Answer:
<u>.</u><u>5</u><u>x</u> <u>></u> <u>1</u><u>0</u>
.5 .5 x<u>></u>2
●________>
<u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>
0 1 2 3 4
actually it will be easier for you to know it if you convert it into decimal.
So
James = 4/5 = 0.8
Eric = 7/9 = 0.78
So James ate more
Answer:

And when we apply the limit we got that:

Step-by-step explanation:
Assuming this complete problem: "The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit . 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2"
We have the following formula in order to find the sum of cubes:

We can express this formula like this:
![\lim_{n\to\infty} \sum_{n=1}^{\infty}i^3 =\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2](https://tex.z-dn.net/?f=%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7Di%5E3%20%3D%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5B%5Cfrac%7Bn%28n%2B1%29%7D%7B2%7D%5D%5E2)
And using this property we need to proof that: 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2
![\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2](https://tex.z-dn.net/?f=%20%5Clim_%7Bn%5Cto%5Cinfty%7D%20%5B%5Cfrac%7Bn%28n%2B1%29%7D%7B2%7D%5D%5E2)
If we operate and we take out the 1/4 as a factor we got this:

We can cancel
and we got

We can reorder the terms like this:

We can do some algebra and we got:

We can solve the square and we got:

And when we apply the limit we got that:

The answer is 1.49942857143.
Or just to sum it up it's 1.499.
Answer:
Rotating Q 180 degrees using the center P has the same effect as reflecting Q over the Line M
Step-by-step explanation:
Rotating Q 180 degrees using the center P has the same effect as reflecting Q over the Line M and this is because Lines L and M are perpendicular lines ( i.e. lines that meet a right angle ( 90° ).