Answer:
∴V=(14-2x)(10-2x)x
where V is the volume of the box in cubic units.
Step-by-step explanation:
Given that,
A rectangle has a length of 14 units and width of 10 units.
4 squares with dimensions x by x units are cut out from the each corner of the rectangle.
The length of the box is= (14-2x) units
The width of the box is =(10-2x) units
The height of the box is = x unit.
The volume of the box is = Length × width×height
=(14-2x)(10-2x)x cubic units.
∴V=(14-2x)(10-2x)x
where V is the volume of the box in cubic units.
If you need an explanation please ask, but basically solve for y with the given equation
Answer:
(3, 2)
Step-by-step explanation:
y = 3x - 7
2x + 5y = 16
Solve for x:
2x + 5(3x - 7) = 16
2x + 15x - 35 = 16
17x = 51
x = 3
Solve for y:
y = 3(3) - 7
y = 9 - 7
y = 2
Answer:
(3, 2)
Answer:
d=rt In this problem we are looking for the distance, but we will have to go about it indirectly. If she's traveling the same exact road going and returning, then the distance traveled both ways is exactly the same. Since d = rt, and d is the same, by the substitution property, if and , then , and . So we need to rt for the trip going, rt for the trip returning and set them equal to each other and solve for t. Going is a rate of 24, and the time is t (since we don't know t), and returning is a rate of 30, and the time is 13 1/3-t. (If the whole trip takes 13 1/3 hours, and t is the time going, then the time returning is the difference between the total time and the going time. That concept is one that baffles most algebra students!). So our r1t1 is 24t, and our r2t2 is 30(13 1/3 - t). Set them equal to each other and that will look like this: That fraction of 40/3 is 13 1/3 made into an improper fraction. Distributing that we will have and 54t = 400. That means that t = 7.407. We have time, and that's great, but we need distance! Go back to one of your equations for distance and sub in t and solve for d. d = 24t, and d = 24(7.407), so d = 177.768 miles.
Answer:
The lateral area of the pyramid is
Step-by-step explanation:
we know that
The lateral area of the pyramid is equal to the area of its three triangular lateral faces
so
where
------> the slant height
substitute the values