1%29%20%20%5Cend%20%7Bpmatrix%7D%20%5Cend%20%7Bpmatrix%7D2)
![L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 - (2+2(i-1))](https://tex.z-dn.net/?f=L_6%20%3D%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%20%5Csum%20%5Cimits%20%5E%7B6%7D_%7Bi%3D1%7D%206%20-%20%282%2B2%28i-1%29%29)
![L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 - (2+2i-2)](https://tex.z-dn.net/?f=L_6%20%3D%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%20%5Csum%20%5Cimits%20%5E%7B6%7D_%7Bi%3D1%7D%206%20-%20%282%2B2i-2%29)
![L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 -2i](https://tex.z-dn.net/?f=L_6%20%3D%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%20%5Csum%20%5Cimits%20%5E%7B6%7D_%7Bi%3D1%7D%206%20-2i)
![L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 - \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 2i](https://tex.z-dn.net/?f=L_6%20%3D%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%20%5Csum%20%5Cimits%20%5E%7B6%7D_%7Bi%3D1%7D%206%20-%20%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%20%5Csum%20%5Cimits%20%5E%7B6%7D_%7Bi%3D1%7D%202i)
![L_6 = \lim_{n \to \infty} \sum \imits ^{6}_{i=1} 6 - 2 \lim_{n \to \infty} \sum \imits ^{6}_{i=1} i](https://tex.z-dn.net/?f=L_6%20%3D%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%20%5Csum%20%5Cimits%20%5E%7B6%7D_%7Bi%3D1%7D%206%20-%202%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%20%5Csum%20%5Cimits%20%5E%7B6%7D_%7Bi%3D1%7D%20i)
Estimating the integrals, we have :
![= 6n - 2 ( \dfrac{n(n-1)}{2})](https://tex.z-dn.net/?f=%3D%206n%20-%202%20%28%20%5Cdfrac%7Bn%28n-1%29%7D%7B2%7D%29)
= 6n - n(n+1)
replacing thevalue of n = 6 (i.e the sub interval number), we have:
= 6(6) - 6(6+1)
= 36 - 36 -6
= -6
Let E=elevation range
Low elevation=1700≤E≤2500<span>
Mid elevation= 2500</span>≤E≤4000<span>
Sulbarine=4000</span>≤E≤6500<span>
Alpine=6500</span>≤E<span>≤14,410</span>
This actually a puzzle, not a math problem.
Write it out with no commas:
101112131415161
it actually says 10, 11, 12, etc.
so the next number would be 718
For this case we have to define root properties:
![\sqrt [n] {a ^ n} = a ^ {\frac {n} {n}} = a](https://tex.z-dn.net/?f=%5Csqrt%20%5Bn%5D%20%7Ba%20%5E%20n%7D%20%3D%20a%20%5E%20%7B%5Cfrac%20%7Bn%7D%20%7Bn%7D%7D%20%3D%20a)
In addition, we know that:
![a ^ {- 1} = \frac {1} {a ^ 1} = \frac {1} {a}](https://tex.z-dn.net/?f=a%20%5E%20%7B-%201%7D%20%3D%20%5Cfrac%20%7B1%7D%20%7Ba%20%5E%201%7D%20%3D%20%5Cfrac%20%7B1%7D%20%7Ba%7D)
On the other hand:
![4 ^ 2 = 16](https://tex.z-dn.net/?f=4%20%5E%202%20%3D%2016)
Thus, we can rewrite the given expression as:
![\sqrt {4 ^ 2 * a ^ 8 * \frac {1} {b ^ 2}} =\\4 ^ {\frac {2} {2}} * a ^ {\frac {8} {2}} * \frac {1} {b ^ {\frac {2} {2}}} =\\4 * a ^ 4 * \frac {1} {b}](https://tex.z-dn.net/?f=%5Csqrt%20%7B4%20%5E%202%20%2A%20a%20%5E%208%20%2A%20%5Cfrac%20%7B1%7D%20%7Bb%20%5E%202%7D%7D%20%3D%5C%5C4%20%5E%20%7B%5Cfrac%20%7B2%7D%20%7B2%7D%7D%20%2A%20a%20%5E%20%7B%5Cfrac%20%7B8%7D%20%7B2%7D%7D%20%2A%20%5Cfrac%20%7B1%7D%20%7Bb%20%5E%20%7B%5Cfrac%20%7B2%7D%20%7B2%7D%7D%7D%20%3D%5C%5C4%20%2A%20a%20%5E%204%20%2A%20%5Cfrac%20%7B1%7D%20%7Bb%7D)
ANswer:
Option B