Y=1/3x+1 is the equation for ur question
<h2>
Answer:</h2>
<u>The expression is</u><u> 5(40+80)</u>
<h2>
Step-by-step explanation:</h2>
The sum of 40 and 80 can be written as
(40+80)
and when we take the 5 times of the sum
we can write as
5(40+80)
Answer:
55.5 square feet
Step-by-step explanation:
so the dog house consists of 4 walls, two roof sections, and two triangle bits to make the roof pointy (I'm sure they have a name but I don't know what it is lol) Of the walls there would be 2 of the longer walls and 2 of the shorther ones. Knowing this we just have to add together the area of each part to find the total surface area.
So starting with the walls:
two of them are 4 feet by 2 feet.
The area of a rectangle is just length x width
so those two would have an area of 8 ft^2
and the other two are 3 feet by 2 feet so they would have an area of 6 ft^2
the roof sections are also rectangles with dimensions of 4 feet by 2.5 feet
4x2.5=10 so each side of the roof has a surface area of 10 ft^2
now the triangle parts have a base of 3 feet and a height of 2.5 feet
the formula for area of a triangle is base x height / 2
3x2.5=7.5
7.5/2= 3.75
now we add them together
8+8+6+6+10+10+3.75+3.75=55.5
so the total surface area is 55.5 square ft
At the standard 0.05 value, this is not significant, however at a 0.01 level, it would be significant
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
