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3 years ago

7

Each number below is a factor of Miguel's number.

2 answers:

BabaBlast [244]3 years ago

6 0

Pretty sure it d.12 bc you were counting from 3’s so the next number up would be 12

ahrayia [7]3 years ago

6 0

36 because all the numbers go into it evenly

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Enter the equation of the circle described below.<br> Center (3, -2), radius = 5

coldgirl [10]

The equation of a circle is <em>(x-h)^2+(y-k)^2=r^2</em>, where <em>(h,k)</em> is the center and <em>r</em> is the radius. Our equation will be (x-3)^2+(y+2)^2=25 (remember to always square the radius).

:)

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3 years ago

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Elizabeth has 14 boxes of cookies each box has 28 cookies how many cookies does she have

aev [14]

We calculate this by multiplying the number of cookies by the number of boxes:

28*14=(20+8)*14=280+80+32=360+32=392 (that's how I like to count it, there are also other methods)

so she has 392 cookies in common.

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

: In his bank account, Karl has a balance of −$80.40. His friends Friedrich and Jung have a bank balance of −$220 and -$40 respe

Anika [276]

The answer is option C. Both Friedrich and Jung have the most debt. Hope I could help! :D

6 0

3 years ago

suppose the circumference of a crop circle is 150.7968 hectometers (hm) what's the radius of the circle? (usett=3.1416)

gulaghasi [49]

Answer:

The radius of the circle is $24\ hm$

Step-by-step explanation:

we know that

The circumference of a circle is equal to

$C=2\pi r$

In this problem we have

$C=150.7968\ hm$

$\pi=3.1416$

substitute and solve for r

$150.7968=2(3.1416)r$

$r=150.7968/(2(3.1416))=24\ hm$

7 0

3 years ago

Read 2 more answers