Answer:
Paul can drive 162.14 miles a day for the budget of 120 dollars.
OB. C=38.93+0.50x
Step-by-step explanation:
120=C
120=38.93+0.50x
81.07=0.50x
162.14=x
We are given
rectangle
length is 6 inches
so, ![L=6inch](https://tex.z-dn.net/?f=%20L%3D6inch%20)
width is 4 inches
so, ![W=4inch](https://tex.z-dn.net/?f=%20W%3D4inch%20)
now, we can find area
![Area=length*width](https://tex.z-dn.net/?f=%20Area%3Dlength%2Awidth%20)
![Area=L*W](https://tex.z-dn.net/?f=%20Area%3DL%2AW%20)
now, we can plug values
![Area=6*4 in^2](https://tex.z-dn.net/?f=%20Area%3D6%2A4%20in%5E2%20)
![Area=24 in^2](https://tex.z-dn.net/?f=%20Area%3D24%20in%5E2%20)
now, we are given scale
![\frac{1}{4} in=2ft](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B4%7D%20in%3D2ft%20)
we can take square both sides
![\frac{1}{4^2} in^2=2^2 ft^2](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B4%5E2%7D%20in%5E2%3D2%5E2%20ft%5E2%20)
![\frac{1}{16} in^2=4 ft^2](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B16%7D%20in%5E2%3D4%20ft%5E2%20)
now, we can find 1 in^2
![1 in^2=16*4 ft^2](https://tex.z-dn.net/?f=%201%20in%5E2%3D16%2A4%20ft%5E2%20)
now, we can multiply both sides by 24
![24*1 in^2=24*16*4 ft^2](https://tex.z-dn.net/?f=%2024%2A1%20in%5E2%3D24%2A16%2A4%20ft%5E2%20)
![24 in^2=1536 ft^2](https://tex.z-dn.net/?f=%2024%20in%5E2%3D1536%20ft%5E2%20)
so,
Area of rectangle is 1536 feet^2..........Answer
Answer:
P=6
Step-by-step explanation:
okay so the problem states that this pentagon has one line of symmetry.
if you know that AE=4x, AB=2x+1, and Bc=x+2, then you know that CD=x+2, and DE=2x+1.
since you know the perimeter is 18 cm, you can make an equation to find x:
4x+2(2x+1)+2(x+2)=18
-->4x+4x+2+2x+4=18
-->10x+6=18
-->10x=12
-->x=1.2
P would be 6
The base will have the greatest area for a given perimeter if it is square. If the edge of the square base has length x (in feet), then the total material requirement in square feet is
.. m = x^2 +(4/x^2)*(4x)
.. m = x^2 +16/x
This will have a minimum where dm/dx = 0.
.. dm/dx = 2x -16/x^2 = 0
.. x^3 = 8 . . . . . . . . . . . . . . . multiply by x^2/2 and add 8
.. x = 2
The tank is 2 feet square and 1 ft high.
_____
You will note that it is half the height of a cube that has double the volume. This is the generic solution to all minimum cost open-top box problems. Actually, the costs of pairs of opposite sides are equal to each other and to the cost of the base. If material costs are not identical in all directions, that is the more generic solution.