Mathefreaker2021

Um diese Frage zu beantworten, müssen wir die Formel für den exponentiellen Zuwachs verwenden. Die Formel lautet:

\(A = P * (1 + r)^t\)

wo A das Endbetrag ist, P der Anfangsbetrag, r der jährliche Zuwachs in Prozent (hier 6%) und t die Anzahl der Zeitperioden (hier alle 2 Tage).

Wir wissen, dass A = 22500000 und P = 11500000. Daher können wir die Formel so anpassen:

\(22500000 = 11500000 * (1 + r)^t\)

Und wir können t berechnen, indem wir beide Seiten der Formel durch 11500000 teilen:

\(t = log((22500000 / 11500000) / (1 + r)) / log(1 + r)\)

Diese Formel kann nicht analytisch gelöst werden, aber sie kann numerisch approximiert werden, z.B. mithilfe einer Taschenrechner-Funktion für den natürlichen Logarithmus oder einer Spreadsheet-Software wie Microsoft Excel oder Google Sheets.

Wenn aber die 

Erdbeschleunigung g = 9,807m/sek² 

Schallgeschwindigkeit c = 343m/sek

ist, dann kann man die Fallstrecke S (in Metern) berechnen, indem man folgende Formel verwendet:

\(S = (1/2)g * t^2\)

Wobei t die Fallzeit in Sekunden ist. Da die Fallzeit in der Aufgabe mit 6,5 Sekunden angegeben ist, kann man die Fallstrecke wie folgt berechnen:

\(S = (1/2) * 9,807 m/sek^2 * (6,5 sek)^2 = (1/2) * 9,807 * 42,25 = (1/2) * 411,974 = 205,987 m\)

Die Fallstrecke der Stahlkugel beträgt also etwa 205,987 Meter.

Um die Fallstrecke der Stahlkugel zu berechnen, müssen wir die Formel für die freie Fallbewegung nutzen. Die Formel für die freie Fallbewegung lautet:

\(h = 1/2 * g * t^2\)

h steht für die Fallstrecke, g ist die Gravitationsbeschleunigung (9,8 m/s^2) und t steht für die Zeit. Da die Zeit 6,5 Sekunden beträgt, können wir die Formel wie folgt berechnen:

\(h = 1/2 * 9,8 * (6,5)^2 = 170,125 m\)

Die Fallstrecke beträgt also 170,125 Meter.

Um die Funktion F(x) zu bestimmen, müssen wir die Funktionswerte für jedes x berechnen und limitieren. Da für jede reelle Zahl x gilt:

\(f_0(x) = x^2 f_1(x) = x^2 + x/2 f_2(x) = x^2 + x/2 + x^2/6 f_3(x) = x^2 + x/2 + x^2/6 + x^3/24\)

usw.

Können wir sehen, dass jede Funktion f_n(x) eine Polynomfunktion von Grad n ist. Da die Funktionen immer höheren Grads sind, kann die Folge (fn) als eine Potenzreihe dargestellt werden:

\(f_n(x) = x^2 + x/2 + x^2/6 + ... + x^n/(n!)\).

Die Potenzreihe kann auch als exponentialfunktion dargestellt werden:

\(f_n(x) = e^x - 1 - x\).

Da jede Potenzreihe für jedes x konvergiert, muss auch die Funktion f_n(x) für jedes x gegen eine bestimmte Funktion konvergieren. Daher ist die gesuchte Funktion F(x) gleich:

\(F(x) = e^x - 1\).

The number of ways of choosing three points from the grid to form an isosceles right triangle is the same as the number of lattice points that are the vertices of an isosceles right triangle.

We consider the four cases for the right angle of the isosceles right triangle:

The right angle is at the origin (0, 0). In this case, the other two points are at (a, 0) and (0, a), where a is a positive integer. There are a total of \(sqrt(2n) - 1\) points of this type for n = 50.

The right angle is at the point (a, a), where a is a positive integer. In this case, the other two points are at (a, 0) and (0, a). There are a total of \(sqrt(2n) - 1\) points of this type for n = 50.

