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3 years ago

Which equation represents a line that passes through (–9, –3) and has a slope of –6?

Mathematics

2 answers:

Helen [10]3 years ago

3 0

Answer:

y+3= -6(x+9) is the answer

Step-by-step explanation:

dem82 [27]3 years ago

3 0

Answer:

$y+3=-6(x+9)$

Step-by-step explanation:

We are given that

Slope of a line=-6

Given point =(-9,-3)

We have to find the equation which represents the line.

The equation of line passing through the given point $(x_1,y_1)$ with slope m is given by

$y-y_1=m(x-x_1)$

Substitute the values then we get

The equation of line passing through the point (-9,-3) with slope -6 is given by

$y-(-3)=-6(x-(-9))$

$y+3=-6(x+9)$

Hence, the equation of line that passes through (-9,-3) and has a slope -6 is given by

$y+3=-6(x+9)$

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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

3 years ago

How do I convert 60° to radian?

Oksi-84 [34.3K]

60° in radians is 1.047

The formula used is 60° × π/180 = 1.047rad

6 0

3 years ago

Explain how 1 is greater than -1!<br> pls help

umka2103 [35]

Answer:

because -1 is before 1

Step-by-step explanation:

4 0

2 years ago

Tuition of $2700 is due when the spring term begins, in 9 months. What amount should a student deposit today, at 11%, to have en

OLga [1]

T= 9 months= 9/12 year = 3/4 year
r=11%
A=2700=P(1+r*t/100)=P(1+11*3/4)=P(1+8.25)

so,P= 2700/9.25 =<span>291.89</span> thats the answer

8 0

3 years ago

Preethi writes fifteen 1s in a row and randomly writes + and - between each pair of consecutive 1s. Find the probability that th

amid [387]

Answer:

The probability that the expression Preeti wrote has the value 7 cannot be easily determined.

Step-by-step explanation:

Given that Preeti writes

111111111111111,

then randomly writes + and - between each pair of consecutive 1's.

It is not specified that the + and - signs are written to be alternating, they are said to be just 'random'. So, there are a lot of possibilities. Things like:

1+1-1+1+1-1+1-1-1-1+1+1-1+1-1 = 1

1-1-1+1+1-1-1+1+1-1+1-1+1+1-1 = -1

1-1+1-1-1+1-1+1+1-1-1+1-1+1+1 = 0

And so on. There is a lot of possibilities.

So, clearly, the probability is difficult to determine, as the expression Preeti wrote could have had the value 7 in many different forms.

Because:

1+1+1+1+1+1+1-1+1-1+1-1+1-1+1 = 7

1+1+1+1-1-1-1-1-1+1-1-1-1+1+1 = 7

And many more.

Therefore, the probability that the expression Preeti wrote has the value 7 cannot be easily determined.

5 0

3 years ago