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Alenkasestr [34]
3 years ago
5

12(2y + 2) Help please!!!

Mathematics
2 answers:
Yakvenalex [24]3 years ago
5 0

Answer:

24y+24

Step-by-step explanation:

Distribute the 12 to what's inside the parenthesis.

puteri [66]3 years ago
5 0
24y+24 bc u distribute the 12 to the 2y (multiply) then u do the same thing to 2
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Jill has 4 bags of marbles. There are 3red, 5green,2 yellow, and 6black marbles I. Each bag. How many marbles does Jill have? Sh
Eddi Din [679]
3+5+2+6=16 in each bag multiply or divide sorry not that much of an answer
6 0
2 years ago
Read 2 more answers
Find a particular solution to <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7B%20d%5E%7B2%7Dy%20%7D%7Bd%20x%5E%7
Digiron [165]
y=x^r
\implies r(r-1)x^r+6rx^r+4x^r=0
\implies r^2+5r+4=(r+1)(r+4)=0
\implies r=-1,r=-4

so the characteristic solution is

y_c=\dfrac{C_1}x+\dfrac{C_2}{x^4}

As a guess for the particular solution, let's back up a bit. The reason the choice of y=x^r works for the characteristic solution is that, in the background, we're employing the substitution t=\ln x, so that y(x) is getting replaced with a new function z(t). Differentiating yields

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dz}{\mathrm dt}
\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac1{x^2}\left(\dfrac{\mathrm d^2z}{\mathrm dt^2}-\dfrac{\mathrm dz}{\mathrm dt}\right)

Now the ODE in terms of t is linear with constant coefficients, since the coefficients x^2 and x will cancel, resulting in the ODE

\dfrac{\mathrm d^2z}{\mathrm dt^2}+5\dfrac{\mathrm dz}{\mathrm dt}+4z=e^{2t}\sin e^t

Of coursesin, the characteristic equation will be r^2+6r+4=0, which leads to solutions C_1e^{-t}+C_2e^{-4t}=C_1x^{-1}+C_2x^{-4}, as before.

Now that we have two linearly independent solutions, we can easily find more via variation of parameters. If z_1,z_2 are the solutions to the characteristic equation of the ODE in terms of z, then we can find another of the form z_p=u_1z_1+u_2z_2 where

u_1=-\displaystyle\int\frac{z_2e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt
u_2=\displaystyle\int\frac{z_1e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt

where W(z_1,z_2) is the Wronskian of the two characteristic solutions. We have

u_1=-\displaystyle\int\frac{e^{-2t}\sin e^t}{-3e^{-5t}}\,\mathrm dt
u_1=\dfrac23(1-2e^{2t})\cos e^t+\dfrac23e^t\sin e^t

u_2=\displaystyle\int\frac{e^t\sin e^t}{-3e^{-5t}}\,\mathrm dt
u_2=\dfrac13(120-20e^{2t}+e^{4t})e^t\cos e^t-\dfrac13(120-60e^{2t}+5e^{4t})\sin e^t

\implies z_p=u_1z_1+u_2z_2
\implies z_p=(40e^{-4t}-6)e^{-t}\cos e^t-(1-20e^{-2t}+40e^{-4t})\sin e^t

and recalling that t=\ln x\iff e^t=x, we have

\implies y_p=\left(\dfrac{40}{x^3}-\dfrac6x\right)\cos x-\left(1-\dfrac{20}{x^2}+\dfrac{40}{x^4}\right)\sin x
4 0
2 years ago
Can someone answer all these for me pls
Tems11 [23]
Okay
1 yes and yes
2 yes and no
3 no and yes
4 0
3 years ago
Please help! What is the slope of the line graphed on the coordinate plane?
Lisa [10]

Answer:

The slope is 3/4

Step-by-step explanation:

You can Y2-Y1/X2-X1 to find the answer to this question

Since the points are (-2,3) and (2,6) it will be..

6-3/2+2   ** the 2 becomes a plus in the denominator because negative and negative is positive

3/4 is the slope

8 0
3 years ago
1) AU (C - B) =
torisob [31]

Answer:

A \cup (C-B) = (A \cup C) \cap  (A \cup B^c)\\A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \,\,\, \text{Here you use double inclusion.}\\A \cup B \cup C =  A \cup (B \cup C)\\B^c \cap (A-C)  = (B^c - C) \cap A\\(B \cap C) - A = (B-A)\cap C\\C-(A\cup B) =  (C - A )\cap B\\A^c \cap (B \cap C) = ( A^c \cap B) \cap C

Step-by-step explanation:

A \cup (C-B) = A \cup (C \cap B^c) = (A \cup C) \cap  (A \cup B^c)\\A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \,\,\, \text{Here you use double inclusion.}\\A \cup B \cup C = (A \cup B ) \cup C  = A \cup (B \cup C)\\B^c \cap (A-C)  = B^c \cap (A \cap C^c) =  (B^c \cap C^c) \cap A =  (B^c - C) \cap A\\(B \cap C) - A = (B \cap C) \cap A^c  = (B \cap C) \cap A^c = (B \cap A^c)\cap C = (B-A)\cap C\\C-(A\cup B) = C \cap (A\cup B)^c = C \cap (A^c \cap B^c ) = (C  \cap A^c )\cap B^c = (C - A )\cap B\\

A^c \cap (B \cap C) = ( A^c \cap B) \cap C

3 0
3 years ago
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