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ZanzabumX [31]
3 years ago
15

Justin is going to rent a truck for one day. There are two companies he can choose from, and they have the following prices. Com

pany A charges $ 124 and allows unlimited mileage. Company B has an initial fee of $ 75 and charges an additional $ 0.70 for every mile driven. For what mileages will Company A charge less than Company B? Use m for the number of miles driven, and solve your inequality for m .
Mathematics
1 answer:
algol [13]3 years ago
4 0
A=124
B=0.7x+75
124<0.7x+75
49<0.7x
70<x
When you drive more than 70 miles, company A is cheaper
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Find a particular solution to y" - y + y = 2 sin(3x)
leonid [27]

Answer with explanation:

The given differential equation is

y" -y'+y=2 sin 3x------(1)

Let, y'=z

y"=z'

\frac{dy}{dx}=z\\\\d y=zdx\\\\y=z x

Substituting the value of , y, y' and y" in equation (1)

z'-z+zx=2 sin 3 x

z'+z(x-1)=2 sin 3 x-----------(1)

This is a type of linear differential equation.

Integrating factor

     =e^{\int (x-1) dx}\\\\=e^{\frac{x^2}{2}-x}

Multiplying both sides of equation (1) by integrating factor and integrating we get

\rightarrow z\times e^{\frac{x^2}{2}-x}=\int 2 sin 3 x \times e^{\frac{x^2}{2}-x} dx=I

I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx -\int \frac{2\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx-\frac{2I}{3}\\\\\frac{5I}{3}=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{5}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{5} dx

8 0
3 years ago
What is the answer for 7/5=g-6/5
Nataly [62]

Answer:

13/5

Step-by-step explanation:

g=7/5+6/5

g=13/5

8 0
3 years ago
Radioactive isotopes are commonly used to estimate artifact ages. Let X be the time in years for a Carbon-14 atom to decay to Ni
Alexeev081 [22]

Answer:

a) 1 / years

b) P ( 5000 < X < 6000) =  0.06222

c) 8267 years

d) T_1/2 = 5730.25 years          

Step-by-step explanation:

Given:

- The decay constant λ = 1 / 8267

- X is an exponential random Variable

Find:

a. What are the units of What are the units of λ?

b. What is the probability the decay time is between 5000 and 6000 years?

c. Compute SD(X).

d. Compute the median decay time. The result is called the half-life of Carbon-14?

Solution:

- The decay constant λ means the rate at which the C-14 atoms decays into N-14 atom. The rate is expressed in units of 1 / year.

- The random variable X follows an exponential distribution which has a probability mass function and cumulative density functions as follows:

                    P ( X = t ) = λ*e^(-λ*t)

                    P ( X = t ) = e^(-t/8267) / 8267

                    P ( X < t ) = 1 - e^(-λ*t) = 1 - e^(-t/8267)

- The probability of the decay time between years 5000 and 6000 years is:

                    P ( 5000 < X < 6000) = 1 - e^(-6000/8267) - 1 + e^(-5000/8267)

                    P ( 5000 < X < 6000) =  0.06222

- The standard deviation of the exponential distribution is given by:

                    SD(X) = 1 / λ = 8267 years

- The median decay time for an exponential distribution is given by:

                    T_1/2 = Ln(2) / λ = Ln(2)*8267

                    T_1/2 = 5730.25 years                        

3 0
3 years ago
Read 2 more answers
The radius of the base of a cylinder is increasing at a rate of 7 millimeters per hour. The height of the cylinder is fixed at 1
Ilya [14]

Answer:

The rate of change of the volume of the cylinder at that instant = 791.28\ mm^3/hr

Step-by-step explanation:

Given:

Rate of increase of base of radius of base of cylinder = 7 mm/hr

Height of cylinder = 1.5 mm

Radius at a certain instant = 12 mm

To find rate of change of volume of cylinder at that instant.

Solution:

Let r represent radius of base of cylinder at any instant.

Rate of increase of base of radius of base of cylinder can be given as:

\frac{dr}{dt}=7\ mm/hr

Volume of cylinder is given by:

V=\pi\ r^2h

Finding derivative of the Volume with respect to time.

\frac{dV}{dt}=\pi\ h\ 2r\frac{dr}{dt}

Plugging in the values given:

\frac{dV}{dt}=\pi\ (1.5)\ 2(12)(7)

\frac{dV}{dt}=252\pi

Using \pi=3.14

\frac{dV}{dt}=252(3.14)

\frac{dV}{dt}=791.28\ mm^3/hr (Answer)

Thus rate of change of the volume of the cylinder at that instant = 791.28\ mm^3/hr

6 0
3 years ago
Yolanda has finished 45 percent of her homework and she has finished 27 problems. How many problems was Yolanda assigned
masya89 [10]
If 45 percent of the homework is only 27 problems, she needs to figure out this:

What is 47 percent of 60?
              or
47% x 60

Which is 27. 

So, Yolanda was assigned 60 problems.


6 0
3 years ago
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