In the above word problem, If she chooses the 8-inch tiles, she will need a quarter as many tiles as she would with 2-inch tiles, Quarter 8-inch tiles will cover the same area as one 2-inches.
<h3>What is the justification for the above?</h3>
Note that the area of the one 2-inch tiles is given as:
A1 = 4in²
The area of the quarter 8-inch tiles is:
A2 = 1/4 x 8 x 8
A2 = 16inch²
Divide both areas
A2/A1= 16/4
= 4
This implies she'll need four 2-inch tiles to cover the same amount of space as a quarter 8-inch tile.
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Full Question:
A homeowner is deciding on the size of tiles to use to fully tile a rectangular wall in her bathroom that is 80 inches by 40 inches. The tiles are squares and come in three side lengths: 8 inches, 4 inches, and 2 inches. State if you agree with each statement about the tiles. Explain your reasoning.
If she chooses the 8-inch tiles, she will need a quarter as many tiles as she would with 2-inch tiles,
Answer:Hmmm I think angry
Step-by-step explanation:
Answer:
-1
Step-by-step explanation:
She walked down 2 flights, then up 3 flights. This would cause her to be 1 floor above her floor. So, she would have to walk down 1 flight (hence the negative in the -1) and she would be on her floor.
Answer:
7.5 ft³/min
Step-by-step explanation:
Let x be the depth below the surface of the water. The height, h of the water is thus h = 10 - x.
Now, the volume of water V = Ah where A = area of isosceles base of trough = 1/2bh' where b = base of triangle = 4 ft and h' = height of triangle = 1 ft. So, A = 1/2 × 4 ft × 1 ft = 2 ft²
So, V = Ah = 2h = 2(10 - x)
The rate of change of volume is thus
dV/dt = d[2(10 - x)]/dt = -2dx/dt
Since dV/dt = 15 ft³/min,
dx/dt = -(dV/dt)/2 = -15 ft³/min ÷ 2 = -7.5 ft³/min
Since the height of the water is h = 10 - x, the rate at which the water level is rising is dh/dt = d[10 - x]/dt
= -dx/dt
= -(-7.5 ft³/min)
= 7.5 ft³/min
And the height at this point when x = 8 inches = 8 in × 1 ft/12 in = 0.67 ft is h = (10 - 0.67) ft = 9.33 ft
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➷ 5 + 4 = 9
4068/9 = 452
4 x 452 = 1808
There are 1808 woman at the stadium
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