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larisa86 [58]
3 years ago
12

Please answer this for me pleaseeeee

Mathematics
1 answer:
kobusy [5.1K]3 years ago
4 0

Answer:

\Huge\boxed{x=71.9}

Step-by-step explanation:

Hello There!

Once again we are going to use trigonometry to solve for x

Here are the <u>Trigonometric Ratios</u>

sin = opposite divided by hypotenuse

cos = adjacent divided by hypotenuse

tan = opposite divided by adjacent

we need to find x and we are given its opposite side length (58) and the adjacent side length (19)

this corresponds with tangent so once again we will be using tangent to solve for x

Looking at tangent we see that its equal to opposite divided by adjacent so we create an equation

tan(x)=\frac{58}{19}\\\frac{58}{19} =3.052631579

now we have

tan(x)=3.052631579

Once again we need to get rid of the tan

to do so we take the inverse of tan ( tan^-1) and apply it to each side

arcsin^-^1(tan(x))=x\\arcsin^-^1(3.052631579)=71.86191784

finally we round to the nearest tenth

we're left with x = 71.9

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430;400

Step-by-step explanation:

So tenth is the 2.. 5 makes it go to 3 430

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3 years ago
Find a particular solution to the nonhomogeneous differential equation y??+4y?+5y=?10x+e^(?x).
Firdavs [7]

Answer:

A) Particular solution:

2x+\frac{1}{2}e^{-x}-\frac{8}{5}

B) Homogeneous solution:

y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))

C) The most general solution is

y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}

Step-by-step explanation:

Given non homogeneous ODE is

y''+4y'+5y=10x+e^{-x}---(1)

To find homogeneous solution:

D^{2}+4D+5=0\\D^{2}+4D+4-4+5=0\\\\(D+2)^{2}=-1\\D+2=\pm iD=-2 \pm i\\y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))---(2)

To find particular solution:

y_{p}=Ax+B+Ce^{-x}\\\\y'_{p}=A-Ce^{-x}\\y''_{p}=Ce^{-x}\\

Substituting y_{p},y'_{p},y''_{p} in (1)

y''_{p}+4y'_{p}+5y_{p}=10x+e^{-x}\\Ce^{-x}+4(A-Ce^{-x})+5(Ax+B+Ce^{-x})=10x+e^{-x}\\

Equating the coefficients

5Ax+2Ce^{-x}+4A+5B=10x+e^{-x}\\5A=10\\A=2\\4A+5B=0\\B=-\frac{4A}{5}B=-\frac{8}{5}2C=1\\C=\frac{1}{2}\\So,\\y_{p}=2x+\frac{1}{2}e^{-x}-\frac{8}{5}---(3)\\

The general solution is

y=y_{h}+y_{p}

from (2) ad (3)

y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}

6 0
3 years ago
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