The right angle is at the point (-a, a), where a is a positive integer. In this case, the other two points are at (0, a) and (-a, 0). There are a total of \(sqrt(2n) - 1\) points of this type for n = 50.

The right angle is at the point (-a, -a), where a is a positive integer. In this case, the other two points are at (0, -a) and (-a, 0). There are a total of \(sqrt(2n) - 1\)  points of this type for n = 50.

Thus, the total number of isosceles right triangles is \(4 * (sqrt(2n) - 1) = 4 * (sqrt(100) - 1) = 4 * 10 - 4 = 36\).

Therefore, there are 36 different ways of choosing three points from the grid to form an isosceles right triangle.

Finding the function F(x) that satisfies the given functional equation can be done through successive substitution. Here's one approach:

First, substitute x = 3 into the equation to get

F(3) + F(0) = 3.

Since F(0) is not defined, we'll replace it with a constant, say a. So, we have

F(3) + a = 3.

Next, substitute x = 2 into the equation to get

F(2) + F((2 * 2 - 3) / 2) = 2.

Substituting the value of F(3) from step 1 into this equation gives us

F(2) + F(1/2) = 2.

Now, substitute x = 1/2 into the equation to get

F(1/2) + F((2 * (1/2) - 3) / (1/2)) = 1/2.

Substituting the value of F(2) from step 2 into this equation gives us

F(1/2) + a = 1/2.

Finally, substituting the values of F(3) and F(1/2) from steps 1 and 3 into the original equation gives us

F(x) + a = x,

for all x ≠ 1 and x ≠ 2.

So, F(x) = x - a.

Since we have two equations in two unknowns (a and F(x)), we can solve for a. From step 1, a = 3 - F(3). And from step 3, a = 1/2 - F(1/2). Equating these two values of a gives us

3 - F(3) = 1/2 - F(1/2).

Substituting the expression for F(x) from step 4 into this equation gives us

3 - (3 - a) = 1/2 - (1/2 - a),

which simplifies to

a = 2.

So, the function F(x) that satisfies the given functional equation is

F(x) = x - 2.

Therefore, F(x) is a rational function with polynomials in the numerator and denominator.

Let's start by finding the prime numbers greater than 10 that Syd could have chosen. The prime numbers between 11 and 100 are 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97.

Since Syd is choosing two different primes, he has 20 options for the first prime and 19 options for the second. The resulting products would be:

11 * 13 = 143

11 * 17 = 187

11 * 19 = 209

...

89 * 97 = 8613

So there are 20 * 19 = 380 different products that Syd could have ended up with.

There are 24 * 23 = 552 possible pairs of tiles that Steve and Ellen can choose.

Out of these, there are 4 * 4 = 16 tiles that are both red.

So, the probability that both tiles are red is 16/552 = 4/138.

The problem can be restated as finding the smallest positive integer n such that there exist 100 distinct subsets of the club, each of size k, where 1 <= k <= n. This is equivalent to finding the smallest positive integer n such that n choose k is greater than or equal to 100 for all values of k from 1 to n.

By the Pigeonhole Principle, if we have n people, then at most n choose k people can review the same book. Hence, we want the smallest positive integer n such that:

C(n,1) >= 100

C(n,2) >= 100

...

C(n,n) >= 100

We can start by computing some of the binomial coefficients and finding the smallest value of n that satisfies the conditions.

C(2,1) = 2 >= 100 (not satisfied)

C(3,1) = 3 >= 100 (not satisfied)

C(3,2) = 3 >= 100 (not satisfied)

C(4,1) = 4 >= 100 (not satisfied)

C(4,2) = 6 >= 100 (not satisfied)

C(4,3) = 4 >= 100 (not satisfied)

C(5,1) = 5 >= 100 (not satisfied)

C(5,2) = 10 >= 100 (not satisfied)

C(5,3) = 10 >= 100 (not satisfied)

C(5,4) = 5 >= 100 (not satisfied)

The smallest n that satisfies all the conditions is 6, so the smallest positive integer n that allows us to avoid having the exact same group of members over all 100 days is 6.