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IrinaK [193]
3 years ago
5

PLEASE HELP! I ONLY HAVE 15 MINS! WILL MARK BRAINLIEST! 20 POINTS! Suppose N coins lie heads up on a table. In one turn, you can

turn over any (N−1) coins. Is it possible that N coins could lay tails up in any number of turns, if
Mathematics
1 answer:
Romashka [77]3 years ago
6 0

Answer: it is only possible if N = 2.

Step-by-step explanation:

if you have N  heads up coins.

turn one, you turn N -1 coins up.

Now you have N - 1 coins tails up and one coin heads up.

Second turn, you turn N - 1 coins, (if you select the same N - 1 coins as before, you will end in the initial situation, so now we will change one coin, this time we will select the coin that is still heads up and we will leave one of the tails up unchanged)

Then we have N - 2 coins heads up and 2 coins tails up.

Now, we can select any N -1 coins to flip, and we will end with with N - 1 coins tail up and 1 coin heads up.

and this cycle will repeat.

Then the only situation is having N = 2.

Becuase when we have N -2 coins heads up and 2 coins tails up we will have: N - 2 = 0, so we have 0 coins heads up and 2 coins tails up.

This means that the statement is only true if N = 2.

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Alinara [238K]

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3 years ago
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Write a formula for rrr in terms of \thetaθtheta for the following figure.
xxMikexx [17]

Answer:  r = 4cm/θ

Step-by-step explanation:

For a circle of radius R, the perimeter is:

P = 2*pi*R

Now, if we have an arc with an angle θ, the cord of this arc is:

C = (θ/2*pi)*2*pi*R

Where you can see that if θ = 2*pi (this is the full circle) we have the perimeter of the circle.

In this case, we can see that the cord for an angle θ is 4cm, then:

C = 4cm = (θ/2*pi)*2*pi*r = θ*r

Now we can solve this for r:

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Step-by-step explanation:

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3 years ago
Find the eigenvalues of the matrix: <br><br> [-43 0 80]<br> [40 -3 80]<br> [24 0 45]
saveliy_v [14]

Answer:

-3

1 + 4 sqrt( 241 )

1 - 4 sqrt( 241 )

Step-by-step explanation:

We need minus lambda on the entries down the diagonal. I'm going to use m instead of the letter for lambda.

[-43-m 0 80]

[40 -3-m 80]

[24 0 45-m]

Now let's find the determinant

(-43-m)[(-3-m)(45-m)-0(80)]

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+80[40(0)-(-3-m)(24)]

Let's simplify:

(-43-m)[(-3-m)(45-m)]

-0

+80[-(-3-m)(24)]

Continuing:

(-43-m)[(-3-m)(45-m)]

+80[-(-3-m)(24)]

I'm going to factor (-3-m) from both terms:

(-3-m)[(-43-m)(45-m)-80(24)]

Multiply the pair of binomials in the brackets and the other pair of numbers;

(-3-m)[-1935-2m+m^2-1920]

Simplify and reorder expression in brackets:

(-3-m)[m^2-2m-3855]

Set equal to 0 to find the eigenvalues

-3-m=0 gives us m=-3 as one eigenvalue

The other is a quadratic and looks scary because of the big numbers.

I guess I will use quadratic formula and a calculator.

(2 +/- sqrt( (-2)^2 - 4(1)(-3855) )/(2×1)

(2 +/- sqrt( 15424 )/(2)

(2 +/- sqrt( 64 )sqrt( 241 )/(2)

(2 +/- 8 sqrt( 241 )/(2)

1 +/- 4 sqrt( 241 )

3 0
3 years ago
Separable differential equation <br> y’ln^2y+ysqrtx=0 y(0)=e
Maksim231197 [3]

By applying the theory of <em>separable ordinary differential</em> equations we conclude that the solution of the <em>differential</em> equation \frac{dy}{dx} \cdot (\ln y)^{2} + y\cdot \sqrt{x} = 0 with y(0) = e is y = e^{\sqrt [3]{-2\cdot x^{\frac{3}{2} }+1}}.

<h3>How to solve separable differential equation</h3>

In this question we must separate each variable on each side of the equivalence, integrate each side of the expression and find an <em>explicit</em> expression (y = f(x)) if possible.

\frac{dy}{dx} \cdot (\ln y)^{2} + y\cdot \sqrt{x} = 0

(\ln y)^{2}\,dy =  -y \cdot \sqrt{x}\, dx

-\frac{(\ln y)^{2}}{y}\, dy = \sqrt{x} \,dx

-\int {\frac{(\ln y)^{2}}{y} } \, dy = \int {\sqrt{x}} \, dx

If u = ㏑ y and du = dy/y, then:

-\int {u^{2}\,du } = \int {x^{\frac{1}{2} }} \, dx

-\frac{1}{3}\cdot u^{3} = \frac{2\cdot x^{\frac{3}{2} }}{3} + C

u^{3} = -2\cdot x^{\frac{3}{2} } + C

(\ln y)^{3} = - 2\cdot x^{\frac{3}{2} } + C

C = (\ln e)^{3}

C = 1

And finally we get the <em>explicit</em> expression:

\ln y = \sqrt [3]{-2\cdot x^{\frac{3}{2} }+ 1}

y = e^{\sqrt [3]{-2\cdot x^{\frac{3}{2} }+1}}

By applying the theory of <em>separable ordinary differential</em> equations we conclude that the solution of the <em>differential</em> equation \frac{dy}{dx} \cdot (\ln y)^{2} + y\cdot \sqrt{x} = 0 with y(0) = e is y = e^{\sqrt [3]{-2\cdot x^{\frac{3}{2} }+1}}.

To learn more on ordinary differential equations: brainly.com/question/14620493

#SPJ1

6 0
2 years ago
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