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alex41 [277]
3 years ago
15

45 POINTS PLEASE HELP

Mathematics
2 answers:
lutik1710 [3]3 years ago
5 0

Hello,

The relationship between the number of hours a plumber works and the total work fee she charges is proportional. Her fee for 5 hours of work is $350.

Which of the following could be combinations of values for the plumber's work hours and total work fee she charges?

Solution:

Find similar ratios to 5/350

Similar Ratios,

1/70

2/140

3/210

3.5/245

4/280

6/420

7.25/507.50

Answers:

B) 3.5 hours and $245

C) 6 hours and $420

D) 7.25 hours and $507.50

Harman [31]3 years ago
5 0

Answer: B, C, D

Step-by-step explanation:      Got it from khan:)

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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

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
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Find the output (y) of the function y=6x+4 if the input (x) is 2
Dmitry_Shevchenko [17]

Answer:

f(2) = 16

or

y = 16

Step-by-step explanation:

Step 1: Write out function

y = 6x + 4

Step 2: Define variable for problem

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Step 3: Plug into function f(x)

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Step 4: Change f(2) to y

y = 16

6 0
3 years ago
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