The volume of the bag of nuts is the amount of nuts it can contain
The volume of the bag of nuts is 84.78 cubic inches
<h3>How to determine the volume of the bag?</h3>
From the figure, we have the following parameters:
- Radius, r = 3 in
- Height, h = 9 in
The volume of a cone is calculated using:
V = 1/3 πr^2h
Substitute the known values
V = 1/3 * 3.14 * 3^2 * 9
Evaluate the product
V = 84.78
Hence, the volume of the bag of nuts is 84.78 cubic inches
Read more about volumes at:
brainly.com/question/1972490
Answer:

Step-by-step explanation:

hope this helps
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Let's convert 8 pounds and 3 ounces, into all ounces.
So we know one pound = 16 ounces.
1 lb = 16 oz.
Now since Omar is 8 pounds + 3 ounces, we multiply 16 by 8, and add on the 3:
16 * 8 = 128
128 + 3 = 131
So Omar weights 131 ounces when he was born.
Now Omar has to gain 5 ounces each week, for 4 weeks. This means in total, he'll have to gain 5*4=20 ounces in 4 weeks.
Add them together:
131 + 20 = 151 oz OR 9 7/16 lb.
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The equation of the line through (-9, -7) and (-6, -3) is 
<u>Solution:</u>
Given, two points are (-9, -7) and (-6, -3)
We have to find that a line that passes through the given two points.
First let us find the slope of the line that passes through given two points.
The slope of the line "m" is given as:


So slope of our line 
Now, let us find the line equation using point slope form:

By substituting the values we get,


Hence, the line equation is 