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SashulF [63]
3 years ago
6

To win at LOTTO in one​ state, one must correctly select 7 numbers from a collection of 50 numbers​ (1 through 46​). The order i

n which the selection is made does not matter. How many different selections are​ possible?
Mathematics
1 answer:
Lerok [7]3 years ago
3 0

3 seletcions are possibleAnswer:

Step-by-step explanation:

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Suppose you invest $400 at an annual interest rate of 7.6% compounded continuously. How much will you have in the account after
likoan [24]
\bf ~~~~~~ \textit{Compounding Continuosly Interest Earned Amount}\\\\
A=Pe^{rt}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to& \$400\\
r=rate\to 7.6\%\to \frac{7.6}{100}\to &0.076\\
t=years\to &1.5
\end{cases}
\\\\\\
A=400e^{0.076\cdot 1.5}\implies A=400e^{0.114}
6 0
3 years ago
Use the divergence theorem to calculate the surface integral s f · ds; that is, calculate the flux of f across s. f(x, y, z) = x
valkas [14]
\mathbf f(x,y,z)=x^4\,\mathbf i-x^3z^2\,\mathbf j+4xy^2z\,\mathbf k
\mathrm{div}(\mathbf f)=\dfrac{\partial(x^4)}{\partial x}+\dfrac{\partial(-x^3z^2)}{\partial y}+\dfrac{\partial(4xy^2z)}{\partial z}=4x^3+0+4xy^2=4x(x^2+y^2)


Let \mathcal D be the region whose boundary is \mathcal S. Then by the divergence theorem,

\displaystyle\iint_{\mathcal S}\mathbf f\cdot\mathrm d\mathbf S=\iiint_{\mathcal D}4x(x^2+y^2)\,\mathrm dV

Convert to cylindrical coordinates, setting

x=r\cos\theta
y=r\sin\theta

and keeping z as is. Then the volume element becomes


\mathrm dV=r\,\mathrm dr\,\mathrm d\theta\,\mathrm dz

and the integral is

\displaystyle\iiint_{\mathcal D}4x(x^2+y^2)\,\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=0}^{z=r\cos\theta+7}4r\cos\theta\cdot r^2\cdot r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta
=\displaystyle4\iiint_{\mathcal D}r^4\cos\theta\,\mathrm dz\,\mathrm dr\,\mathrm d\theta
=\dfrac{2\pi}3
4 0
3 years ago
Which point is located at 2 7/10 PLEASE ANSWER QUICK
german

C, with the ten segments and moving over seven of them you will land on C.

4 0
2 years ago
A tank with a capacity of 1600 L is full of a mixture of water and chlorine with a concentration of 0.0125 g of chlorine per lit
Veronika [31]

Answer:

y(t) = 20 [1600^(-5/3)] x (1600-24t)^ (5/3)

Step-by-step explanation:

1) Identify the problem

This is a differential equation problem

On this case the amount of liquid in the tank at time t is 1600−24t. (When the process begin, t=0 ) The reason of this is because the liquid is entering at 16 litres per second and leaving at 40 litres per second.

2) Define notation

y = amount of chlorine in the tank at time t,

Based on this definition, the concentration of chlorine at time t is y/(1600−24t) g/ L.

Since liquid is leaving the tank at 40L/s, the rate at which chlorine is leaving at time t is 40y/(1600−24t) (g/s).

For this we can find the differential equation

dy/dt = - (40 y)/ (1600 -24 t)

The equation above is a separable Differential equation. For this case the initial condition is y(0)=(1600L )(0.0125 gr/L) = 20 gr

3) Solve the differential equation

We can rewrite the differential equation like this:

dy/40y = -  (dt)/ (1600-24t)

And integrating on both sides we have:

(1/40) ln |y| = (1/24) ln (|1600-24t|) + C

Multiplying both sides by 40

ln |y| = (40/24) ln (|1600 -24t|) + C

And simplifying

ln |y| = (5/3) ln (|1600 -24t|) + C

Then exponentiating both sides:

e^ [ln |y|]= e^ [(5/3) ln (|1600-24t|) + C]

with e^c = C , we have this:

y(t) = C (1600-24t)^ (5/3)

4) Use the initial condition to find C

Since y(0) = 20 gr

20 = C (1600 -24x0)^ (5/3)

Solving for C we got

C = 20 / [1600^(5/3)] =  20 [1600^(-5/3)]

Finally the amount of chlorine in the tank as a function of time, would be given by this formula:

y(t) = 20 [1600^(-5/3)] x (1600-24t)^ (5/3)

7 0
3 years ago
-8x^2+4x+5=0 using quadratic formula
daser333 [38]
So you want to solve for x?

It would be nice if this would easily factor:
(-4x + 5)(2x +1)  = 0    This will not work!

So you need to use the quadratic formula:
a = -8, b = 4, c = 5

x = \frac{-b+/-  \sqrt{b^{2}-4ac} }{2a}

x = (-4 +/- \sqrt{16-4(-8)(5)})/2(-8)
   = (-4 +/- \sqrt{16 + 160})/-16
   = (-4 +\sqrt{176})/-16
   = 1/4 - \sqrt{11}/4

6 0
3 years ago
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