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3 years ago

Find the approximate side length of a square game board with an area of 192 in2.

Mathematics

2 answers:

padilas [110]3 years ago

8 0

Squares have 4 equal sides, thus, the answer is 48 in^2.

brilliants [131]3 years ago

5 0

Look it's a square therefor it will have 4 sides hence the answer will be 48 /2 which is 16

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What is the sum of a 56-term arithmetic sequence where the first term is 6 and the last term is 391?

Sati [7]

$S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\a_1=6;\ n=56;\ a_{56}=391\\\\subtitute\\\\S_{56}=\dfrac{6+391}{2}\cdot56=\dfrac{397}{2}\cdot56=397\cdot28=11,116$

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2 years ago

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An oil rig has a height of 74.98 m visible above water and extends 518.16 m below the surface to the sea floor. What is the tota

pashok25 [27]

Answer:

So you know that the height that is visible above x=0 is 74.98 m
From x=0 below it has a total height of 518.16m to the sea floor

That tells us that we have two parts account for

74.98+518.16m = 593.14 m

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2 years ago

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A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows