The volume of aggregate that she needs to build the patio is 13.18 ft^3
<h3>How many cubic feet of aggregate would she need to build the walkway?</h3>
We know that the backyard is 14ft long, 6 feet wide and 4 inches deep.
Rememeber that:
1ft = 12 in
Then:
4 in = (4/12) ft = (1/3) ft
Then the volume of the backyard is:
V = (14ft)*(6ft)*(1/3 ft) = 28ft^3
Now we know that we have a mix of 4 parts aggregate to 4.5 parts loose cement that need to fill these 28 cubic feet.
Note that:
4 + 4.5 = 8.5
How many times we have 8.5 in 28?
28/8.5 = 3.294
Now for each of these we have 4 parts of aggregate, then the volume of aggregate that we need is:
4*3.294 ft^3 = 13.18 ft^3
The volume of aggregate that she needs to build the patio is 13.18 ft^3
If you want to learn more about volumes:
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Answer:
200
Step-by-step explanation:
Answer:
150,595
Step-by-step explanation:
First, convert miles to feet:
4.1 mi × (5280 ft/mi) = 21648 ft
The area that the crowd takes up is:
2 × 8 ft × 21648 ft = 346368 ft²
The number of people is:
346368 ft² × (1 person / 2.3 ft²) = 150594.8
Rounding to the nearest integer, there are approximately 150,595 people in the crowd.
Answer:
Plane R
Plane DEG
Plane FDG
Plane FEG
Step-by-step explanation:
You can name the plane by the script letter "R", and you can name the plane with any three co-linear points on the plane
<h3>
Answer: Choice B) 122%</h3>
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Work Shown:
A = 1250 is the initial amount
B = 2780 is the new amount
C = percent change = unknown for now
The formula to use is
C = [ (B-A)/A ] * 100
to calculate the percent change.
Basically we calculate the change (B-A) first and then divide that over the original amount A to compute the decimal form of the answer, which is then converted over to percentage form. The "100" tacked on at the end is what converts from decimal to percent form.
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C = [ (B-A)/A ] * 100
C = [ (2780 - 1250)/1250 ] * 100
C = (1530/1250)*100
C = 1.224 * 100
C = 122.4%
C = 122%
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The positive C value indicates we have a percent increase. If C were negative, then we'd have a percent decrease.
