Answer:
53 cupcakes
Step-by-step explanation:
Take the amount of frosting and divide by the amount needed per cupcake
431/8
53.875
We round down since we cannot frost part of a cupcake
53 cupcakes
The first equation is linear:
![x\dfrac{\mathrm dy}{\mathrm dx}-y=x^2\sin x](https://tex.z-dn.net/?f=x%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D-y%3Dx%5E2%5Csin%20x)
Divide through by
![x^2](https://tex.z-dn.net/?f=x%5E2)
to get
![\dfrac1x\dfrac{\mathrm dy}{\mathrm dx}-\dfrac1{x^2}y=\sin x](https://tex.z-dn.net/?f=%5Cdfrac1x%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D-%5Cdfrac1%7Bx%5E2%7Dy%3D%5Csin%20x)
and notice that the left hand side can be consolidated as a derivative of a product. After doing so, you can integrate both sides and solve for
![y](https://tex.z-dn.net/?f=y)
.
![\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1xy\right]=\sin x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5B%5Cdfrac1xy%5Cright%5D%3D%5Csin%20x)
![\implies\dfrac1xy=\displaystyle\int\sin x\,\mathrm dx=-\cos x+C](https://tex.z-dn.net/?f=%5Cimplies%5Cdfrac1xy%3D%5Cdisplaystyle%5Cint%5Csin%20x%5C%2C%5Cmathrm%20dx%3D-%5Ccos%20x%2BC)
![\implies y=-x\cos x+Cx](https://tex.z-dn.net/?f=%5Cimplies%20y%3D-x%5Ccos%20x%2BCx)
- - -
The second equation is also linear:
![x^2y'+x(x+2)y=e^x](https://tex.z-dn.net/?f=x%5E2y%27%2Bx%28x%2B2%29y%3De%5Ex)
Multiply both sides by
![e^x](https://tex.z-dn.net/?f=e%5Ex)
to get
![x^2e^xy'+x(x+2)e^xy=e^{2x}](https://tex.z-dn.net/?f=x%5E2e%5Exy%27%2Bx%28x%2B2%29e%5Exy%3De%5E%7B2x%7D)
and recall that
![(x^2e^x)'=2xe^x+x^2e^x=x(x+2)e^x](https://tex.z-dn.net/?f=%28x%5E2e%5Ex%29%27%3D2xe%5Ex%2Bx%5E2e%5Ex%3Dx%28x%2B2%29e%5Ex)
, so we can write
![(x^2e^xy)'=e^{2x}](https://tex.z-dn.net/?f=%28x%5E2e%5Exy%29%27%3De%5E%7B2x%7D)
![\implies x^2e^xy=\displaystyle\int e^{2x}\,\mathrm dx=\frac12e^{2x}+C](https://tex.z-dn.net/?f=%5Cimplies%20x%5E2e%5Exy%3D%5Cdisplaystyle%5Cint%20e%5E%7B2x%7D%5C%2C%5Cmathrm%20dx%3D%5Cfrac12e%5E%7B2x%7D%2BC)
![\implies y=\dfrac{e^x}{2x^2}+\dfrac C{x^2e^x}](https://tex.z-dn.net/?f=%5Cimplies%20y%3D%5Cdfrac%7Be%5Ex%7D%7B2x%5E2%7D%2B%5Cdfrac%20C%7Bx%5E2e%5Ex%7D)
- - -
Yet another linear ODE:
![\cos x\dfrac{\mathrm dy}{\mathrm dx}+\sin x\,y=1](https://tex.z-dn.net/?f=%5Ccos%20x%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%2B%5Csin%20x%5C%2Cy%3D1)
Divide through by
![\cos^2x](https://tex.z-dn.net/?f=%5Ccos%5E2x)
, giving
![\dfrac1{\cos x}\dfrac{\mathrm dy}{\mathrm dx}+\dfrac{\sin x}{\cos^2x}y=\dfrac1{\cos^2x}](https://tex.z-dn.net/?f=%5Cdfrac1%7B%5Ccos%20x%7D%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%2B%5Cdfrac%7B%5Csin%20x%7D%7B%5Ccos%5E2x%7Dy%3D%5Cdfrac1%7B%5Ccos%5E2x%7D)
![\sec x\dfrac{\mathrm dy}{\mathrm dx}+\sec x\tan x\,y=\sec^2x](https://tex.z-dn.net/?f=%5Csec%20x%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%2B%5Csec%20x%5Ctan%20x%5C%2Cy%3D%5Csec%5E2x)
![\dfrac{\mathrm d}{\mathrm dx}[\sec x\,y]=\sec^2x](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5B%5Csec%20x%5C%2Cy%5D%3D%5Csec%5E2x)
![\implies\sec x\,y=\displaystyle\int\sec^2x\,\mathrm dx=\tan x+C](https://tex.z-dn.net/?f=%5Cimplies%5Csec%20x%5C%2Cy%3D%5Cdisplaystyle%5Cint%5Csec%5E2x%5C%2C%5Cmathrm%20dx%3D%5Ctan%20x%2BC)
