What is the solution to the system of equations below?

HELP ASAP!!!!!! WILL MARK BRAINLIEST

Lana71 [14]

Answer:

Its the one at the bottom on the left

Step-by-step explanation:

It is that one because one its a polygon and has angles and is congruent

4 0

3 years 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

3 years ago

What ratio shows 20:12 written in simplest form?

arsen [322]

It is 5:3 or 5/3

Divide it by 4

4 0

3 years ago

Read 2 more answers

How to solve using substitution y=3/4X+5 4x-3y=-1

Pachacha [2.7K]

Please, use parentheses to enclose each fraction:
y=3/4X+5 should be written as y=(3/4)X+5

Let's eliminate the fraction 3/4 by multiplying the above equation through by 4:

4[y] = 4[(3/4)x + 5]
Then 4y = 3x + 20
(no fraction here)

Let 's now solve the system

4y=3x + 20
4x-3y=-1

We are to solve this system using subtraction. To accomplish this, multiply the first equation by 3 and the second equation by 4. Here's what happens:

12y = 9x + 60 (first equation)
16x-12y = -4, or -12y = -4 - 16x (second equation)

Then we have

12y = 9x + 60
-12y =-16x - 4

If we add here, 12y-12y becomes zero and we then have 0 = -7x + 56.
Solving this for x: 7x = 56; x=8

We were given equations
y=3/4X+5
4x-3y=-1

We can subst. x=8 into either of these eqn's to find y. Let's try the first one:

y = (3/4)(8)+5 = 6+5=11

Then x=8 and y=11.

You should check this result. Subst. x=8 and y=11 into the second given equation. Is this equation now true?

5 0

3 years ago

You are drawing two cards, without replacement, from a standard deck of 52 cards. What is the

Serga [27]

Answer:

Dear student,

Answer to your query is provided below

Probability of drawing a 2 at first is (4/52) = (1/13)

Probability of drawing a face card in second try without replacement = 12/51

Step-by-step explanation:

Total no. Of cards = 52

Total no. Of cards of 2 = 4

Prob. Of drawing 2 = 4/52

Total no. Of face card = 12

Total no.of cards left in deck = 51

Prob. Of drawing face card = 12/51

5 0

3 years ago