Mathefreaker2021

The number of ways the baking club can form an executive committee with people is given by the binomial coefficient$C(n,k)$, where n is the total number of people in the baking club (including Mark) and k is the number of people on the executive committee.

So, the number of ways the baking club can form an executive committee with people is $C(n,k) = n! / (k! (n-k)!)$.

To solve this problem, we can use similar triangles. Let the side length of the cube be s, and let the midpoints of AB, AC, and BC be P, Q, and R, respectively. Then OP, OQ, and OR are the diagonals of the faces of the cube, and so they are equal to $$\sqrt{2}s$$.

Consider triangle AOB. By the Pythagorean theorem, we have $$OA^2 + OB^2 = 1 + 1 = 2$$, so $$AO^2 = 2$$. Since angle AOB is 90 degrees, we can use similarity to see that triangle OAP is similar to triangle OAB, so $$\frac{OA}{OP} = \frac{OB}{OP}$$. Solving, we find that $$OP = \frac{\sqrt{2}}{2}$$.

Using similar triangles, we can find that $$OQ = \frac{\sqrt{2}}{2}$ and $OR = \frac{\sqrt{2}}{2}$$. Since the sum of the squares of the lengths of the diagonals of a cube is equal to three times the square of the side length, we have

$$$(OP^2 + OQ^2 + OR^2) = 3s^2$$$

Substituting the values we found for $OP$, $OQ$, and $OR$, we have

$$$\left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 = 3s^2$$$

Simplifying and solving for $$s$$, we find that $$s = \frac{1}{3}$$. Thus, the side length of the cube inscribed in the tetrahedron is $$\frac{1}{3}$$.

For part (a), we can use the AM-GM (Arithmetic Mean-Geometric Mean) inequality, which states that for any positive real numbers a_1, a_2, ..., a_n, their arithmetic mean is always greater than or equal to their geometric mean:

$(a_1 + a_2 + ... + a_n) / n >= sqrt[(a_1 * a_2 * ... * a_n)^(1/n)]$

Using this inequality repeatedly, we can prove the given inequality as follows:

$(a^2 + b^2)^2 (c^4 + d^4)(e^4 + f^4) = [(a^2 + b^2)^2][(c^4 + d^4)(e^4 + f^4)]$

By AM-GM,

$[(a^2 + b^2)^2][(c^4 + d^4)(e^4 + f^4)] >= [(a^2 + b^2)^2][sqrt[(c^4 * d^4 * e^4 * f^4)^(1/4)]]$

And by AM-GM again,

$[(a^2 + b^2)^2][sqrt[(c^4 * d^4 * e^4 * f^4)^(1/4)]] >= [(a^2 + b^2)^2][sqrt[(c^2 * d^2 * e^2 * f^2)^(1/2)]]$

Finally, by AM-GM once more,

$[(a^2 + b^2)^2][sqrt[(c^2 * d^2 * e^2 * f^2)^(1/2)]] >= [(a^2 + b^2)^(1/2)][(c * d * e * f)^(1/2)]^2$

Applying AM-GM to the expression inside the square root on the right-hand side, we have

$[(a^2 + b^2)^(1/2)][(c * d * e * f)^(1/2)]^2 >= [(a^2 + b^2)^(1/2)][2(abcdef)^(1/6)]^2$

Finally, squaring both sides, we get

$[(a^2 + b^2)^2][(c * d * e * f)^(1/2)]^2 >= 32abcdef$, as desired.

For part (b), we can use the AM-GM inequality directly:

$(a^2 + b^2)(c^2 + d^2)(e^2 + f^2) = (a^2 + b^2) * [(c^2 + d^2) * (e^2 + f^2)]$

By AM-GM,

$(a^2 + b^2) * [(c^2 + d^2) * (e^2 + f^2)] >= (a^2 + b^2) * 2sqrt[(c^2 * d^2)(e^2 * f^2)]$

By AM-GM again,

$(a^2 + b^2) * 2sqrt[(c^2 * d^2)(e^2 * f^2)] >= 2(a^2$

We can use the formula for average speed, which is the total distance traveled divided by the total time taken. Let's call the total distance traveled "d".

Since Carissa took the same route both ways, the distance traveled uphill is the same in both directions, and the same is true for the distance traveled downhill and across flat ground. So, the total distance traveled uphill is 2d/3, the total distance traveled across flat ground is d/3, and the total distance traveled downhill is 2d/3.

The time taken to travel uphill at 100 meters per hour is 2d/3 * (1 hour/100 meters) = 2d/(300 meters), the time taken to travel across flat ground at 120 meters per hour is d/3 * (1 hour/120 meters) = d/(360 meters), and the time taken to travel downhill at 180 meters per hour is 2d/3 * (1 hour/180 meters) = 2d/540 meters.

The total time taken is 2d/(300 meters) + d/(360 meters) + 2d/540 meters = (12d + 12d + 10d)/(180 meters) = 24d/180 meters = 4d/30 meters.

The average speed during the entire round trip is d/(4d/30 meters) = 30/4 = 7.5 meters per hour.

(a) No, 1/7 is not in the set S.

(b) Yes, 1/4 is in the set S. The number 1/4 can be expressed as 1/3^2 + 1/3^4 + 1/3^5 + ..., which is a series that converges to 1/4.

Wir haben beide die gleiche Lösung gehabt 

(a) The left-hand side counts the number of ways to choose k elements from a set of n - 1 elements. The first term, C(n - 1, k - 1), counts the number of ways to choose k - 1 elements, and the second term, C(n - 1, k), counts the number of ways to choose k elements. To form the set of k elements in C(n, k), we simply add an element n to each of these two sets. So, C(n - 1, k - 1) + C(n - 1, k) = C(n, k).

