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2 years ago

Which represents the inverse of the function f(x) = 4x?

Mathematics

1 answer:

Degger [83]2 years ago

3 0

It's D<span>

the inverse is basically the operation that will undo your original function
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If using the method of completing the square to solve the quadratic equation

zheka24 [161]

Answer: 25

Step-by-step explanation: The number that is added to both sides is the number that is needed to create a perfect square trinomial.

The question is, what is that number?

Well, it comes from a formula. The number that is added to both sides always comes from half the coefficient of the middle term squared.

So half of -10 squared or -5 squared which is 25.

So 25 is added to both sides.

8 0

2 years ago

Solve. −2x 3. 25=0. 5x 4. 75 x=−0. 6 x=−1 x=−3. 2 I don't know.

worty [1.4K]

Answer:

x=-0.6

Step-by-step explanation:

5 0

1 year ago

A mountain climber found that the relationship between his altitude and the surrounding air

sukhopar [10]

Answer:- 1/100 degrees Fahrenheit

Step-by-step explanation:

The directions say decreasing so it has to be negative, and the slope is showing how much it is decreasing every meter. y=mx+b. The mx is the slope which is where the answer is located.

5 0

2 years ago

Use the method of undetermined coefficients to find the general solution to the de y′′−3y′ 2y=ex e2x e−x

djverab [1.8K]

I'll assume the ODE is

$y'' - 3y' + 2y = e^x + e^{2x} + e^{-x}$

Solve the homogeneous ODE,

$y'' - 3y' + 2y = 0$

The characteristic equation

$r^2 - 3r + 2 = (r - 1) (r - 2) = 0$

has roots at $r=1$ and $r=2$ . Then the characteristic solution is

$y = C_1 e^x + C_2 e^{2x}$

For nonhomogeneous ODE (1),

$y'' - 3y' + 2y = e^x$

consider the ansatz particular solution

$y = axe^x \implies y' = a(x+1) e^x \implies y'' = a(x+2) e^x$

Substituting this into (1) gives

$a(x+2) e^x - 3 a (x+1) e^x + 2ax e^x = e^x \implies a = -1$

For the nonhomogeneous ODE (2),

$y'' - 3y' + 2y = e^{2x}$

take the ansatz

$y = bxe^{2x} \implies y' = b(2x+1) e^{2x} \implies y'' = b(4x+4) e^{2x}$

Substitute (2) into the ODE to get

$b(4x+4) e^{2x} - 3b(2x+1)e^{2x} + 2bxe^{2x} = e^{2x} \implies b=1$

Lastly, for the nonhomogeneous ODE (3)

$y'' - 3y' + 2y = e^{-x}$

take the ansatz

$y = ce^{-x} \implies y' = -ce^{-x} \implies y'' = ce^{-x}$

and solve for $c$ .

$ce^{-x} + 3ce^{-x} + 2ce^{-x} = e^{-x} \implies c = \dfrac16$

Then the general solution to the ODE is

$\boxed{y = C_1 e^x + C_2 e^{2x} - xe^x + xe^{2x} + \dfrac16 e^{-x}}$

6 0

1 year ago

263/4 options 65r2 64r7 65 65r3

Reika [66]

Answer:

65R3

Step-by-step explanation:

260/4 is a whole number, so then there would be a remainder 3. The only one with a r3 is D.

Hope this helps :D

4 0

2 years ago