Split up the interval [1, 9] into <em>n</em> subintervals of equal length (9 - 1)/<em>n</em> = 8/<em>n</em> :
[1, 1 + 8/<em>n</em>], [1 + 8/<em>n</em>, 1 + 16/<em>n</em>], [1 + 16/<em>n</em>, 1 + 24/<em>n</em>], …, [1 + 8 (<em>n</em> - 1)/<em>n</em>, 9]
It should be clear that the left endpoint of each subinterval make up an arithmetic sequence, so that the <em>i</em>-th subinterval has left endpoint
1 + 8/<em>n</em> (<em>i</em> - 1)
Then we approximate the definite integral by the sum of the areas of <em>n</em> rectangles with length 8/<em>n</em> and height
:
![\displaystyle \int_1^9 (x^2-4x+6) \,\mathrm dx \approx \sum_{i=1}^n \frac8n\left(\left(1+\frac8n(i-1)\right)^2-4\left(1+\frac8n(i-1)\right)+6\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_1%5E9%20%28x%5E2-4x%2B6%29%20%5C%2C%5Cmathrm%20dx%20%5Capprox%20%5Csum_%7Bi%3D1%7D%5En%20%5Cfrac8n%5Cleft%28%5Cleft%281%2B%5Cfrac8n%28i-1%29%5Cright%29%5E2-4%5Cleft%281%2B%5Cfrac8n%28i-1%29%5Cright%29%2B6%5Cright%29)
Take the limit as <em>n</em> approaches infinity and the approximation becomes exact. So we have
![\displaystyle \int_1^9 (x^2-4x+6) \,\mathrm dx = \lim_{n\to\infty} \sum_{i=1}^n \frac8n\left(\left(1+\frac8n(i-1)\right)^2-4\left(1+\frac8n(i-1)\right)+6\right) \\\\ = \lim_{n\to\infty} \frac8n \sum_{i=1}^n \left(1+\frac{16}n(i-1)+\frac{64}{n^2}(i-1)^2-4-\frac{32}n(i-1)+6\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \sum_{i=1}^n \left(64(i-1)^2-16n(i-1)+3n^2\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \sum_{i=0}^{n-1} \left(64i^2-16ni+3n^2\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \left(64\sum_{i=0}^{n-1}i^2 - 16n\sum_{i=0}^{n-1}i + 3n^2\sum{i=0}^{n-1}1\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \left(\frac{64(2n-1)n(n-1)}{6} - \frac{16n^2(n-1)}{2} + 3n^3\right) \\\\= \lim_{n\to\infty} \frac8{n^3} \left(\frac{49n^3}3-24n^2+\frac{32n}3\right) \\\\= \lim_{n\to\infty} \frac{8\left(49n^2-72n+32\right)}{3n^2} = \boxed{\frac{392}3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_1%5E9%20%28x%5E2-4x%2B6%29%20%5C%2C%5Cmathrm%20dx%20%3D%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Csum_%7Bi%3D1%7D%5En%20%5Cfrac8n%5Cleft%28%5Cleft%281%2B%5Cfrac8n%28i-1%29%5Cright%29%5E2-4%5Cleft%281%2B%5Cfrac8n%28i-1%29%5Cright%29%2B6%5Cright%29%20%5C%5C%5C%5C%20%3D%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cfrac8n%20%5Csum_%7Bi%3D1%7D%5En%20%5Cleft%281%2B%5Cfrac%7B16%7Dn%28i-1%29%2B%5Cfrac%7B64%7D%7Bn%5E2%7D%28i-1%29%5E2-4-%5Cfrac%7B32%7Dn%28i-1%29%2B6%5Cright%29%20%5C%5C%5C%5C%3D%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cfrac8%7Bn%5E3%7D%20%5Csum_%7Bi%3D1%7D%5En%20%5Cleft%2864%28i-1%29%5E2-16n%28i-1%29%2B3n%5E2%5Cright%29%20%5C%5C%5C%5C%3D%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cfrac8%7Bn%5E3%7D%20%5Csum_%7Bi%3D0%7D%5E%7Bn-1%7D%20%5Cleft%2864i%5E2-16ni%2B3n%5E2%5Cright%29%20%5C%5C%5C%5C%3D%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cfrac8%7Bn%5E3%7D%20%5Cleft%2864%5Csum_%7Bi%3D0%7D%5E%7Bn-1%7Di%5E2%20-%2016n%5Csum_%7Bi%3D0%7D%5E%7Bn-1%7Di%20%2B%203n%5E2%5Csum%7Bi%3D0%7D%5E%7Bn-1%7D1%5Cright%29%20%5C%5C%5C%5C%3D%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cfrac8%7Bn%5E3%7D%20%5Cleft%28%5Cfrac%7B64%282n-1%29n%28n-1%29%7D%7B6%7D%20-%20%5Cfrac%7B16n%5E2%28n-1%29%7D%7B2%7D%20%2B%203n%5E3%5Cright%29%20%5C%5C%5C%5C%3D%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cfrac8%7Bn%5E3%7D%20%5Cleft%28%5Cfrac%7B49n%5E3%7D3-24n%5E2%2B%5Cfrac%7B32n%7D3%5Cright%29%20%5C%5C%5C%5C%3D%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Cfrac%7B8%5Cleft%2849n%5E2-72n%2B32%5Cright%29%7D%7B3n%5E2%7D%20%3D%20%5Cboxed%7B%5Cfrac%7B392%7D3%7D)
It is 5 then because 5×5 LIKE five times will at leaste be five i think
Answer:
A
Step-by-step explanation:
180-105=75
180=75+55+x
x=50
Start by taking out a common factor.
x^10
x^10(x^6 - 2x^5 - x^4 + 4x^3 - x^2 - 2x + 1) = 0
So far we know that D won't work.
x - 1 is a factor because putting 1 in for the xs in the expression inside the brackets gives 0 Now you need to do a division. I'm going to assume you can do that.
Using division, I get
x^5 - x^4 - 2x^3 + 2x^2 + x - 1 Now divide x - 1 into this mess again. You get
x^4 - 2x^2 + 1 which factors into
(x^2 - 1)(x^2 - 1) which factors into
(x - 1)(x + 1)(x - 1)(x + 1) = 0
The first two divisions add (x -1)(x - 1)
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So we have (x - 1) with a multiplicity of 4 and (x + 1) with a multiplicity of 2
or x = 1 with a multiplicity of 4 and x = -1 with a multiplicity of 2 and x = 0
with a multiplicity of 10
That's the answer.
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C <<<< ===== Answer.