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sdas [7]
3 years ago
12

Travis was attempting to make muffins to take to a neighbor. The recipe that he was using required 3/4 cup of sugar and 1/8 cup

of butter. Travis accidentally put a whole cup of butter in the mix.
A.What is the ratio of the sugar to butter in the original recipe?

B.What amount of sugar does Travis need to mix to have the same ratio of sugar to butter that the original recipe calls for?

C.if Travis wants to keep the ratios the same as they are in the original recipe, how will the amounts of all the other ingredients for this new mixture compare to the amounts given in the recipe?
Mathematics
1 answer:
bogdanovich [222]3 years ago
8 0

A. The recipe  required 3/4 cup of sugar and 1/8 cup of butter.

The ratio of the sugar to butter in the original recipe is

\dfrac{\frac{3}{4}}{\frac{1}{8}}=\dfrac{3}{4}\cdot \dfrac{8}{1}=\dfrac{6}{1}.

B. If Travis accidentally put a whole cup of butter in the mix, then he needs to put

\dfrac{3}{4}\cdot \dfrac{6}{1}=\dfrac{9}{2}=4\dfrac{1}{2} cup of sugar.

C. Travis has to put 1/8 cup of butter, but he put 1 cup of butter. The ratio is

\dfrac{1}{\frac{1}{8}}=\dfrac{1}{1}\cdot \dfrac{8}{1}=\dfrac{8}{1}.

If Travis wants to keep the ratios the same as they are in the original recipe, he needs to put 8 times more of all the other ingredients.

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Find a particular solution to the nonhomogeneous differential equation y′′+4y=cos(2x)+sin(2x).
I am Lyosha [343]
Take the homogeneous part and find the roots to the characteristic equation:

y''+4y=0\implies r^2+4=0\implies r=\pm2i

This means the characteristic solution is y_c=C_1\cos2x+C_2\sin2x.

Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form y_p=ax\cos2x+bx\sin2x. Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.

With y_1=\cos2x and y_2=\sin2x, you're looking for a particular solution of the form y_p=u_1y_1+u_2y_2. The functions u_i satisfy

u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\int\frac{y_1(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx

where W(y_1,y_2) is the Wronskian determinant of the two characteristic solutions.

W(\cos2x,\sin2x)=\begin{bmatrix}\cos2x&\sin2x\\-2\cos2x&2\sin2x\end{vmatrix}=2

So you have

u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx
u_1=-\dfrac x4+\dfrac18\cos^22x+\dfrac1{16}\sin4x

u_2=\displaystyle\frac12\int(\cos2x(\cos2x+\sin2x))\,\mathrm dx
u_2=\dfrac x4-\dfrac18\cos^22x+\dfrac1{16}\sin4x

So you end up with a solution

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but since \cos2x is already accounted for in the characteristic solution, the particular solution is then

y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x

so that the general solution is

y=C_1\cos2x+C_2\sin2x-\dfrac14x\cos2x+\dfrac14x\sin2x
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3 years ago
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VARVARA [1.3K]

Answer:

8 sand bags can be carried along.

Step-by-step explanation:

First you want to subtract 300-140. That subtracts the pilots weight giving you the maximum weight capacity left. (160) Then you want to divide 160 (the weight that can be carried) by 20 (the weight of the sand bags) after doing that you will get 8. And that is how i got 8 sand bags as my answer.

7 0
3 years ago
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