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2 years ago

Si 409 es el 100%, cuanto porciento es 220?

Mathematics

1 answer:

saw5 [17]2 years ago

3 0

Answer:

53.79%

Step-by-step explanation:

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the sum of three numbers is 131. The second of three numbers is seven more than twice the first. The third number is 12 less tha

spin [16.1K]

No. 1 = x
No. 2 = 2x+7
No. 3 = x-12

Answer:

x+(2x+7)+(x-12) = 131

From there you solve to get the numbers:

34, 75 and 22

3 0

2 years ago

Needing help with getting the slope

bixtya [17]

Answer:

The answer is 2/5

Step-by-step explanation:

I found 2 points and did y2-y1/ x2-x1 and got 2/5. And then I tested it and it worked.

6 0

3 years ago

Read 2 more answers

5(3a+b)−2(3a+b) in simplest form<br> I NEED HELP!

nordsb [41]

Answer:

$9a + 3b$

Is the answer

Step-by-step explanation:

$5(3a + b) - 2(3a + b) \\ = 15a + 5b - 6a - 2b \\ = 9a + 3b$

$( - ) \times ( + ) = ( - ) \\$

That is why

$( - 2) \times b = - 2$

And

$( - 2) \times 3a = ( - 6)$

7 0

2 years ago

An electrician earns $110 after his first hour of working for a client. His total pay based on the number of hours worked can be

Sveta_85 [38]

Add the rest of the question

5 0

2 years ago

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$

8 0

2 years ago