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ser-zykov [4K]
3 years ago
11

While shopping for​ clothes, Consuelo spent ​$3 less than twice what Brenda spent. Consuelo spent ​$73. How much did Brenda​ spe

nd
Mathematics
1 answer:
ryzh [129]3 years ago
4 0
Brenda spended $70 on clothes
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a

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Is y=x squared +2 a linear function
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No, this would not be a linear function. Being linear means it creates a straight line, but this function has an x squared. Anytime x^2 is in a function, the graph is a u shape, called a parabola. A simple linear function is y=x :D
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Only number one please help ​
Alex73 [517]

Answer:

C) 75 + 50x = 25 + 70x

Step-by-step explanation:

ABC Tech:

$50 an hour. So,for x hours= 50 * x = 50x

Service charge = 75 + 50x

Tech Squad:

$70 an hour. So,for x hours= 70 * x = 70x

Service charge = 25 + 70x

C) 75 + 50x = 25 + 70x

50x - 70x = 25 - 75

-20x = -50

x = -50/-20

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3 years ago
How do I solve this?
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1 i think 1 is the answer
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3 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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