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arccos(0.9359059) =~0.36 radians

\(PQ = \dfrac{10 \sqrt{10}}{3}\)

By similar triangles, UV = 40/3.

[3 C 1] + [3 C 2] + [3 C 3] ==[3 + 3 + 1] ==7 / 2^3 ==7 / 8 - probability of at least 1 head.

The probability that the first card has a certain number is 2/20​. The probability that the second card has the same number is 1/19​. Therefore, the probability that the first two cards have the same number is 2/20​×1/19​ = 1/190​​.

There can be 0, 1, 2, or 3 heads.  Therefore, the probability that at least one of the coins comes up heads is 3/4.

Can you walk me through you're process. This seems incorrect. 

The answer is $2,589.71.

1. The volume of a square pyramid is given by the formula V=1/3*​bh, where b is the length of the base and h is the height. In this case, the base is a square with side length 12, and the height is the distance from the apex of the pyramid to the center of the base. To find the height, we can use the Pythagorean theorem. The hypotenuse of the right triangle formed by the height, the slant height, and half the base is 12, and the slant height is 12. Therefore, the height is sqrt(12^2−(12/2)^2)​=6*sqrt(2)​.

Plugging in b=12 and h=6*sqrt(2)​, we get V=1/3*(12)(6*sqrt(2)​)=24*sqrt(2).

I will say 0.

Thank you so much!!!

Thank you so much!!!

I get it now

The sum of the digits of a six-digit number is divisible by 10 if and only if the sum of the digits of its last three digits is divisible by 5.

There are 9 choices for the first digit of the number, 10 choices for each of the next three digits, and 2 choices for the last digit (4 or 9).

Therefore, there are 9×10^3×2=180000​ six-digit numbers where the sum of the digits is divisible by 10.

Here's a more detailed explanation:

The first digit can be any digit from 1 to 9, inclusive.

The second, third, and fourth digits can be any digit from 0 to 9, inclusive.

The fifth digit must be either 4 or 9, so that the sum of the digits is divisible by 10.

Therefore, there are 9×10^3×2=180000​ six-digit numbers where the sum of the digits is divisible by 10.

There are 2 combinations as follows:

145555 ==6!/4! ==30 permutations

225555==6!/4!2!==15 permutations

Total == 30  +  15 ==45 different sequences of rolls. 

Catherine rolls a standard 6-sided die six times. If the product of her rolls is \(2500\), then how many different sequences of rolls could there have been? (The order of the rolls matters.)

a=tens digit,  b=ones digit

10*a + b ==10*b + a + 45...............(1)

a ==2*b - 1......................................(2), solve for a, b

a==11 and b==6 - this solution does not work since a is a 2-digit number.

Try: b=2*a - 1 which gives:

a=-4  and  b ==-9

[-4 * 10 - 9] == - 49 - which appears to be Sophie's favorite number.

Check: -49 - ( - 94) ==45 - difference between them

              [2 * - 4] == - 8 which is 1 greater than - 9, NOT one less !

Therefore, I see NO SOLUTION for the 2nd part of the question.

You included the first term while adding, keep in mind that you only start to add 5 from the second term onwards. #2 has it correct, where the general formula for the nth term of an arithmetic sequence with first term a and common difference d would be a + (n-1)*d.

set the integers equal to x, x+1, x+2, x+3, x+4, x+5, x+6

sum of all integers = 7x+21

21*4 = 84

7x + 21 = 84(x+6)

7x + 21 = 84x + 504

-483 = 77x

x is not an integer, thus there is no solution. However, as I interpreted * to signal multiplication, but the word "times" is written out immediately afterwards, I believe the question is incorrectly formatted.

Check: (16 / 15, 17 / 16, 18 / 17, 19 / 18, 20 / 19, 21 / 20, 22 / 21, 23 / 22, 24 / 23, 25 / 24, 26 / 25, 27 / 26, 28 / 27, 29 / 28, 30 / 29, 31 / 30, 32 / 31, 33 / 32, 34 / 33, 35 / 34)==7/3 - your solution

Check: (11 / 10, 12 / 11, 13 / 12, 14 / 13, 15 / 14, 16 / 15, 17 / 16, 18 / 17, 19 / 18, 20 / 19, 21 / 20, 22 / 21, 23 / 22, 24 / 23, 25 / 24, 26 / 25, 27 / 26, 28 / 27, 29 / 28, 30 / 29)==3 - the one above you.

The first fraction that Levans wrote has a numerator that is 1 greater than the denominator. This means that the numerator is n and the denominator is n−1. The next fraction that Levans wrote has a numerator of n+1 and a denominator of n. This pattern continues for 20 fractions.

The product of all 20 fractions is 3. This means that the numerators and denominators of the fractions must cancel each other out, leaving only 3. The only way for this to happen is if the numerators and denominators are consecutive integers.

The first fraction that Levans wrote has a numerator of n and a denominator of n−1. The next fraction has a numerator of n+1 and a denominator of n. This means that the numerators of the fractions are n,n+1,n+2,...,n+19. The denominators of the fractions are n−1,n,n+1,...,n+18.

The product of the numerators is n(n+1)(n+2)...(n+19). The product of the denominators is (n−1)n(n+1)...(n+18). The numerators and denominators cancel each other out, leaving only n+19.

The value of the first fraction that Levans wrote is n−1n​. The value of the product of all 20 fractions is n−1n+19​. The value of the product of all 20 fractions is 3. This means that (n+19)/(n-1)​=3. Solving for n, we find that n=16.

Therefore, the value of the first fraction that Levans wrote is 16/15.

1 - 0.0340741737109317132502568002711

2 - 1.1305261922200515915484062278955

3 - 1.751839807478977396407519406377

4 - 1.3235708547255705556209938932339

5 - arccos(1.3235708547255705556209938932339)

6 - 0.244715 - radians

Well Georgemichels, it looks like you are going to be banned....

An arithmetic sequence begins: 3,8,13,18...  

what is the 61st term of the sequence?  

The sequence is, you add 5 to the previous number. 

2^nd term,  add 1 x 5 to 3  

3^rd term,  add 2 x 5 to 3  

4^th term,  add 3 x 5 to 3  

So you see, for each n^th term, add (n – 1) x 5 to 3  

For the 61st term, add (61 – 1) * 5 to 3  

                                     (60 * 5) + 3  =  303  

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