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koban [17]
3 years ago
10

Number 3 please explain and show work ☺☺

Mathematics
2 answers:
vlabodo [156]3 years ago
7 0
There are 16 ounces in a pound, so 28 pounds is 448 ounces. Divide 448 by 8 which equals 56 ounces of dog food.
Allisa [31]3 years ago
6 0
Buddy, you cant ask every question on your hw. there are 16 ounces in a pound so 16 times 28 which is a large number. then divide it by 8 and that will give you your number
multi choice:
A.62
B.98.777
C.56
D.48
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From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

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y_1=e^x\cos x
y_2=e^x\sin x

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

y_p=u_1y_1+u_2y_2

where

u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx

and W(y_1,y_2) is the Wronskian of the fundamental solutions. We have

W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}

and so

u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx
u_1=\dfrac13\ln|\cos x|

u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx
u_2=\dfrac13x

Therefore the particular solution is

y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x

so that the general solution to the ODE is

y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x
7 0
3 years ago
What is the answer for (8x - 5)
Svetradugi [14.3K]

Answer:

-40 i think

Step-by-step explanation:

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2 years ago
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Mice21 [21]

Answer:

Subtract 2 from the term number

Step-by-step explanation:

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