Suppose a building has 5 floors (1-5) which are occupied by officesThe ground floor (0) is not

Which table of ordered pairs represents a proportional relationship?

grin007 [14]

Answer:

It is the second one

Step-by-step explanation:

A proportional relationship happens when you get the same answer every time that you divide the y by the x.

(2,10) (4,20) (6,30)

10/2 = 5

20/4 = 5

30/6 = 5

That is not true for the other tables.

7 0

1 year ago

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

2 years ago

The following list details the numbers of tickets sold for a talent show grade 9:22; grade 10:32; grade 11: 41; and grade 12:30.

Montano1993 [528]

Answer: C. domain: {9, 10, 11, 12); range: (22, 32, 41, 30)

Step-by-step explanation:

The data set is:

(9, 22)

(10,32)

(11, 41)

(12, 30).

In the usual notation, the number at the left is the input (belons to the domain) and the number in the right is the output (belongs to the range).

Then the domain would be:

{9, 10, 11, 12}

and the range:

{22, 32, 41, 30}

The correct option is C

7 0

2 years ago

Which unit would you use to measure the weight of a car <br> A. mg <br> B. kg <br> C. g<br> D. km

kobusy [5.1K]

Answer:

A. Mg

Step-by-step explanation:

Brainliest plezzzzzzzzz

4 0

3 years ago

Read 2 more answers

A textbook store sold a combined total of 290 math and history textbooks in a week. The number of math textbooks sold was 64 mor

11Alexandr11 [23.1K]

Answer:

113

Step-by-step explanation:

290 - 64 = 226

226 ÷ 2 = 113

4 0

2 years ago