1 equals 0.46 (make a line over the 6 to show it is repeated)
2 equals 0.4 (make a line over the 4 to show it is repeated)
3 equals -0.6 (make a line over the 6 to show it is repeated)
4 equals -0.857142 (make a line over all numbers past the decimal to show it is repeated)
5 equals 3.3409 (make a line over the 09 at the end to show it is repeated)
6 equals -1.7727 (make a line over the 27 at the end to show it is repeated)
7 equals 66.5%
8 equals 3.16 (make a line over the 6 to show it is repeated)
9 equals -9 over 10
10 equals -17 over 20
11 equals -3 and 4 over 5
Answer:
Given definite integral as a limit of Riemann sums is:
![\lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]](https://tex.z-dn.net/?f=%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Csum%5E%7Bn%7D%20_%7Bi%3D1%7D3%5B%5Cfrac%7B9%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B%5Cfrac%7B36%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%2B%5Cfrac%7B97%7D%7B2n%7Di%2B22%5D)
Step-by-step explanation:
Given definite integral is:
Substituting (2) in above
![f(x_{i})=\frac{1}{2}(4+\frac{3}{n}i)+(4+\frac{3}{n}i)^{3}\\\\f(x_{i})=(2+\frac{3}{2n}i)+(64+\frac{27}{n^{3}}i^{3}+3(16)\frac{3}{n}i+3(4)\frac{9}{n^{2}}i^{2})\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{3}{2n}i+\frac{144}{n}i+66\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{291}{2n}i+66\\\\f(x_{i})=3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]](https://tex.z-dn.net/?f=f%28x_%7Bi%7D%29%3D%5Cfrac%7B1%7D%7B2%7D%284%2B%5Cfrac%7B3%7D%7Bn%7Di%29%2B%284%2B%5Cfrac%7B3%7D%7Bn%7Di%29%5E%7B3%7D%5C%5C%5C%5Cf%28x_%7Bi%7D%29%3D%282%2B%5Cfrac%7B3%7D%7B2n%7Di%29%2B%2864%2B%5Cfrac%7B27%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B3%2816%29%5Cfrac%7B3%7D%7Bn%7Di%2B3%284%29%5Cfrac%7B9%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%29%5C%5C%5C%5Cf%28x_%7Bi%7D%29%3D%5Cfrac%7B27%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B%5Cfrac%7B108%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%2B%5Cfrac%7B3%7D%7B2n%7Di%2B%5Cfrac%7B144%7D%7Bn%7Di%2B66%5C%5C%5C%5Cf%28x_%7Bi%7D%29%3D%5Cfrac%7B27%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B%5Cfrac%7B108%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%2B%5Cfrac%7B291%7D%7B2n%7Di%2B66%5C%5C%5C%5Cf%28x_%7Bi%7D%29%3D3%5B%5Cfrac%7B9%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B%5Cfrac%7B36%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%2B%5Cfrac%7B97%7D%7B2n%7Di%2B22%5D)
Riemann sum is:
![= \lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]](https://tex.z-dn.net/?f=%3D%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Csum%5E%7Bn%7D%20_%7Bi%3D1%7D3%5B%5Cfrac%7B9%7D%7Bn%5E%7B3%7D%7Di%5E%7B3%7D%2B%5Cfrac%7B36%7D%7Bn%5E%7B2%7D%7Di%5E%7B2%7D%2B%5Cfrac%7B97%7D%7B2n%7Di%2B22%5D)
The volume of the first cube is (5h^2)^3, while the volume of the second cube is (3k)^3, so their total volume is (5h^2)^3 + (3k)^3. We can use the special formula for factoring a sum of two cubes:
x^3 + y^3 = (x + y)(x^2 - xy + y^2)
(5h^2)^3 + (3k)^3 = (5h^2 + 3k)((5h^2)^2 - (5h^2)(3k) + (3k)^2)
= (5h^2 + 3k)(25h^4 - 15(h^2)(k) + 9k^2)
This is the second of the given choices.
Answer:
Javier has 27, Tommy has 16
Step-by-step explanation:
43-11=32
32/2=16
16+11=27
Check: 27+16=43
