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irina1246 [14]
3 years ago
12

What is 60% of 500??

Mathematics
1 answer:
tino4ka555 [31]3 years ago
4 0

300! happy to answer!

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A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and
goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}

Solve for S(t):

\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}

e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}

The left side is the derivative of a product:

\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}

Integrate both sides:

e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt

e^{t/800}S(t)=64e^{t/800}+C

S(t)=64+Ce^{-t/800}

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

0=64+C\impleis C=-64

and so (b) the amount of sugar in the tank at time t is

S(t)=64\left(1-e^{-t/800}\right)

As t\to\infty, the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

7 0
3 years ago
Please help me. It’s due in 20 minutes.
egoroff_w [7]

Answer:whatever the first number was times 2

Step-by-step explanation:

3 0
3 years ago
What is the simplified answer to 5/2
Yanka [14]
Maybe it the mixed number form. 2 1/2?
8 0
3 years ago
If f(x) = x + 4 and g(x)=x^2-1, what is m(g o f)(x)?
Andreas93 [3]

Until now, given a function  f(x), you would plug a number or another variable in for x. You could even get fancy and plug in an entire expression for x. For example, given  f(x) = 2x + 3, you could find f(y2 – 1) by plugging y2 – 1 in for x to get f(y2 – 1) = 2(y2 – 1) + 3 = 2y2 – 2 + 3 = 2y2 + 1.

In function composition, you're plugging entire functions in for the x. In other words, you're always getting "fancy". But let's start simple. Instead of dealing with functions as formulas, let's deal with functions as sets of (x, y) points:

Let f = {(–2, 3), (–1, 1), (0, 0), (1, –1), (2, –3)} and  

let g = {(–3, 1), (–1, –2), (0, 2), (2, 2), (3, 1)}.  

 

Find (i) f (1), (ii) g(–1), and (iii) (g o f )(1).

(i) This type of  exercise is meant to emphasize that the (x, y) points are really (x, f (x)) points. To find  f (1), I need to find the (x, y) point in the set of (x, f (x)) points that has a first coordinate of x = 1. Then f (1) is the y-value of that point. In this case, the point with x = 1 is (1, –1), so:

8 0
3 years ago
Read 2 more answers
What's the expression for the sum of two and the quotation of a number x and five ??
7nadin3 [17]
X + x = 2
1 × 5 = 5
So it would be blank plus blank gives you x=2
Then blank times blank gives you x=5
3 0
2 years ago
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