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2 years ago

PLEASEEE HELP ! I WOULD APPRECIATE IT VERY MUCH :)

Mathematics

1 answer:

azamat2 years ago

8 0

hi!

so the original equation is (x^2 -121) / x+ 11.

Those two equations look similar...

well, we know that the top equation looks like a^2 - b^2, and that equation iis equivalent to this equation: (a+b)(a-b).

So if we factor that out, we get:

( (x + 11) * (x-11) )/ (x + 11)

we can cancel the x+11 on the top and the x+11 on the bottom out.

that leaves us with x-11.

Hope this helped!

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A plane flying horizontally at an altitude of 3 miles and a speed of 500 mi/h passes directly over a radar station. Find the rat

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Answer:

The rate at which the distance from the plane to the station is increasing is 331 miles per hour.

Step-by-step explanation:

We can find the rate at which the distance from the plane to the station is increasing by imaging the formation of a right triangle with the following dimensions:

a: is one side of the triangle = altitude of the plane = 3 miles

b: is the other side of the triangle = the distance traveled by the plane when it is 4 miles away from the station and an altitude of 3 miles

h: is the hypotenuse of the triangle = distance between the plane and the station = 4 miles

First, we need to find b:

$a^{2} + b^{2} = h^{2}$ (1)

$b = \sqrt{h^{2} - a^{2}} = \sqrt{(4 mi)^{2} - (3 mi)^{2}} = \sqrt{7} miles$

Now, to find the rate we need to find the derivative of equation (1) with respect to time:

$\frac{d}{dt}(a^{2}) + \frac{d}{dt}(b^{2}) = \frac{d}{dt}(h^{2})$

$2a\frac{da}{dt} + 2b\frac{db}{dt} = 2h\frac{dh}{dt}$

Since "da/dt" is constant (the altitude of the plane does not change with time), we have:

$0 + 2b\frac{db}{dt} = 2h\frac{dh}{dt}$

And knowing that the plane is moving at a speed of 500 mi/h (db/dt):

$\sqrt{7} mi*500 mi/h = 4 mi*\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{\sqrt{7} mi*500 mi/h}{4 mi} = 331 mi/h$

Therefore, the rate at which the distance from the plane to the station is increasing is 331 miles per hour.

I hope it helps you!

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