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3 years ago

9

If 14 g of a radioactive substance are present initially and 3 yr later only 7 g remain, how much of the substance will be pres

ent after 6 yr?

1 answer:

marishachu [46]3 years ago

5 0

It would be 48 bc if you add it all up and subtract it and add again and multiply and divide and multiply within a and the substance of present after 6 yr you will see that i have wasted your time and that is not the answer

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Of the total amount of available water on the earth, what percent is usable?

prohojiy [21]

Only 20%of earths water is useable

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4 years ago

Examine triangle ABC.<br><br> What is the value of x?<br><br> 37<br> 48<br> 127<br> 138

Flura [38]

The outside angle of B needs to equal 180 - 79 = 101 degrees.

The 3 outside angles of a triangle need to equal 360 degrees.

Set up the equation with what we know:

X + (x+5) + 101 = 360

Simplify:

2x + 106 = 360

Subtract 106 from both sides:

2x = 254

Divide both sides by 2:

x = 254 / 2

x = 127

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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.

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3 years ago

What is the slope of this line?

ArbitrLikvidat [17]

Answer:

The slope is 1 because its rise over run is 1/1 which equals to 1. Its also a positive slope. Hope this helps!

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3 years ago

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What is the horizontal asymptote of f(x)=-2x/x+1<br> A.-2<br> B.-1<br> C.0<br> D.1

Mrrafil [7]

A horizontal asymptote of a function f(x) is given by y = lim f(x) as x --> ∞ and x --> –∞. In this case,
$\lim_{x \to \infty} \frac{-2x}{x+1} = \lim_{x \to \infty} \frac{-2x}{x(1+ \frac{1}{x} )}\\
=\lim_{x \to \infty} \frac{-2}{1+ \frac{1}{x} }\\
= -2
$

Thus, the horizontal asymptote of f(x) is y = –2.

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3 years ago

Read 2 more answers