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1 year ago

Given the points A (-4,-2) and B (4, 10), find the

Mathematics

1 answer:

mafiozo [28]1 year ago

8 0

Answer:

P = (2, 7)

Step-by-step explanation:

You want to find coordinates of P on segment AB such that P is 3/4 is of the way from A to B.

<h3>Equation for P</h3>

For some fraction q of the distance from A to B, the point P that lies at that fraction of the distance is given by ...

P = A +q(B -A) = (1 -q)A +qB

<h3>Application</h3>

For q = 3/4, the location of P is ...

P = (1 -3/4)A + 3/4B = (A +3B)/4

Using the given point coordinates, we have ...

P = ((-4, -2) +3(4, 10))/4 = (-4 +12, -2 +30)/4 = (8, 28)/4

P = (2, 7)

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There are 7.48 gallons in a cubic foot. If residents of the apartment building are using the water from the tower at an average

lorasvet [3.4K]

Answer:

Time it will take to drain the entire tower = 2.8minutes

Step-by-step explanation:

The question is incomplete as the volume of the tower was not indicated.

Let's consider the following question:

If there are 7.48 gallons in a cubic foot, and the volume of the tower is around 36000in cubed. Residents of the apartment building are using the water from the tower at an average rate of 56 gallons per minute, determine how long it will take to drain the entire tower.

Solution:

Volume = 36000in³

Conversion of in³ to ft³

1 inch = 0.0833 feet

12 inch = 1 ft

1 ft³ = 1ft × 1ft × 1ft

= 12 in x 12 in x 12 in = 1728 in³

36000in³ × [(1ft³)/(1728 in³) = (36000/1728)ft³

= 20.833ft³

Volume = 20.833ft³

There are 7.48 gallons in a cubic foot

In 20.833ft³ = 20.833ft³× (7.48 gallons/1ft³)

= 20.833× 7.48gallons

Volume = 155.83 gallons

The rate of usage = 56 gallons per minute

The rate of usage for 155.83 gallons = 155.83 gallons × (1min/56gallons)

= (155.83/56)minute

= 2.8minutes

Time it will take to drain the entire tower = 2.8minutes

5 0

3 years ago

Find a particular solution to y" - y + y = 2 sin(3x)

leonid [27]

Answer with explanation:

The given differential equation is

y" -y'+y=2 sin 3x------(1)

Let, y'=z

y"=z'

$\frac{dy}{dx}=z\\\\d y=zdx\\\\y=z x$

Substituting the value of , y, y' and y" in equation (1)

z'-z+zx=2 sin 3 x

z'+z(x-1)=2 sin 3 x-----------(1)

This is a type of linear differential equation.

Integrating factor

$=e^{\int (x-1) dx}\\\\=e^{\frac{x^2}{2}-x}$

Multiplying both sides of equation (1) by integrating factor and integrating we get

$\rightarrow z\times e^{\frac{x^2}{2}-x}=\int 2 sin 3 x \times e^{\frac{x^2}{2}-x} dx=I$

$I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx -\int \frac{2\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx-\frac{2I}{3}\\\\\frac{5I}{3}=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{5}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{5} dx$