![\implies y=\cos x\tan x+C\cos x](https://tex.z-dn.net/?f=%5Cimplies%20y%3D%5Ccos%20x%5Ctan%20x%2BC%5Ccos%20x)
![y=\sin x+C\cos x](https://tex.z-dn.net/?f=y%3D%5Csin%20x%2BC%5Ccos%20x)
- - -
In case the steps where we multiply or divide through by a certain factor weren't clear enough, those steps follow from the procedure for finding an integrating factor. We start with the linear equation
![a(x)y'(x)+b(x)y(x)=c(x)](https://tex.z-dn.net/?f=a%28x%29y%27%28x%29%2Bb%28x%29y%28x%29%3Dc%28x%29)
then rewrite it as
![y'(x)=\dfrac{b(x)}{a(x)}y(x)=\dfrac{c(x)}{a(x)}\iff y'(x)+P(x)y(x)=Q(x)](https://tex.z-dn.net/?f=y%27%28x%29%3D%5Cdfrac%7Bb%28x%29%7D%7Ba%28x%29%7Dy%28x%29%3D%5Cdfrac%7Bc%28x%29%7D%7Ba%28x%29%7D%5Ciff%20y%27%28x%29%2BP%28x%29y%28x%29%3DQ%28x%29)
The integrating factor is a function
![\mu(x)](https://tex.z-dn.net/?f=%5Cmu%28x%29)
such that
![\mu(x)y'(x)+\mu(x)P(x)y(x)=(\mu(x)y(x))'](https://tex.z-dn.net/?f=%5Cmu%28x%29y%27%28x%29%2B%5Cmu%28x%29P%28x%29y%28x%29%3D%28%5Cmu%28x%29y%28x%29%29%27)
which requires that
![\mu(x)P(x)=\mu'(x)](https://tex.z-dn.net/?f=%5Cmu%28x%29P%28x%29%3D%5Cmu%27%28x%29)
This is a separable ODE, so solving for
![\mu](https://tex.z-dn.net/?f=%5Cmu)
we have
![\mu(x)P(x)=\dfrac{\mathrm d\mu(x)}{\mathrm dx}\iff\dfrac{\mathrm d\mu(x)}{\mu(x)}=P(x)\,\mathrm dx](https://tex.z-dn.net/?f=%5Cmu%28x%29P%28x%29%3D%5Cdfrac%7B%5Cmathrm%20d%5Cmu%28x%29%7D%7B%5Cmathrm%20dx%7D%5Ciff%5Cdfrac%7B%5Cmathrm%20d%5Cmu%28x%29%7D%7B%5Cmu%28x%29%7D%3DP%28x%29%5C%2C%5Cmathrm%20dx)
![\implies\ln|\mu(x)|=\displaystyle\int P(x)\,\mathrm dx](https://tex.z-dn.net/?f=%5Cimplies%5Cln%7C%5Cmu%28x%29%7C%3D%5Cdisplaystyle%5Cint%20P%28x%29%5C%2C%5Cmathrm%20dx)
![\implies\mu(x)=\exp\left(\displaystyle\int P(x)\,\mathrm dx\right)](https://tex.z-dn.net/?f=%5Cimplies%5Cmu%28x%29%3D%5Cexp%5Cleft%28%5Cdisplaystyle%5Cint%20P%28x%29%5C%2C%5Cmathrm%20dx%5Cright%29)
and so on.
Answer:
Step-by-step explanation:
A. The leading coefficient is the number on the highest degree term. We have a 2nd degree polynomial because of the power of 2 on the x. That's what leads this polynomial off. So the leading coefficient is the number stuck to that x-squared term. Our leading coefficient is a -.2, not 5. 5 is the linear term.
B. The constant is the number that doesn't have a variable stuck to it. Our constant, then, is -8, not 2.
C. The expression does in fact have 3 terms. Each term is separated from the other terms by either a plus sign or a minus sign. A polynomial with 3 terms is called a trinomial.
D. The expression actually factors to x = 1.718... and x = 23.281... Neither one of those is the factor given in the statement.
The only one that is true is C.
D.
This is because the probability of getting a number greater than 2 is 4/6.
Therefore, simplifying the expression gives us the fraction 2/3.
The correct answer is 13.86.
In order to find this, we must first note that the tangent of any angle is equal to the opposite side length over the adjacent side length. If we use that formula and the 60 degree angle, we can find y.
Tan(angle) = Opp/Adj
Tan(60) = y/8
8Tan(60) = y
13.86 = y