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3 years ago

Change 12.39% to a decimal

2 answers:

6 0

To change a number to percent, move the
decimal point 2 places that way ==> .

To change a percent to a number, move the
decimal point 2 places this way <== .

12.39% = 0.1239 .

PSYCHO15rus [73]3 years ago

5 0

Answer:

The answar is 0.1239

Step-by-step explanation:

The percent sign tells un to divide 12.39 by a hundred(12.39/100). Divide 12.39 by hundred and you get 0.1239

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WHATS THE ANSWERS PLS TELL MEH PLS

Ray Of Light [21]

Answer: 3) 9 7/10 (C)

4) 26 1/12 pounds (A)

5) 17 5/6 miles (D)

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3 years ago

Read 2 more answers

Which ordered pair in the form(x,y) is a solution of this equation?

Gnom [1K]

The ordered pair is (-2, -3)

Step-by-step explanation:

Step 1: To find whether an ordered pair is a solution, substitute values of x and y and see whether it satisfies the equation.

(-3, -2) ⇒ 7 × -3 - 5 × -2 = -21 + 10 = -11 ≠ 1

(-2, -3) ⇒ 7 × -2 - 5 × -3 = -14 + 15 = 1

(0, 4) ⇒ 0 - 5 × 4 = 20 ≠ 1

(4, 0) ⇒ 7 × 4 - 0 = 28 ≠ 1

So the ordered pair is (-2, -3)

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3 years ago

What is the solution to the equation fraction 4 over 5 n minus fraction 3 over 5 equals fraction 1 over 5 n?

Andreas93 [3]

Answer:

The solution is n=1

Step-by-step explanation:

I hope you mean

4/5n-3/5=1/5n

if so then here's the solution

5n×4/5n-3/5×5n=1/5n×5n

4-3n=1

4-1=3n

3n=3

n=1

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3 years ago

Read 2 more answers

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

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

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1 year ago

There are 36 workers on a crew of electric kins and 29 of them came to work what percentage of them showed up for work

Ahat [919]

29/36 =0.805555556 = 81% if you need to round up

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3 years ago