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Vedmedyk [2.9K]
3 years ago
12

Which choice is equivalent to the expression below?

Mathematics
1 answer:
Mars2501 [29]3 years ago
4 0
There is no picture or expression
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Blank x blank equal parts
Brilliant_brown [7]

Answer:

wait what????

Step-by-step explanation:

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3 years ago
Tanya writes the number 7.05. Bryce says he can write a different number using the same digits. Is he correct?
Bad White [126]
Yes, actually a couple:
different number with same digits (ex):

7.05
7.50
5.07
5.70
0.57
0.75
705
70.5
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etc

hope this helps
8 0
3 years ago
How much acetic acid is in a 5 liter container that is marked 80% acetic acid. How much is water
Vinvika [58]

Answer:

1 liter of water

Step-by-step explanation:

If 80 % is acid that's means 4 liters of the 5 is acid. (0.8 * 5)

The remaining amount is 1

8 0
3 years ago
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What does f(6)=24 mean?
d1i1m1o1n [39]

Answer:

we're is the answer

Step-by-step explanation:

were

8 0
2 years ago
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

u_1y_1+u_2y_2=\dfrac18\cos2x-\dfrac14x\cos2x+\dfrac14x\sin2x

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
7 0
3 years ago
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