Answer:
Area of rectangle possible. : 6ft² ; 10 ft², 12 ft²
Step-by-step explanation:
Feets of fencing = 14
Perimeter = 14
Let x = Length y = width
Perimeter = 2(x + y)
For x = 1
2(1 + y) = 14
1 + y = 7
y =6
Area of rectangle ; x *y = 1 * 6 = 6 ft²
For x = 2
2(2 + y) = 14
2 + y = 7
y = 5
Area of rectangle ; x *y = 2 * 5 = 10 ft²
For x = 3
2(3 + y) = 14
3 + y = 7
y =4
Area of rectangle ; x *y = 3 * 4 = 12 ft²
For x = 4
2(1 + y) = 14
4 + y = 7
y = 3
Area of rectangle ; x *y = 4 * 3 =12 ft²
Hence, Area of rectangle possible. : 6ft² ; 10 ft², 12 ft²
Answer:
293.38 pounds
Step-by-step explanation:
We are given that
Distance between poles=35 feet
Weight of cable=10.4 per linear foot
We have to find the weight of the cable.
Differentiate w.r.t
Let 
![s=\frac{2}{0.0225}\times\frac{2}{3}[t^{\frac{3}{2}}]^{17.5}_{0}](https://tex.z-dn.net/?f=s%3D%5Cfrac%7B2%7D%7B0.0225%7D%5Ctimes%5Cfrac%7B2%7D%7B3%7D%5Bt%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%5D%5E%7B17.5%7D_%7B0%7D)
![s=2\times \frac{2}{3\times0.0225}[(1+0.0255x)^{\frac{3}{2}]^{17.5}_{0}](https://tex.z-dn.net/?f=s%3D2%5Ctimes%20%5Cfrac%7B2%7D%7B3%5Ctimes0.0225%7D%5B%281%2B0.0255x%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%5D%5E%7B17.5%7D_%7B0%7D)
s=28.21
Weight of cable=
pound
Answer:
30 inches squared
Step-by-step explanation:
First, find the area of the entire frame using length times height. 9*6 is 54 inches squared. Then, find the area of the middle space the same way. 6*4 is 24 inches squared. Lastly, subtract the middle from the entire frame's area. 54-24 is 30 inches squared.
Yes, he is correct, using the distributive property you get the same answer
