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One six-sided die tossed 7 times should give you this many permutations:

6^7 = 279,936 possible sequences.

To prove that the inequality [\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \ge \sqrt{b^2 + bc + c^2}] holds [1], we can use the Cauchy-Schwarz inequality. This states that for any three real numbers a, b, and c, we have [(a^2 + b^2 + c^2)(a + b + c) \ge (ab + bc + ca)^2.]

Now, let $a = \sqrt{a^2 - ab + b^2}$, $b = \sqrt{b^2 + bc + c^2}$, and $c = \sqrt{a^2 - ac + c^2}$. Then, substituting these values into the Cauchy-Schwarz inequality gives us the desired result.

Equality occurs when the Cauchy-Schwarz equality is met, which happens when $a + b + c = 0$ and $ab + bc + ca = 0$. This is equivalent to saying that $a = -b - c$, $b = -a - c$, and $c = -a - b$.

Let's consider the value of i^x for all possible values of x:

i^1 = i i^2 = -1 i^3 = -i i^4 = 1

As we can see, the value of i^x starts repeating after every 4th power. This means that for any positive integer n, the value of i^n can be expressed as i^(n % 4).

Since 1 ≤ x < y ≤ 100, we have four possibilities for the value of (x % 4) and (y % 4):

(x % 4) = 1, (y % 4) = 3

(x % 4) = 2, (y % 4) = 2

(x % 4) = 3, (y % 4) = 1

(x % 4) = 0, (y % 4) = 0

For the first case, we have 25 pairs of (x, y) where x = 1, 5, 9, ..., 97 and y = 3, 7, 11, ..., 99.

For the second case, we have 25 pairs of (x, y) where x = 2, 6, 10, ..., 98 and y = 2, 6, 10, ..., 98.

For the third case, we have 25 pairs of (x, y) where x = 3, 7, 11, ..., 99 and y = 1, 5, 9, ..., 97.

For the fourth case, we have 24 pairs of (x, y) where x = 4, 8, 12, ..., 100 and y = 4, 8, 12, ..., 100.

Therefore, the total number of pairs of integers (x, y) with 1 ≤ x < y ≤ 100 such that i^x + i^y is a real number is 25 + 25 + 25 + 24 = 99.

-1701.87356322

Hi! The answer is 8.1

No body knows...

I mean, in a real game, you would have 2775 points, but this equation looks really weird, so idk.

If a dice is rolled, it is generally a 1/6 chance. If 3 are rolled, it is a 1/2 chance. So the chances of rolling a 6 is a 1/2 chance.

a = (20-10)/2(2)
a=10/4
a=2.5

Hope that helps!

1) Decreasing, odd, invertible

2) Increasing, even

3) Even, Invertible

4) Increasing, Even, Invertible

First off, the chances of getting any red sock is a 1/3 chance because there are 2 red socks and 6 socks in total. If you were to pull to random socks, every day for 3 days, the chances of you getting matching pairs for alll of the days is a 1/5832 chance.

I think. I'm not that smart.

can anybody help me with english\

Let the number of adults ==A

Then, the number of children ==310  -  A

12A  +  7.50[310  -  A] ==3,180, solve for A

A==190 adults visited the amusement park

310  -  190 ==120 children visited the amusement park.

It has been solved already

https://www.algebra.com/algebra/homework/word/travel/Travel_Word_Problems.faq.question.1133985.html

Hi, this was helpful, but I think you used the wrong numbers..

a=listfor(n, 1, 11, (11+n) nCr 4);print a, "=", sum a

(495, 715, 1001, 1365, 1820, 2380, 3060, 3876, 4845, 5985, 7315) = 32857 - total number of solutions.