we know that
1 ft is equal to 12 in
1 cubic yard is equal to 27 cubic feet
Step 1
<u>Find the area of the circular border of uniform width around the pool</u>
Let
x---------> the uniform width around the pool
we know that
The diameter of the circular pool measures 10 feet
so
the radius r=5 ft
the area of the circular border is equal to
![A=\pi *(5+x)^{2}- \pi *5^{2} \\A= \pi *[(x+5) ^{2}-5^{2} ] \\ A= \pi * [x^{2} +10x]](https://tex.z-dn.net/?f=A%3D%5Cpi%20%2A%285%2Bx%29%5E%7B2%7D-%20%5Cpi%20%2A5%5E%7B2%7D%20%5C%5CA%3D%20%5Cpi%20%2A%5B%28x%2B5%29%20%5E%7B2%7D-5%5E%7B2%7D%20%5D%20%5C%5C%20A%3D%20%5Cpi%20%2A%20%5Bx%5E%7B2%7D%20%2B10x%5D)
step 2
volume of the concrete to be used to create a circular border is equal to
V=1 yd^{3}-------> convert to ft^{3}
V=27 ft^{3} -------> equation 1
the depth is equal to 4 in-------> convert to ft
depth=4/12=(1/3) ft
volume of the concrete to be used to create a circular border is also equal to
V=Area of the circular border*Depth
![V= \pi * [x^{2} +10x]*(1/3)](https://tex.z-dn.net/?f=V%3D%20%5Cpi%20%2A%20%5Bx%5E%7B2%7D%20%2B10x%5D%2A%281%2F3%29)
-------> equation 2
equate equation 1 and equation 2
![27=\pi * [x^{2} +10x]*(1/3) \\ x^{2} +10x- \frac{81}{\pi }=0](https://tex.z-dn.net/?f=27%3D%5Cpi%20%2A%20%5Bx%5E%7B2%7D%20%2B10x%5D%2A%281%2F3%29%20%5C%5C%20x%5E%7B2%7D%20%2B10x-%20%5Cfrac%7B81%7D%7B%5Cpi%20%7D%3D0)
using a graph tool------> to resolve the second order equation
see the attached figure
the solution is the point
x=2.126 ft
therefore
<u>the answer is</u>
The uniform width around the circular pool border is 2.126 ft
Answer:
3.278 * 10 ^ 2
Step-by-step explanation:
To change a number to scientific notation you:
Move the decimal point to the left to make the number have 1 number before the decimal point.
From there you multiply by 10 to the power 2 the number of decimal points you moved.
So the answer is 3.278 * 10^2
HOPE IT HELPED
I think it’s called an Heptadecagon
Answer:
The first quartile of the strengths of this alloy is 9.055 GPa.
Step-by-step explanation:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
The strength of an aluminum alloy is normally distributed with mean 10 gigapascals (GPa) and standard deviation 1.4 GPa.
This means that 
What is the first [lower] quartile of the strengths of this alloy?
This is the 100/4 = 25th percentile, which is X when Z has a pvalue of 0.25, so X when Z = -0.675.
The first quartile of the strengths of this alloy is 9.055 GPa.
Answer:
<em>4x-3y=8</em>
<em>Simplifies to x=3/4y+2</em>
2x+5y=-9
Simplifies to -5/2y+-9/2