(b) The left-hand side counts the number of ways to choose 0 to n elements from a set of n elements. For each positive integer i from 0 to n, C(n, i) counts the number of ways to choose i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of n elements, which is 2^n.

(c) The left-hand side counts the number of ways to choose 0 to k elements from a set of n + k elements. Each term C(n + i, i) counts the number of ways to choose i elements from a set of n + i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of n + k elements, which is C(n + k + 1, k).

(d) The left-hand side counts the number of ordered pairs (X, Y) where X is a set of r elements selected from a set of m elements and Y is a set of r elements selected from a set of n elements. For each positive integer i from 0 to r, C(m, i) C(n, r - i) counts the number of ordered pairs where X has i elements and Y has r - i elements. The sum on the left-hand side is equivalent to counting the number of possible combinations of r elements from the set of m + n elements, which is C(m + n, r).

(e) The left-hand side counts the number of ordered pairs (a, b) of positive integers where 1 <= a, b <= n. The expression (n + 1)^2 - 2n - 1 counts the total number of ordered pairs of positive integers where 1 <= a, b <= n + 1 and then subtracts the ordered pairs (n + 1, n + 1) and (n + 1, i) for 1 <= i <= n, as well as (i, n + 1) for 1 <= i <= n. This gives us the total number of ordered pairs of positive integers where 1 <= a, b <= n.

(f) The left-hand side counts the number of ordered triples (a, b, c) of positive integers where 1 <= a, b, c <= n. The expression (n + 1)^3 - 3n^2 - 3n - 1 counts the total number of ordered triples of positive integers where 1 <= a, b, c <= n + 1 and then subtracts the ordered triples (n + 1, n + 1, n + 1), (n + 1, n + 1, i), (n + 1, i, n + 1), and (i, n + 1, n + 1) for 1 <= i <= n, as well as all ordered triples where at least one element is equal to n + 1. This gives us the total number of ordered triples of positive integers where 1 <= a, b, c <= n.

This inequality can be proven using the Cauchy-Schwarz inequality.

Let $x = sqrt(a^2 + ab + b^2), y = sqrt(a^2 + ac + c^2)$, and $z = sqrt(b^2 + bc + c^2)$, then we can rewrite the inequality as:

$x + y + z >= sqrt(3) (sqrt(ab) + sqrt(ac) + sqrt(bc))$

Squaring both sides, we get:

$(x + y + z)^2 >= 3 (sqrt(ab) + sqrt(ac) + sqrt(bc))^2$

Expanding the left-hand side, we get:

$x^2 + y^2 + z^2 + 2xy + 2xz + 2yz >= 3 (sqrt(ab) + sqrt(ac) + sqrt(bc))^2$

Expanding the right-hand side, we get:

$x^2 + y^2 + z^2 + 2xy + 2xz + 2yz >= 3 (ab + ac + bc + 2 sqrt(ab) sqrt(ac) + 2 sqrt(ab) sqrt(bc) + 2 sqrt(ac) sqrt(bc))$

Using the Cauchy-Schwarz inequality, we have:

$2xy + 2xz + 2yz >= 2 (sqrt(ab) sqrt(ac) + sqrt(ab) sqrt(bc) + sqrt(ac) sqrt(bc))$

Substituting this into the previous inequality, we get:

$x^2 + y^2 + z^2 + 2xy + 2xz + 2yz >= 3 (ab + ac + bc) + 2 (sqrt(ab) sqrt(ac) + sqrt(ab) sqrt(bc) + sqrt(ac) sqrt(bc))$

Using the definition of x, y, and z, we get:

$(a^2 + ab + b^2) + (a^2 + ac + c^2) + (b^2 + bc + c^2) + 2 sqrt((a^2 + ab + b^2)(a^2 + ac + c^2)) + 2 sqrt((a^2 + ab + b^2)(b^2 + bc + c^2)) + 2 sqrt((a^2 + ac + c^2)(b^2 + bc + c^2)) >= 3 (ab + ac + bc) + 2 (sqrt(ab) sqrt(ac) + sqrt(ab) sqrt(bc) + sqrt(ac) sqrt(bc))$

Combining like terms and rearranging, we get:

$2 (sqrt((a^2 + ab + b^2)(a^2 + ac + c^2)) + sqrt((a^2 + ab + b^2)(b^2 + bc + c^2)) + sqrt((a^2 + ac + c^2)(b^2 + bc + c^2))) >= 2 (sqrt(ab) sqrt(ac) + sqrt(ab) sqrt(bc) + sqrt(ac) sqrt(bc))$

So, the inequality is proven.

Equality occurs when x = y = z, which means$a^2 + ab + b^2 = a^2 + ac + c^2 =$

Yes, there is a theorem that states that$a^2 + b^2 > ab$ for any real numbers a and b. This theorem is known as the Cauchy-Schwarz inequality.

The Cauchy-Schwarz inequality states that for any two sequences of real numbers $(a_1, a_2, ..., a_n)$ and $(b_1, b_2, ..., b_n)$, we have:

$(a_1^2 + a_2^2 + ... + a_n^2)(b_1^2 + b_2^2 + ... + b_n^2) >= (a_1 b_1 + a_2 b_2 + ... + a_n b_n)^2$

Taking $n = 2$, we have:

$(a^2 + b^2)(b^2 + b^2) >= (ab + ab)^2$

Expanding the right-hand side and cancelling the $b^2$ terms, we get:

$a^2 + b^2 > ab$

So, for any real numbers a and b,$a^2 + b^2 > ab$.

It's a very useful and important theorem in mathematics and has many applications in various fields such as linear algebra, functional analysis, and optimization.