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3 years ago

4(5x – 6) = 2(8x + 10) Which of the following has all correct justifications Bradley used to solve this equation?

Mathematics

1 answer:

Citrus2011 [14]3 years ago

4 0

The answer would most likely be C

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Are these ratios equivalent?

Debora [2.8K]

<h3>Answer: Yes</h3>

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Explanation

The ratio 8:10 simplifies to 4:5 when you divide both parts by 2.

The ratio 16:20 simplifies to 4:5 when you divide both parts by 4

Therefore the two ratios 8:10 and 16:20 are both equal 4:5, so they are equal to one another.

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Put another way,

(8 large)/(10 small) = (16 large)/(20 small)

8/10 = 16/20

8*20 = 10*16 ... cross multiply

160 = 160

We get a true equation, so the first equation is true as well.

This shows the ratios are equivalent.

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Or you could have...

(8 large)/(16 large) = (10 small)/(20 small)

8/16 = 10/20

8*20 = 16*10

160 = 160

We get the same conclusion as before.

3 0

3 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

Distribute and simplify these radicals. 23-(V+ 3) O 2V5 + 6 02/15 O 30 O 6/2+6

zhannawk [14.2K]

Answer:

Step-by-step explanation:

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3 years ago

A parallelogram is ______ a rhombus. never, always, sometimes

Nataly [62]

Answer:

never

Step-by-step explanation:

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3 years ago

Is 7/11 a irrational number

frozen [14]

Answer: No, it’s rational number

5 0

4 years ago

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