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sergiy2304 [10]

3 years ago

15

A Shark Weighs 405 kilograms and 68 grams a second shark weighs 324 kilograms and 75 grams how much more does the first shark we

igh in grams than the second shark

1 answer:

never [62]3 years ago

3 0

Answer:

7 grams

Step-by-step explanation:

If you're asking me grams specifically, and not kilograms, subtract 68 from 75 to get your answer. If it's only kilograms subtract 324 from 405. If it's both of them added together subtract 399 from 473, you're welcome!

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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

BIG POINT FOR WHOEVER HELPS ME! PLEASE!!!!!Simplify and determine the coefficient of (-2/3X)(3Y).-X)

rewona [7]

You multiply like terms first, which means -2/3x*-x. It becomes 2/3x*3y. The coefficient of x is 2/3. The coefficient of y is 3. You can’t combine them because they are not like terms with the variables.

6 0

3 years ago

I need help graphing 2x - 6y = 42 I'm just really lazy

solong [7]

Answer:

We can use slope intercept form to get the points needed. Y= -7+1/3x The points are (0,-7) and (3,-6)

Step-by-step explanation:

Subtract 2x from the left side and place it over to the right side with the 42. Now we have -6y= 42-2x. From here we divide by -6 and we get y= -7+1/3x. We know that are slope is 1/3 which the one is the rise and the 3 is the run. We also know that our y intercept is -7. We plot the points at (0,-7) and (3,-6)

3 0

3 years ago

i When 20 is added to a certain number and the sun is divided by 3, the result is 3 times the number, find the number.

Bingel [31]

Answer:

$let \: the \: number \: be \: x \\ \frac{20 + x}{3} = 3x \\ \frac{3(20 + x)}{3} = (3x)3 \\ 20 + x = 9x \\ 20 = 9x - x \\ 20 = 8x \\ \frac{20}{8} = \frac{8x}{8} \\ x = 2.5$

Step-by-step explanation:

Hope that this is helpful .

Have a great day.

3 0

3 years ago

(2-cos^2A)(1+2cot^2A)=(2+Cot^2A)(2-sin^2A)

fomenos

Answer:

2 sin (2A) cos (2A)^4 + 2cos (2A)

---------------------------------------------------

sin (2A)

Step-by-step explanation:

7 0

3 years ago