Answer: 10 times
Step-by-step explanation:
1 litre = 1000 ml
6 litres = 6000 ml
Lawnmover holds 600 ml
So 6000 ml will fill
6000/600 = 10 times
<span>You could set up the relation as a table of ordered pairs. Then, test to see if each element in the domain is matched with exactly one element in the range. If so, you have a function! Watch this tutorial to see how you can determine if a relation is a function.</span>
Substitute what you know y is equal to into the equation:
y = x + 2
3/2x - 2 = x + 2
Then you can simplify your equation by multiplying both sides by 2 to eliminate the fraction which will give you:
3x - 4 = 2x + 4
Add the -4 to the other side to get 8 and subtract the 2x from the 3x so you have:
x = 8 Now solve for y by substituting your value for x into the other equation:
y = 3/2(8) - 2 Simplify by combining your numbers and you get:
y = 10 Check your answer by substituting y and x into the original equation:
(10) = (8) - 2
Correctly done! So your answers are x = 8 and y = 10
Answer:
48cm
Step-by-step explanation:
Answer:
Step-by-step explanation:
The definite integral of a continuous function <em>f</em> over the interval [a,b] denoted by 
, is the limit of a Riemann sum as the number of subdivisions approaches infinity. That is,
where 
and 
To evaluate the integral
you must:
Find 
Find 
Therefore,
![\lim_{n \to \infty}\frac{2}{n} \sum_{i=1}^{n} 7(-2+\frac{2i}{n})^{2} +7(-2+\frac{2i}{n})\\\\\lim_{n \to \infty}\frac{2}{n} \sum_{i=1}^{n} 7[(-2+\frac{2i}{n})^{2} +(-2+\frac{2i}{n})]\\\\\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} (-2+\frac{2i}{n})^{2} +(-2+\frac{2i}{n})](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7B2%7D%7Bn%7D%20%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%207%28-2%2B%5Cfrac%7B2i%7D%7Bn%7D%29%5E%7B2%7D%20%2B7%28-2%2B%5Cfrac%7B2i%7D%7Bn%7D%29%5C%5C%5C%5C%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7B2%7D%7Bn%7D%20%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%207%5B%28-2%2B%5Cfrac%7B2i%7D%7Bn%7D%29%5E%7B2%7D%20%2B%28-2%2B%5Cfrac%7B2i%7D%7Bn%7D%29%5D%5C%5C%5C%5C%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7B14%7D%7Bn%7D%20%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20%28-2%2B%5Cfrac%7B2i%7D%7Bn%7D%29%5E%7B2%7D%20%2B%28-2%2B%5Cfrac%7B2i%7D%7Bn%7D%29)
![\lim_{n \to \infty}\frac{14}{n} \sum_{i=1}^{n} \frac{4i^2}{n^2}-\frac{6i}{n}+2\\\\\lim_{n \to \infty}\frac{14}{n}[ \sum_{i=1}^{n} \frac{4i^2}{n^2}-\sum_{i=1}^{n}\frac{6i}{n}+\sum_{i=1}^{n}2]\\\\\lim_{n \to \infty}\frac{14}{n}[ \frac{4}{n^2}\sum_{i=1}^{n}i^2 -\frac{6}{n}\sum_{i=1}^{n}i+\sum_{i=1}^{n}2]](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7B14%7D%7Bn%7D%20%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20%5Cfrac%7B4i%5E2%7D%7Bn%5E2%7D-%5Cfrac%7B6i%7D%7Bn%7D%2B2%5C%5C%5C%5C%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7B14%7D%7Bn%7D%5B%20%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20%5Cfrac%7B4i%5E2%7D%7Bn%5E2%7D-%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%5Cfrac%7B6i%7D%7Bn%7D%2B%5Csum_%7Bi%3D1%7D%5E%7Bn%7D2%5D%5C%5C%5C%5C%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7B14%7D%7Bn%7D%5B%20%5Cfrac%7B4%7D%7Bn%5E2%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7Di%5E2%20-%5Cfrac%7B6%7D%7Bn%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7Di%2B%5Csum_%7Bi%3D1%7D%5E%7Bn%7D2%5D)
We can use the facts that
![\lim_{n \to \infty}\frac{14}{n}[ \frac{4}{n^2}\cdot \frac{n(n+1)(2n+1)}{6}-\frac{6}{n}\cdot \frac{n(n+1)}{2}+2n]\\\\\lim_{n \to \infty}\frac{14}{n}[-n+\frac{2\left(n+1\right)\left(2n+1\right)}{3n}-3]\\\\\lim_{n \to \infty}\frac{14\left(n^2-3n+2\right)}{3n^2}](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7B14%7D%7Bn%7D%5B%20%5Cfrac%7B4%7D%7Bn%5E2%7D%5Ccdot%20%5Cfrac%7Bn%28n%2B1%29%282n%2B1%29%7D%7B6%7D-%5Cfrac%7B6%7D%7Bn%7D%5Ccdot%20%20%5Cfrac%7Bn%28n%2B1%29%7D%7B2%7D%2B2n%5D%5C%5C%5C%5C%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7B14%7D%7Bn%7D%5B-n%2B%5Cfrac%7B2%5Cleft%28n%2B1%5Cright%29%5Cleft%282n%2B1%5Cright%29%7D%7B3n%7D-3%5D%5C%5C%5C%5C%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%5Cfrac%7B14%5Cleft%28n%5E2-3n%2B2%5Cright%29%7D%7B3n%5E2%7D)
Thus,
