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3 years ago

The cost C (in dollars) for the care and maintenance of a horse and carriage is C=15x+2000 where x is the number of rides.

Mathematics

1 answer:

MissTica3 years ago

6 0

Answer: 100 rides are needed to break even.

Step-by-step explanation:

We have given the cost function

C(x)=15x+2000 where x is the number of rides.

And rides cost 35$

⇒ revenue function would be 35 times x

i.e.R(x)=35 x , where x is the number of rides.

Break even point of a firm occurs when at a certain point x the total cost equals to the total revenue.

i.e. at break even point

total revenue=total cost

⇒35x=15x+2000

⇒35-15x=2000[subtract 15x from both sides]

⇒20x=2000 [simplify]

⇒x=2000/20[dividing both sides with 20]

⇒x=100

∴ 100 rides are needed to break even.

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A metal beam was brought from the outside cold into a machine shop where the temperature was held at 65degreesF. After 5 min, t

ivolga24 [154]

Answer:

The beam initial temperature is 5 °F.

Step-by-step explanation:

If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

$T(t)=T_a+(T_0-T_a)e^{-kt}$

where $T_a$ is the ambient temperature, $T_0$ is the initial temperature, $t$ is the time and $k$ is a constant yet to be determined.

The goal is to determine the initial temperature of the beam, which is to say $T_0$

We know that the ambient temperature is $T_a=65$ , so

$T(t)=65+(T_0-65)e^{-kt}$

We also know that when $t=5 \:min$ the temperature is $T(5)=35$ and when $t=10 \:min$ the temperature is $T(10)=50$ which gives:

$T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}$

$T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}$

Rearranging,

$35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}$

$50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}$

If we divide these two equations we get

$\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}$

$\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}$

Now, that we know the value of $k$ we can use it to find the initial temperature of the beam,

$35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5$

so the beam started out at 5 °F.

6 0

3 years ago

The quantities x and y are proportional.

Ksenya-84 [330]

Answer:

r=3,r=25/27,r=3

Step-by-step explanation:

In equation y=rx

make r the subject of the formula by dividing both sides by x

r=y/x

1). when y=15 and x=5

r=15/5

r=3

2). when y=25 and x=81/3

r=25/81/3

r=25÷81/3

r=25*3/81

r=75/27

r=25/27

3). when y=33 and x=11

r=33/11

r=3

4 0

3 years ago

PLEASE HELP!!!!!! which line is a linear model for the data.

Aleksandr [31]

Answer:

you could go watch you tube videos to help you with that buddy.

Step-by-step explanation:

6 0

3 years ago

anthony's school has 692 kids.the cafeteria orders milk cartons for the week if each kid drinks 1 carton a day for 5 days how ma

amid [387]

Answer:

so at Anthony's school it is 692 kids pluss him

one school week has five days

so 692 ×5

600×5=3000

90×5=450

2×5=10

3000

450

=3460 cartons of milk should the cafetaria order for one week

6 0

3 years ago

Simplify 6 - 2³ + (-9 + 5) · 2.

oksian1 [2.3K]

Answer: your answer is -10

Step-by-step explanation:j

if you dont believe me check your answer on a math website

8 0

3 years ago

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