Which graph correctly represents this 2y+x=-5and y+3x=0

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 is up guys!

melisa1 [442]

Answer:

Yes, No, Yes

Step-by-step explanation:

Use the Pythagorean theorem:

a² + b² = c²

Verify this for each line:

1.

28² + 45² = 53²
2809 = 2809
Yes

2.

33² + 56² = 64²
4225 ≠ 4096
No

3.

36² + 77² = 85²
7225 = 7225
Yes

5 0

2 years ago

WILL MARK BRAINLIEST! PLEASE HELP! Write 206 in base 7

Anna71 [15]

Answer:

413_7

Step-by-step explanation:

Convert the following to base 7:

206_10

Hint: | Starting with zero, raise 7 to increasingly larger integer powers until the result exceeds 206.

Determine the powers of 7 that will be used as the places of the digits in the base-7 representation of 206:

Power | \!$\*SuperscriptBox[\(Base$, $Power$]\) | Place value

3 | 7^3 | 343

2 | 7^2 | 49

1 | 7^1 | 7

0 | 7^0 | 1

Hint: | The powers of 7 (in ascending order) are associated with the places from right to left.

Label each place of the base-7 representation of 206 with the appropriate power of 7:

Place | | | 7^2 | 7^1 | 7^0 |

| | | ↓ | ↓ | ↓ |

206_10 | = | ( | __ | __ | __ | )_(_7)

Hint: | Divide 206 by 7 and find the remainder. The remainder is the first digit from the right.

Determine the value of the first digit from the right of 206 in base 7:

206/7=29 with remainder 3

Place | | | 7^2 | 7^1 | 7^0 |

| | | ↓ | ↓ | ↓ |

206_10 | = | ( | __ | __ | 3 | )_(_7)

Hint: | Divide the whole number part of the previous quotient, 29, by 7 and find the remainder. The remainder is the next digit.

Determine the value of the next digit from the right of 206 in base 7:

29/7=4 with remainder 1

Place | | | 7^2 | 7^1 | 7^0 |

| | | ↓ | ↓ | ↓ |

206_10 | = | ( | __ | 1 | 3 | )_(_7)

Hint: | Divide the whole number part of the previous quotient, 4, by 7 and find the remainder. The remainder is the last digit.

Determine the value of the last remaining digit of 206 in base 7:

4/7=0 with remainder 4

Place | | | 7^2 | 7^1 | 7^0 |

| | | ↓ | ↓ | ↓ |

206_10 | = | ( | 4 | 1 | 3 | )_(_7)

Hint: | Express 206_10 in base 7.

The number 206_10 is equivalent to 413_7 in base 7.

Answer: 206_10 =413_7

5 0

2 years ago

Assume that y varies inversely with x. if y=7 when x=-2, find y when x=7

melisa1 [442]

Answer:

y=-2

Step-by-step explanation:

The formula for inverse variation is

xy =k

if y=7 when x=-2, we can substitute these numbers in to find k

(-2)(7) =k

-14 =k

The equation becomes

xy = -14

Let x =7

7y = -14

Divide each side by 7

7y/7 = -14/7

y = -2

8 0

3 years ago

If you change the sign of a point’s -coordinate, how will the location of the point change?

Elena L [17]

It would go to the opposite side

7 0

3 years ago