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SCORPION-xisa [38]
3 years ago
9

¿Cuantos triángulos hay?

Mathematics
2 answers:
artcher [175]3 years ago
8 0

Answer:

18

Step-by-step explanation:

Fórmula:

\frac{a(a + 1)(b)}{2}  =  \frac{3(3 + 1)(3)}{2}  =  \frac{3(4)(3)}{2}  =  \frac{36}{2}  = 18

nalin [4]3 years ago
5 0

Answer:

Yo veo 16. tal vez sea mas...

Step-by-step explanation:

You might be interested in
1. Look for a pattern in the first three equations. Then, fill in the missing numbers in the rest of the
Alla [95]

Answer:

1=1 times 2/2 1 + 2=2 times 3/2 1 +2+3=3 times 4/2 1+2+3+4=4 times 5/2 1+2+3+4+5=5 times 6/2

4 0
2 years ago
Is 5361 divisible by three how do you know just by looking at it
sleet_krkn [62]

Answer:

Yes!

Step-by-step explanation:

Here is a trick. If you add up all of the numbers in the number, and if that is divisible by three, the number is divisible by three. So...

5+3+6+1=15

15 is divisible by 3,

so 5361 is divisible by 3

(if you wanted to know, it is 1787)

8 0
2 years ago
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

4 0
3 years ago
Please help....And please show steps..
MrRissso [65]

Answer:

1476:1845

Step-by-step explanation:

4+5=9

4÷9 ×3321

And

5÷9=3321

1476:1845

4 0
2 years ago
10 meters wide with a surface area of 80 square meters. ​
Vlada [557]

Answer:

8

Step-by-step explanation:

5 0
3 years ago
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