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3 years ago

Given that the area of rectangle IJKL is 500 units?

Mathematics

1 answer:

raketka [301]3 years ago

6 0

Answer:

d. 17.5 and 28 units

Step-by-step explanation:

An area of a rectangle is length times width. 8x times 5x = 40x^2. 40x^2 is eqaul to 500. After dividing by 40 and taking the square root of the quotient, x is approximately 3.54. 5 times 3.54 and 8 times 3.54 will yield approximately 17.5 and 28 units.

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Daily low temperatures in Columbus, OH in January 2014 were approximately normally distributed with a mean of 15.45 and a standa

Alona [7]

Answer: 12.10

Step-by-step explanation:

Given : Mean : $\mu = 15.45$

Standard deviation : $\sigma = 13.70$

The formula to calculate the z-score :-

$z=\dfrac{x-\mu}{\sigma}$

For x= 5 degrees

$z=\dfrac{5-15.45}{13.70}=-0.7627737226\approx-0.76$

For x= 10 degrees

$z=\dfrac{10-15.45}{13.70}=-0.397810218\approx-0.40$

The P-value : $P(-0.76$

$=0.3445783-0.2236273=0.120951\approx0.1210$

In percent , $0.1210\times100=12.10\%$

Hence, the percentage of days had a low temperature between 5 degrees and 10 degrees = 12.10%

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

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

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