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Archy [21]
3 years ago
5

Math help plz

Mathematics
1 answer:
bija089 [108]3 years ago
8 0

Answer:

I guess this is the answer hope it helps

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Can someone answer this for me please
Nana76 [90]

Answer:

32

Step-by-step explanation:

6 0
3 years ago
A wise women once said, “400 reduced by twice my age is 264.” What is her age?
KonstantinChe [14]

Answer:

My age is 78 years.

Explanation:

Define my age first as a variable.

Let my age be  

x

years.

Reduced by means subtracted.

Subtract twice my age from 400 to get 244:

400

−

2

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244

400

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244

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←

make x positive and isolate it.

156

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x

78

=

x

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
If the first and the last terms of an arithmetic series are 10 and 62, show that the sum of the series varies directly as the nu
Nitella [24]

Answer:

10+^@=72_^) + the series

Step-by-step explanation:


6 0
3 years ago
A swimming class has 20 students. Only 8 students know how to dive off
Allisa [31]

Answer: 40%  

Step-by-step explanation:

To write 8/20 as a percent have to remember that 1 equal 100% and that what you need to do is just to multiply the number by 100 and add at the end symbol % . 8/20 * 100 = 0.4 * 100 = 40% And finally we have: 8/20 as a percent equals 40%

6 0
3 years ago
3y''-6y'+6y=e*x sexcx
Simora [160]
From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

3r^2-6r+6=0\iff r^2-2r+2=0

which has roots at r=1\pm i. This admits the two fundamental solutions

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
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