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inessss [21]
3 years ago
12

A force of 22 pounds presses on a membrane. The average pressure on the membrane is 2.75 pounds per square inch (psi). What is t

he area of the membrane in square inches? The equation for calculating pressure is where P is pressure, F is force, and A is area.
Mathematics
2 answers:
adell [148]3 years ago
7 0

Im stuck on this one too! Its really tough!

Viefleur [7K]3 years ago
3 0

Answer:

8 in^2.

Step-by-step explanation:

F = P * A

22 = 2.75 * A

A = 22/ 2.75

A = 8 in^2.

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The shaded region in the figure shows recommended weight and height combinations. Determine the recommended weight range for a p
vitfil [10]
5h - w = 170
for h = 77
w = 5h - 170
w = 5*77 - 170
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<span>w = 215

Shoutout to @himanshuchaturvedi 

Your answer is : w = 215 

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8 0
3 years ago
Abby Mia wants to know how much must be deposited in her local bank today so that she will receive yearly payments of $18,000 fo
muminat
We are not given tables, so will just use the amortization formula.
P=\frac{A*((1+i)^n-1)}{i*(1+i)^n}
where 
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6 0
3 years ago
PLEASE HELPPPPPPPPPPPPPPPPPPPPPPPPPPPP
Andrew [12]

Answer:

(a) 1825 = 2.25x + (2.25-1)(x -108)

(b) 560 mi/h

Step-by-step explanation:

(a) distance = speed·time

The first plane's speed is x. The distance it travels in 2.25 hours is 2.25x.

The second plane's speed is x-108. It travels only 1.25 hours (since it started an hour later). The distance it travels is then (2.25 -1)(x -108).

The problem statement tells us the total of the distances traveled by the two planes is 1825 miles, so we can write the equation ...

... 1825 = 2.25x + (2.25 -1)(x -108)

(b) Simplifying the equation gives ...

... 1825 = 3.50x -135

To solve this 2-step equation, we add 135, then divide by 3.50.

.. 1960 = 3.50x

... 1960/3.50 = x = 560

The first airplane's speed is 560 mph.

<u>Check</u>

In 2.25 hours, the first plane travels (560 mi/h)·(2.25 h) = 1260 mi.

In 1.25 hours, the second plane travels (452 mi/h)·(1.25 h) = 565 mi.

Then 2.25 hours after the first plane leaves, the planes are 1260 +565 = 1825 miles apart, as given in the problem statement.

3 0
3 years ago
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The histogram is shown below.

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The rest of the bar heights are nearly the same, so the remaining portion of the distribution is nearly uniform. A uniform distribution has every bar the same height. Collectively, all the bars of a uniform distribution forms a larger rectangle.

8 0
3 years ago
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