Let width = w
Let length = l
Let area = A
3w+2l=1200
2l=1200-3w
l=1200-3/2
A=w*l
A=w*(1200-3w)/2
A=600w-(3/2)*w^2
If I set A=0 to find the roots, the maximum will be at wmax=-b/2a which is exactly 1/2 way between the roots-(3/2)*w^2+600w=0
-b=-600
2a=-3
-b/2a=-600/-3
-600/-3=200
w=200
And, since 3w+2l=1200
3*200+2l=1200
2l = 600
l = 300
The dimensions of the largest enclosure willbe when width = 200 ft and length = 300 ft
check answer:
3w+2l=1200
3*200+2*300=1200
600+600=1200
1200=1200
and A=w*l
A=200*300
A=60000 ft2
To see if this is max area change w and l slightly but still make 3w+2l=1200 true, like
w=200.1
l=299.85
A=299.85*200.1
A=59999.985
Answer:
a) Linear
b) Linear
c) Linear
d) Neither
See explanation below.
Step-by-step explanation:
a) ![\frac{dy}{dx} +e^x y = x^2 y^2](https://tex.z-dn.net/?f=%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%2Be%5Ex%20y%20%3D%20x%5E2%20y%5E2%20)
For this case the differential equation have the following general form:
![y' +p(x) y = q(x) y^n](https://tex.z-dn.net/?f=%20y%27%20%2Bp%28x%29%20y%20%3D%20q%28x%29%20y%5En)
Where
and
and since n>1 we can see that is a linear differential equation.
b) ![y + sin x = x^3 y'](https://tex.z-dn.net/?f=%20y%20%2B%20sin%20x%20%3D%20x%5E3%20y%27)
We can rewrite the following equation on this way:
![y' -\frac{1}{x^3} y= \frac{sin (x)}{x^3}](https://tex.z-dn.net/?f=%20y%27%20-%5Cfrac%7B1%7D%7Bx%5E3%7D%20y%3D%20%5Cfrac%7Bsin%20%28x%29%7D%7Bx%5E3%7D)
For this case the differential equation have the following general form:
![y' +p(x) y = q(x) y^n](https://tex.z-dn.net/?f=%20y%27%20%2Bp%28x%29%20y%20%3D%20q%28x%29%20y%5En)
Where
and
and since n=0 we can see that is a linear differential equation.
c) ![ln x -x^2 y =xy'](https://tex.z-dn.net/?f=ln%20x%20-x%5E2%20y%20%3Dxy%27)
For this case we can write the differential equation on this way:
![y' +xy = \frac{ln(x)}{x}](https://tex.z-dn.net/?f=%20y%27%20%2Bxy%20%3D%20%5Cfrac%7Bln%28x%29%7D%7Bx%7D)
For this case the differential equation have the following general form:
![y' +p(x) y = q(x) y^n](https://tex.z-dn.net/?f=%20y%27%20%2Bp%28x%29%20y%20%3D%20q%28x%29%20y%5En)
Where
and
and since n=0 we can see that is a linear differential equation.
d) ![\frac{dy}{dx} + cos y = tan x](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%20%2B%20cos%20y%20%3D%20tan%20x)
For this case we can't express the differential equation in terms:
![y' +p(x) y = q(x) y^n](https://tex.z-dn.net/?f=%20y%27%20%2Bp%28x%29%20y%20%3D%20q%28x%29%20y%5En)
So the is not linear, and since we can separate the variables in order to integrate is not separable. So then the answer for this one is neither.
Answer: Ratio of black horses to all horses = 59/62
Step-by-step explanation:
Total number of horses = 62
Total number of white horses = 3
Assuming that the rest horses are black. That means the total number of black horses would be 62-3 =59. So there are 59 black horses and there are 3 white horses.
Ratio of black horses to all horses will be the total number of black horses / the total number of horses.
Ratio of black horses to all horses
= 59/62 or 59:62
It can also be expressed in decimals which will become
0.952
Well to set up a substitution equation, you need to make sure you have something you can sub in that's identical in both equations.
we have
3x-y=5
x-4y=-24
what's in common? x and y.
take one of the equations and solve for x or y like so (i'm doing y)
3x-5=y
then substitute the whole 3x-5 equation for y in the second equation. it is quite similar to how you would normally substitute a number like if y=5, then you substitute 5 for y in the second equation, but instead the "5" is an equation "3x-5"
x-4y=-24
x-4(3x-5) = -24
Now that you got it set up, you just need to solve for x and once you got x value, plug the x value in the original equations to find y.