X = 1 , y = - 2
Hope this helped
Answer:
Step-by-step explanation:
I can not see your answers to choose but with this equation for 2 variables, you can replace the values of x and y and see if that is right or not.
For example, to know (-2,5) is a solution, you replace x= -2 and y= 5
into this equation: -3x -y = 6
and you have: -3*(-2) -5 =6
or 6-5 =6, and you find that is impossible, so (-2.5) is not a solution.
Hope you understand it
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)
For determining the period of a function, you pick two recognizable tops of the graph and calculate the time difference between them. In this situation the period is 5-1=4, by picking the first two tops
