Answer:
30450.
Step-by-step explanation:
Not sure what you meant by simplify? Hope this helps.
Answer:
![y=-\frac{4}{9}x +\frac{2}{3}](https://tex.z-dn.net/?f=y%3D-%5Cfrac%7B4%7D%7B9%7Dx%20%2B%5Cfrac%7B2%7D%7B3%7D)
Step-by-step explanation:
I'm going to presume you want this linear equation in slope-intercept form, so start with:
![6=4x+9y](https://tex.z-dn.net/?f=6%3D4x%2B9y)
Subtract
from both sides of the equation:
![-9y+6=4x](https://tex.z-dn.net/?f=-9y%2B6%3D4x)
Subtract
from both sides of the equation:
![-9y=4x-6](https://tex.z-dn.net/?f=-9y%3D4x-6)
Divide both sides of the equation by the coefficient of
, which is
:
![y=\frac{4}{-9}x +\frac{-6}{-9}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B4%7D%7B-9%7Dx%20%2B%5Cfrac%7B-6%7D%7B-9%7D)
Simplify:
![y=-\frac{4}{9}x +\frac{2}{3}](https://tex.z-dn.net/?f=y%3D-%5Cfrac%7B4%7D%7B9%7Dx%20%2B%5Cfrac%7B2%7D%7B3%7D)
To help you with the equation you should know that 40 is halfway between 35 and 45. That would mean that orange juice would make half of the mixture. It would also be the same for grapefruit juice. That would make it 100 liters of orange juice and grapefruit juice.
Given:
Monday: 6 laps around a 0.25-mile track
Tuesday: 7 laps around a 0.25-mile track
Total miles jogged on both days.
0.25(6 + 7) = 0.25(13) = 3.25 miles
Mike jogged a total of 3.25 miles on Monday and Tuesday combined.
Euler's method uses the recurrence relation
![y_{n+1}=y_n+hf(x_n,y_n)](https://tex.z-dn.net/?f=y_%7Bn%2B1%7D%3Dy_n%2Bhf%28x_n%2Cy_n%29)
to approximate the value of the solution
![y(x)](https://tex.z-dn.net/?f=y%28x%29)
to the ODE
![y'=f(x,y)](https://tex.z-dn.net/?f=y%27%3Df%28x%2Cy%29)
.
![xy\dfrac{\mathrm dy}{\mathrm dx}=1+\ln(x^2)\implies y'=\dfrac{1+\ln(x^2)}{xy}=f(x,y)](https://tex.z-dn.net/?f=xy%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%3D1%2B%5Cln%28x%5E2%29%5Cimplies%20y%27%3D%5Cdfrac%7B1%2B%5Cln%28x%5E2%29%7D%7Bxy%7D%3Df%28x%2Cy%29)
With a step size of
![h=0.5](https://tex.z-dn.net/?f=h%3D0.5)
, there will only be two steps necessary to find the approximate value of
![y(3)](https://tex.z-dn.net/?f=y%283%29)
based on the initial point
![y(2)=5](https://tex.z-dn.net/?f=y%282%29%3D5)
. See the attached table below for the computation results.
To demonstrate how the table is generated: Since
![y(2)=5](https://tex.z-dn.net/?f=y%282%29%3D5)
, you are using
![(x_0,y_0)=(2,5)](https://tex.z-dn.net/?f=%28x_0%2Cy_0%29%3D%282%2C5%29)
.
![y_1=y_0+hf(x_0,y_0)](https://tex.z-dn.net/?f=y_1%3Dy_0%2Bhf%28x_0%2Cy_0%29)
![y_1=5+0.5\times\dfrac{1+\ln(2^2)}{2\times5}](https://tex.z-dn.net/?f=y_1%3D5%2B0.5%5Ctimes%5Cdfrac%7B1%2B%5Cln%282%5E2%29%7D%7B2%5Ctimes5%7D)
![y_1\approx5.1193](https://tex.z-dn.net/?f=y_1%5Capprox5.1193)
The next point then uses
![x_1=x_0+0.5=2.5](https://tex.z-dn.net/?f=x_1%3Dx_0%2B0.5%3D2.5)
![y_2=y_1+hf(x_1,y_1)](https://tex.z-dn.net/?f=y_2%3Dy_1%2Bhf%28x_1%2Cy_1%29)
![y_1=y_1+0.5\times\dfrac{1+\ln(2.5^2)}{2.5y_1}](https://tex.z-dn.net/?f=y_1%3Dy_1%2B0.5%5Ctimes%5Cdfrac%7B1%2B%5Cln%282.5%5E2%29%7D%7B2.5y_1%7D)