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laiz [17]
3 years ago
5

If a paper plane is thrown from the top of a 147-foot cliff into the water below, the height of the plane at any given time can

be determined by the formula h = -3t² + 147, where h is the height of the plane at t seconds. After how many seconds will the plane be at a height of exactly 20.25 feet?a. 7 seconds b. -6.5 seconds c. 6.5 seconds d. 7 seconds
Mathematics
1 answer:
sasho [114]3 years ago
6 0

Answer:

c. 6.5 seconds.

Step-by-step explanation:

We have been given that a a paper plane is thrown from the top of a 147-foot cliff into the water below, the height of the plane at any given time can be determined by the formula h = -3t^2+147, where h is the height of the plane at t seconds.

To find the time, when plane will be at a height of exactly 20.25 feet, we will substitute h=20.25 in our given formula and solve for t as:

20.25= -3t^2+147

-3t^2+147=20.25

-3t^2+147-147=20.25-147

-3t^2=-126.75

\frac{-3t^2}{-3}=\frac{-126.75}{-3}

t^2=42.25

Take square root of both sides:

t=\pm \sqrt{42.25}

t=\pm 6.5

Since time cannot be negative, therefore, the plane will be at a height of exactly 20.25 feet after 6.5 seconds and option 'c' is the correct choice.

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A 200-gal tank contains 100 gal of pure water. At time t = 0, a salt-water solution containing 0.5 lb/gal of salt enters the tan
Artyom0805 [142]

Answer:

1) \frac{dy}{dt}=2.5-\frac{3y}{2t+100}

2) y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}

3) 98.23lbs

4) The salt concentration will increase without bound.

Step-by-step explanation:

1) Let y represent the amount of salt in the tank at time t, where t is given in minutes.

Recall that: \frac{dy}{dt}=rate\:in-rate\:out

The amount coming in is 0.5\frac{lb}{gal}\times 5\frac{gal}{min}=2.5\frac{lb}{min}

The rate going out depends on the concentration of salt in the tank at time t.

If there is y(t) pounds of  salt and there are 100+2t gallons at time t, then the concentration is: \frac{y(t)}{2t+100}

The rate of liquid leaving is is 3gal\min, so rate out is =\frac{3y(t)}{2t+100}

The required differential equation becomes:

\frac{dy}{dt}=2.5-\frac{3y}{2t+100}

2) We rewrite to obtain:

\frac{dy}{dt}+\frac{3}{2t+100}y=2.5

We multiply through by the integrating factor: e^{\int \frac{3}{2t+100}dt }=e^{\frac{3}{2} \int \frac{1}{t+50}dt }=(50+t)^{\frac{3}{2} }

to get:

(50+t)^{\frac{3}{2} }\frac{dy}{dt}+(50+t)^{\frac{3}{2} }\cdot \frac{3}{2t+100}y=2.5(50+t)^{\frac{3}{2} }

This gives us:

((50+t)^{\frac{3}{2} }y)'=2.5(50+t)^{\frac{3}{2} }

We integrate both sides with respect to t to get:

(50+t)^{\frac{3}{2} }y=(50+t)^{\frac{5}{2} }+ C

Multiply through by: (50+t)^{-\frac{3}{2}} to get:

y=(50+t)^{\frac{5}{2} }(50+t)^{-\frac{3}{2} }+ C(50+t)^{-\frac{3}{2} }

y(t)=(50+t)+ \frac{C}{(50+t)^{\frac{3}{2} }}

We apply the initial condition: y(0)=0

0=(50+0)+ \frac{C}{(50+0)^{\frac{3}{2} }}

C=-12500\sqrt{2}

The amount of salt in the tank at time t is:

y(t)=(50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}

3) The tank will be full after 50 mins.

We put t=50 to find how pounds of salt it will contain:

y(50)=(50+50)- \frac{12500\sqrt{2} }{(50+50)^{\frac{3}{2} }}

y(50)=98.23

There will be 98.23 pounds of salt.

4) The limiting concentration of salt is given by:

\lim_{t \to \infty}y(t)={ \lim_{t \to \infty} ( (50+t)- \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }})

As t\to \infty, 50+t\to \infty and \frac{12500\sqrt{2} }{(50+t)^{\frac{3}{2} }}\to 0

This implies that:

\lim_{t \to \infty}y(t)=\infty- 0=\infty

If the tank had infinity capacity, there will be absolutely high(infinite) concentration of salt.

The salt concentration will increase without bound.

6 0
3 years ago
There are 4 red golf balls and 2 green golf balls in a bucket. You select a ball at random, keep it, and select again. Find the
Alona [7]

Answer:

C

Step-by-step explanation:

It should be 2/3 because there are 6 balls and two of them are green which gives the fraction 2/6 and can be reduced to 1/3

6 0
3 years ago
Read 2 more answers
PLEASE HELP!! Trevor is writing a paper that must be 50,000 words long. He had already written 23,210 words. Write an inequality
almond37 [142]

Answer:

50,000 ≤ 23,210 + 5x

x = Average number of words per week

Step-by-step explanation:

Start off with 50,000 because that's how many words the paper needs to have.

Use "≤" because Trevor needs to write at least 50,000 words. (He can write more).

Trevor already wrote 23,210 words plus x words per week for 5 weeks.

(Solving for x will help you find the average number of words Trevor must write per week for 5 weeks).

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3 years ago
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katrin2010 [14]

Answer:

14201.63

Step-by-step explanation:

14187.63

+   14.50

=14201.63

7 0
3 years ago
1000000000000000000000000000000000000000000000X100/1/222X1
jok3333 [9.3K]

Answer:

4.5045045 x 10^44

Step-by-step explanation:

I just put it into a calculator idk if its even right

5 0
3 years ago
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