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Delicious77 [7]

2 years ago

13

Carmen is taking a road trip. After 25 miles, she reaches a stretch of highway where she can drive at a constant speed of 65 mil

es per hour. Carmen's friend in the passenger seat is trying to figure out the minimum number of hours Carmen will need to drive to reach over 350 total miles for the trip, assuming she keeps a steady pace.
Carmen's friend creates the inequality 65t + 25 > 350, where t is the driving time, in hours.

What statement is the most accurate?

Carmen can meet her goal in less than 5 hours.

Carmen can meet her goal in less than 4 hours.

Carmen needs more than 5 hours to meet her goal.

Carmen needs more than 6 hours to meet her goal.

2 answers:

garri49 [273]2 years ago

7 0

Answer:

t > 5

Step-by-step explanation:

Step-by-step explanation:

350 - 25 = 325

325 / 65 = 5

It said to reach over 350 miles, so it’s greater than 5

t > 5

abruzzese [7]2 years ago

7 0

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3 years ago

A tank with capactity 500 gal originally contains 200 gal water with 100 lbs of salt mixed into it. Water containing 1 lb of sal

prisoha [69]

Answer:

The amount of salt in the tank at any moment $t$ is

$f(t)=-\frac {4\times 10^6}{(200+t)^2}+200+t$

The concentration of salt in the tank when it is at the point of overflowing is $0.968$ .

The theoretical limiting concentration of an infinite tank is $1$ lb per gallon.

Step-by-step explanation:

Let $f(t)$ be the amount of salt in the tank at any time $t.$

Then, its time rate of change, $f'(t)$ , by (balance law).

Since three gallons of salt water runs in the tank per minute, containing $1$ lb of salt, the salt rate is

$3.1=3$

The amount of water in the tank at any time $t$ is.

$200+(3-2)t=200+t,$

Now, the outflow is $2$ gal of the solution in a minute. That is $\frac 2{200+t}$ of the total solution content in the tank, hence $\frac 2{200+t}$ of the salt salt content $f(t)$ , that is $\frac{2f(t)}{200+t}$ .

Initially, the tank contains $100$ lb of salt,

Therefore we obtain the initial condition $f(0)=100$

Thus, the model is

$f'(t)=3-\frac{2f(t)}{200+t}, f(0)=100$

$\Rightarrow f'(t)+\frac{2}{200+t}f(t)=3, f(0)=100$

$p(t)=\frac{2}{200+t} \;\;\text{and} \;\;q(t)=3$ Linear ODE.

so, an integrating factor is

$e^{\int p dt}=e^{2\int \frac{dt}{200+t}=e^{\ln(200+t)^2}=(200+t)^2$

and the general solution is

$f(t)(200+t)^2=\int q(200+t)^2 dt+c$

$\Rightarrow f(t)=\frac 1{(200+t)^2}\int 3(200+t)^2 dt+c$

$\Rightarrow f(t)=\frac c{(200+t)^2}+200+t$

Now using the initial condition and find the value of $c$ .

$100=f(0)=\frac c{(200+0)^2}+200+0\Rightarrow -100=\frac c{200^2}$

$\Rightarrow c=-4000000=-4\times 10^6$

$\Rightarrow f(t)=-\frac {4\times 10^6}{(200+t)^2}+200+t$

is the amount of salt in the tank at any moment $t.$

Initially, the tank contains $200$ gal of water and the capacity of the tank is $500$ gal. This means that there is enough place for

$500-200=300$ gal

of water in the tank at the beginning. As concluded previously, we have one new gal in the tank at every minute. hence the tank will be full in $30$ min.

Therefore, we need to calculate $f(300)$ to find the amount of salt any time prior to the moment when the solution begins to overflow.

$f(300)=-\frac{4\times 10^6}{(200+300)^2}+200+300=-16+500=484$

To find the concentration of salt at that moment, divide the amount of salt with the amount of water in the tank at that moment, which is $500$ L.

$\text{concentration at t}=300=\frac{484}{500}=0.968$

If the tank had an infinite capacity, then the concentration would be

$\lim\limits_{t \to \infty} \frac{f(t)}{200+t}= \lim\limits_{t \to \infty}\left(\frac{\frac{3\cdot 10^6}{(200+t)^2}+(200+t)}{200+t}\right)$

$= \lim\limits_{t \to \infty} \left(\frac{4\cdot 10^6}{(200+t)^3}+1\right)$

$=1$

Hence, the theoretical limiting concentration of an infinite tank is $1$ lb per gallon.

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4 years ago

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fredd [130]

Answer:

89.6

Step-by-step explanation:

2 divided by 4 = 0.5

0.5 x 3 = 3.5

3.5+ 8=12.3

12.3 + 5 = 12.8

12.8 x 7= 89.6

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Answer:

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