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Sholpan [36]
3 years ago
7

Y = 2 - 4x input output

Mathematics
1 answer:
tester [92]3 years ago
4 0

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 x, while the output, also known as dependent variable, is often indicated with the letter y.

With this notation, we write y = f(x), 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 y = f(x) = 2-4x, 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 x = 2 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 x = 0 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.

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The first blue option <em>or </em> 8x - 17 = 5x + 19

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    They are vertical angles, so they will be set equal to each other.

Have a nice day!

    I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

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5d-9

Step-by-step explanation:

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7x - 5y = -24<br> -9x + 5y = 18
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Answer:

Nothing further can be done with this topic. Please check the expression entered or try another topic.

7x−5y=−24−9x+5y=187x-5y=-24-9x+5y=18

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Use the substitution x = et to transform the given Cauchy-Euler equation to a differential equation with constant coefficients.
Anika [276]

Answer:

\boxed{\sf \ \ \ ax^2+bx^{-10} \ \ \  }

Step-by-step explanation:

Hello,

let's follow the advise and proceed with the substitution

first estimate y'(x) and y''(x) in function of y'(t), y''(t) and t

x(t)=e^t\\\dfrac{dx}{dt}=e^t\\y'(t)=\dfrac{dy}{dt}=\dfrac{dy}{dx}\dfrac{dx}{dt}=e^ty'(x)y'(x)=e^{-t}y'(t)\\y''(x)=\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}(e^{-t}\dfrac{dy}{dt})=-e^{-t}\dfrac{dt}{dx}\dfrac{dy}{dt}+e^{-t}\dfrac{d}{dx}(\dfrac{dy}{dt})\\=-e^{-t}e^{-t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}\dfrac{dt}{dx}=-e^{-2t}\dfrac{dy}{dt}+e^{-t}\dfrac{d^2y}{dt^2}e^{-t}\\=e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt})

Now we can substitute in the equation

x^2y''(x)+9xy'(x)-20y(x)=0\\ e^{2t}[ \ e^{-2t}(\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}) \ ] + 9e^t [ \ e^{-t}\dfrac{dy}{dt} \ ] -20y=0\\ \dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}+ 9\dfrac{dy}{dt}-20y=0\\ \dfrac{d^2y}{dt^2}+ 8\dfrac{dy}{dt}-20y=0\\

so the new equation is

y''(t)+ 8y'(t)-20y(t)=0

the auxiliary equation is

x^2+8x-20=0\\ x^2-2x+10x-20=0\\x(x-2)+10(x-2)=0\\(x+10)(x-2)=0\\ x=-10\text{ or }x=2

so the solutions of the new equation are

y(t)=ae^{2t}+be^{-10t}

with a and b real

as

x(t)=e^t\\ t(x)=ln(x)

y(x)=ae^{2ln(x)}+be^{-10ln(x)}=ax^2+bx^{-10}

hope this helps

do not hesitate if you have any questions

8 0
4 years ago
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