Answer:
solution: x = -8
Step by step:
1) 5x-5+5 = -45+5 : Add 5 to both sides
2) 5x= -40 : Simplify
3) 5x/5 = -40/5 : Divide both sides
Hope that helps
Every function is a rule which tells you how to associate inputs and outputs. The input, also known as independent variable, is often indicated with the letter
, while the output, also known as dependent variable, is often indicated with the letter
.
With this notation, we write
, read "y is a function of x", in the sense that the value of the variable y depends on the value of the variable x, and f is the function that tells you how y depends on x.
In your example, you have
, which means "subtract four times the input (4x) from 2"
So, it doesn't matter which input you chose (i.e. the value for x), because you will always have to behave this way:
- Pick an input value, x
- Multiply it by four to get 4x
- Subtract this number from 2: 2-4x
Here are some examples of explicit calculations: if I choose
and input, the workflow will be
- Pick an input value, 2
- Multiply it by four to get 8
- Subtract this number from 2: 2-8=-6
So, if the input is 2, the output is -6
Similarly, if we choose
as input, we have:
- Pick an input value, 0
- Multiply it by four to get 0
- Subtract this number from 2: 2-0=2
So, if the input is 0, the output is 2. And so on: for every possible value for x you have the correspondant value for y, with the function f telling you how to associate one with the other.
Answer:
The United States was founded on "equality". Our Declaration of Independence was created to ensure the rights given to us as American citizens and state what must be upheld to protect these rights. Equality comes from a very valued form of Government in America called Democracy. Americans are given the right to freedom of speech, and this represents traditional American values.
If
is the cumulative distribution function for
, then

Then the probability density function for
is
:

The
th moment of
is
![E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy](https://tex.z-dn.net/?f=E%5BY%5En%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20y%5Enf_Y%28y%29%5C%2C%5Cmathrm%20dy%3D%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_0%5E%5Cinfty%20y%5E%7Bn-1%7De%5E%7B-%5Cfrac12%28%5Cln%20y%29%5E2%7D%5C%2C%5Cmathrm%20dy)
Let
, so that
and
:
![E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du](https://tex.z-dn.net/?f=E%5BY%5En%5D%3D%5Cdisplaystyle%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7Bnu%7De%5E%7B-%5Cfrac12u%5E2%7D%5C%2C%5Cmathrm%20du%3D%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7Bnu-%5Cfrac12u%5E2%7D%5C%2C%5Cmathrm%20du)
Complete the square in the exponent:

![E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du](https://tex.z-dn.net/?f=E%5BY%5En%5D%3D%5Cdisplaystyle%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B%5Cfrac12%28n%5E2-%28u-n%29%5E2%29%7D%5C%2C%5Cmathrm%20du%3D%5Cfrac%7Be%5E%7B%5Cfrac12n%5E2%7D%7D%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-%5Cfrac12%28u-n%29%5E2%7D%5C%2C%5Cmathrm%20du)
But
is exactly the PDF of a normal distribution with mean
and variance 1; in other words, the 0th moment of a random variable
:
![E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1](https://tex.z-dn.net/?f=E%5BU%5E0%5D%3D%5Cdisplaystyle%5Cfrac1%7B%5Csqrt%7B2%5Cpi%7D%7D%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20e%5E%7B-%5Cfrac12%28u-n%29%5E2%7D%5C%2C%5Cmathrm%20du%3D1)
so we end up with
![E[Y^n]=e^{\frac12n^2}](https://tex.z-dn.net/?f=E%5BY%5En%5D%3De%5E%7B%5Cfrac12n%5E2%7D)
The power(P) of the circuits is the product of the voltage (V) in volts and current(I) in amperes.
P = IV
Substituting the known values to the equation,
P = (250 amps) x (40 volts) = 10,000 Watts
Thus, the answer is 10,000 Watts.