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3 years ago

What comes next 2 7 8 3 12 9

Mathematics

2 answers:

lana [24]3 years ago

8 0

I think it would be 6

algol133 years ago

6 0

2, 3

7, 8 

12, 13 

9, 10 

10 is next, then 13
the answer is 19 becuase thats what comes next.

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1.3 ½ days=______hrs.

spayn [35]

1) 84 hours
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3) 4800 seconds
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8 0

2 years ago

Find the smaller root for x^2 -9 =-5

Charra [1.4K]

X^2-9=-5
x^2-4=0
(x+2)(x-2)=0 [Difference of Two Squares]
x=-2,2
-2 is the smaller root.

6 0

3 years ago

If 0.9x-1.3=-3.1, what is the value of x-0.8?

Butoxors [25]

−2.8
Okay so "0.9x-1.3=-3.1" x=-2

0.9x−1.3=−3.1

add 1.3 to both sides

0.9x−1.3+1.3=−3.1+1.3

0.9x=−1.8

divide both sides by 0.9

0.9x/0.9=−1.8/0.9

x=−2

So if you plug in -2 into "x-0.8" you get −2.8

=−2−0.8

=−2-0.8

=−2.8

Hope this helps.

3 0

3 years ago

Read 2 more answers

A swimming pool with a volume of 30,000 liters originally contains water that is 0.01% chlorine (i.e. it contains 0.1 mL of chlo

SpyIntel [72]

Answer:

$R_{in}=0.2\dfrac{mL}{min}$

$C(t)=\dfrac{A(t)}{30000}$

$R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

$A(t)=300+2700e^{-\dfrac{t}{1500}},$ A(0)=3000$

Step-by-step explanation:

The volume of the swimming pool = 30,000 liters

(a) Amount of chlorine initially in the tank.

It originally contains water that is 0.01% chlorine.

0.01% of 30000=3000 mL of chlorine per liter

A(0)= 3000 mL of chlorine per liter

(b) Rate at which the chlorine is entering the pool.

City water containing 0.001%(0.01 mL of chlorine per liter) chlorine is pumped into the pool at a rate of 20 liters/min.

$R_{in}=$ (concentration of chlorine in inflow)(input rate of the water)

$=(0.01\dfrac{mL}{liter}) (20\dfrac{liter}{min})\\R_{in}=0.2\dfrac{mL}{min}$

Volume of the pool =30,000 Liter

$Concentration, C(t)= \dfrac{Amount}{Volume}\\C(t)=\dfrac{A(t)}{30000}$

(d) Rate at which the chlorine is leaving the pool

$R_{out}=$ (concentration of chlorine in outflow)(output rate of the water)

$= (\dfrac{A(t)}{30000})(20\dfrac{liter}{min})\\R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}$

(e) Differential equation representing the rate at which the amount of sugar in the tank is changing at time t.

$\dfrac{dA}{dt}=R_{in}-R_{out}\\\dfrac{dA}{dt}=0.2- \dfrac{A(t)}{1500}$

We then solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{1500}=0.2\\$The integrating factor: e^{\int \frac{1}{1500}dt} =e^{\frac{t}{1500}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{1500}}+\dfrac{A}{1500}e^{\frac{t}{1500}}=0.2e^{\frac{t}{1500}}\\(Ae^{\frac{t}{1500}})'=0.2e^{\frac{t}{1500}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{1500}})'=\int 0.2e^{\frac{t}{1500}} dt\\Ae^{\frac{t}{1500}}=0.2*1500e^{\frac{t}{1500}}+C, $(C a constant of integration)\\Ae^{\frac{t}{1500}}=300e^{\frac{t}{1500}}+C\\$Divide all through by e^{\frac{t}{1500}}\\A(t)=300+Ce^{-\frac{t}{1500}}$

Recall that when t=0, A(t)=3000 (our initial condition)

$3000=300+Ce^{0}\\C=2700\\$Therefore:\\A(t)=300+2700e^{-\dfrac{t}{1500}}$

3 0

3 years ago

I need help on this I don’t get it

levacccp [35]

Answer:

picture: the one with the triangle in red on the base (first picture of box)

Equation: a^2 + 11^2 = c^2

Step-by-step explanation:

as we can see for the first one, the equation we see is 20^2 + 8^2 = c^2, so you look for the picture with those numbers on the triangle.

and to find the equation we look at the picture and see that in the triangle the numbers are 11 and a. And we know the equation has to be 11^2 + a^2 = c^2 And see if that equation is there

5 0

3 years ago