Answer:
A) coefficient
Step-by-step explanation:
A linear graph is represented by a straight line.
The equation of the line could be represented with: ![\mathbf{y =-4x + 3}](https://tex.z-dn.net/?f=%5Cmathbf%7By%20%3D-4x%20%2B%203%7D)
From the graph (see attachment), we have the following highlights:
- The graph has a negative slope
- The graph has a positive y-intercept
A linear equation is represented as:
![\mathbf{y = mx + b}](https://tex.z-dn.net/?f=%5Cmathbf%7By%20%3D%20mx%20%2B%20b%7D)
Where: m and b are the slopes and the intercepts of the graph
The above highlights mean that:
m < 0 and b > 0
From the list of given options, the only option that fit the above highlights is:
![\mathbf{y =-4x + 3}](https://tex.z-dn.net/?f=%5Cmathbf%7By%20%3D-4x%20%2B%203%7D)
Hence, the equation of the line could be represented with: ![\mathbf{y =-4x + 3}](https://tex.z-dn.net/?f=%5Cmathbf%7By%20%3D-4x%20%2B%203%7D)
Read more about linear equations at:
brainly.com/question/11897796
Answer:
Owlgebra
Heard this math joke before!! LOL
<em>Have a nice day!!:D</em>
![t\dfrac{\mathrm dy}{\mathrm dt}-y^2\ln t+y=0](https://tex.z-dn.net/?f=t%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dt%7D-y%5E2%5Cln%20t%2By%3D0)
Divide both sides by
:
![ty^{-2}\dfrac{\mathrm dy}{\mathrm dt}-\ln t+y^{-1}=0](https://tex.z-dn.net/?f=ty%5E%7B-2%7D%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dt%7D-%5Cln%20t%2By%5E%7B-1%7D%3D0)
Substitute
, so that
.
![-t\dfrac{\mathrm dv}{\mathrm dt}-\ln t+v=0](https://tex.z-dn.net/?f=-t%5Cdfrac%7B%5Cmathrm%20dv%7D%7B%5Cmathrm%20dt%7D-%5Cln%20t%2Bv%3D0)
![t\dfrac{\mathrm dv}{\mathrm dt}-v=\ln t](https://tex.z-dn.net/?f=t%5Cdfrac%7B%5Cmathrm%20dv%7D%7B%5Cmathrm%20dt%7D-v%3D%5Cln%20t)
Divide both sides by
:
![\dfrac1t\dfrac{\mathrm dv}{\mathrm dt}-\dfrac1{t^2}v=\dfrac{\ln t}{t^2}](https://tex.z-dn.net/?f=%5Cdfrac1t%5Cdfrac%7B%5Cmathrm%20dv%7D%7B%5Cmathrm%20dt%7D-%5Cdfrac1%7Bt%5E2%7Dv%3D%5Cdfrac%7B%5Cln%20t%7D%7Bt%5E2%7D)
The left side can be condensed as the derivative of a product:
![\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac1tv\right]=\dfrac{\ln t}{t^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dt%7D%5Cleft%5B%5Cdfrac1tv%5Cright%5D%3D%5Cdfrac%7B%5Cln%20t%7D%7Bt%5E2%7D)
Integrate both sides. The integral on the right side can be done by parts.
![\displaystyle\int\frac{\ln t}{t^2}\,\mathrm dt=-\frac{\ln t}t+\int\frac{\mathrm dt}{t^2}=-\frac{\ln t}t-\frac1t+C](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint%5Cfrac%7B%5Cln%20t%7D%7Bt%5E2%7D%5C%2C%5Cmathrm%20dt%3D-%5Cfrac%7B%5Cln%20t%7Dt%2B%5Cint%5Cfrac%7B%5Cmathrm%20dt%7D%7Bt%5E2%7D%3D-%5Cfrac%7B%5Cln%20t%7Dt-%5Cfrac1t%2BC)
![\dfrac1tv=-\dfrac{\ln t}t-\dfrac1t+C](https://tex.z-dn.net/?f=%5Cdfrac1tv%3D-%5Cdfrac%7B%5Cln%20t%7Dt-%5Cdfrac1t%2BC)
![v=-\ln t-1+Ct](https://tex.z-dn.net/?f=v%3D-%5Cln%20t-1%2BCt)
Now solve for
.
![y^{-1}=-\ln t-1+Ct](https://tex.z-dn.net/?f=y%5E%7B-1%7D%3D-%5Cln%20t-1%2BCt)
![\boxed{y(t)=\dfrac1{Ct-\ln t-1}}](https://tex.z-dn.net/?f=%5Cboxed%7By%28t%29%3D%5Cdfrac1%7BCt-%5Cln%20t-1%7D%7D)
Answer:
1794-40=x?
Step-by-step explanation:
this is genuine answer but a little bit unsure