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andreyandreev [35.5K]

3 years ago

12

The aircraft A330 flies at an average speed of 400 miles per hour. The expression, 400t gives the distance traveled by the aircr

aft in t hours. Find the distance traveled by the A330 in 3 hours.

2 answers:

olga nikolaevna [1]3 years ago

5 0

We're told that distance traveled in t hours is 400t. So the distance traveled in 3 hours is 400(3), or 1200 miles.

matrenka [14]3 years ago

4 0

the distances traveled is 1,200.

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3 years ago

5. A tank initially contains 20 lb of salt dissolved in 200-gallon of water. Starting at time t = 0, a solution containing 0.5 l

sladkih [1.3K]

Answer:

(a) Amount of salt as a function of time

$A(t)=100-80e^{-0.02t}$

(b) The time at which the amount of salt in the tank reaches 50 lb is 23.5 minutes.

(c) The amount of salt when t approaches to +inf is 100 lb.

Step-by-step explanation:

The rate of change of the amount of salt can be written as

$\frac{dA}{dt} =rate\,in\,-\,rate\,out\\\\\frac{dA}{dt}=C_i*q_i-C_o*q_o=C_i*q_i-\frac{A(t)}{V}*q_o\\\\\frac{dA}{dt}=0.5*4-\frac{A(t)}{200}*4\\\\\frac{dA}{dt} =\frac{100-A(t)}{50}=-(\frac{A(t)-100}{50})$

Then we can rearrange and integrate

$\frac{dA}{dt}= -(\frac{A(t)-100}{50})\\\\\int \frac{dA}{A-100}=-\frac{1}{50} \int dt\\\\ln(A-100)=-\frac{t}{50}+C_0\\\\ A-100=Ce^{-0.02*t}\\\\\\A(0)=20 \rightarrow 20-100=Ce^0=C\\C=-80\\\\A=100-80e^{-0.02t}$

Then we have the model of A(t) like

$A(t)=100-80e^{-0.02t}$

(b) The time at which the amount of salt reaches 50 lb is

$A(t)=100-80e^{-0.02t}=50\\\\e^{-0.02t}=(50-100)/(-80)=0.625\\\\-0.02*t=ln(0.625)=-0.47\\\\t=(-0.47)/(-0.02)=23.5$

(c) When t approaches to +infinit, the term e^(-0.02t) approaches to zero, so the amount of salt in the solution approaches to 100 lb.

7 0

3 years ago

Simplify -3/4 ÷ 2/-8. (1 point)

Katarina [22]

Answer:

=3/1 or 3

Step-by-step explanation:

-3/4÷2/-8

=-3 × -8/4 × 2

=24/8

=24 ÷ 8/8 ÷ 8

=3/1

5 0

4 years ago