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3 years ago

Work out the height of this triangle with base, b = 8mm and area, A= 218.8mm2.

2 answers:

5 0

You have to devide 218.8 with the base
h= 218.8 devided by 8mm gives us 27.35 mm and if you multiply 8 by 27.35 it gives you the same answer for the area

Vinil7 [7]3 years ago

3 0

Answer:

54.7

Step-by-step explanation:

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In rectangle ABCD with side AD = 6 cm, point M is the midpoint of side

sineoko [7]

So if the measure of angle AMB = 90 so the another right triangle is formed which is ADM, since the it is a right triangle the legs are equal, then the lenght AD = DM and we can solve the length of AM
AM = sqrt( AD^2 + DM^2)
AM = sqrt( 6^2 + 6^2)
AM = 6sqrt(2)
now we can solve the length of AB
AB = sqrt ( AM^2 + MB^2)
AB = sqrt ( 6sqrt(2)^2 + 6sqrt(2)^2)
AB = 12

so the perimeter = 2(6) + 2(12) = 36

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kupik [55]

Answer:

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Step-by-step explanation:

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3 years ago

If a right circular cone is intersected by a plane only at its vertex, as in the

HACTEHA [7]

The answer is E. A circle

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3 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

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3 years ago