Answer:
Step-by-step explanation:
Since the bag of rice is 4.5 pounds, and there are 16 ounces per pound, multiply 16 x 4 = 64. Now what is half of 16? 8! So divide 5/8 = 0.625.
So in the end, yes the bag of rice will have enough rice. Here's my math:
16 x 4 = 64
5 / 8 = 0.625
So, you will have 64.625 ounces of rice which is plenty for your recipe!
-
-
-
I really hope this helps you!
It’s neither of the graphs that are showed in the photo
Answer:

The interval of convergence is:
Step-by-step explanation:
Given


The geometric series centered at c is of the form:

Where:
first term
common ratio
We have to write

In the following form:

So, we have:

Rewrite as:


Factorize

Open bracket

Rewrite as:

Collect like terms

Take LCM


So, we have:

By comparison with: 



At c = 6, we have:

Take LCM

r = -\frac{1}{3}(x + \frac{11}{3}+6-6)
So, the power series becomes:

Substitute 1 for a


Substitute the expression for r

Expand
![\frac{9}{3x + 2} = \sum\limits^{\infty}_{n=0}[(-\frac{1}{3})^n* (x - \frac{7}{3})^n]](https://tex.z-dn.net/?f=%5Cfrac%7B9%7D%7B3x%20%2B%202%7D%20%3D%20%20%5Csum%5Climits%5E%7B%5Cinfty%7D_%7Bn%3D0%7D%5B%28-%5Cfrac%7B1%7D%7B3%7D%29%5En%2A%20%28x%20-%20%5Cfrac%7B7%7D%7B3%7D%29%5En%5D)
Further expand:

The power series converges when:

Multiply both sides by 3

Expand the absolute inequality

Solve for x

Take LCM


The interval of convergence is: