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Let l, w, and h be the length, width, and height of the rectangular prism, respectively. We know that the total surface area is 56, so we can use the formula for the surface area of a rectangular prism to get the following equation:

2lw+2wh+2lh=56

We also know that the sum of all the edges is 64, so we can use the formula for the sum of the edges of a rectangular prism to get the following equation:

2l+2w+2h=64

We can solve the first equation for l to get:

l=2w56−2wh−2lh​

We can then substitute this into the second equation to get:

2w56−2wh−2lh​+2w+2h=64

We can then simplify this equation to get:

2wh+2lh−56=0

We can then factor this equation to get:

(w−4)(2h−14)=0

We can then solve for w and h to get:

w=4 h=214​=7

We can then use these values to find the length of the diagonal by using the Pythagorean theorem:

d2=l2+w2+h2=42+72+72=108

d=sqrt(108)​=6*sqrt(3)​

Therefore, the length of the diagonal joining one corner of the prism to the opposite corner is 6*sqrt(3).

if we have n people and we want to do groups of 5, the total number of different combinations is:
find the smallest n such c > 365.
let's do it by brute force:
if n = 5, we have:
if n = 6
if n = 7
if n = 8
if n = 9
so 9 is not enough, let's see N = 10
C = 10!/(5!*5!)  = 252
10 is not enough, let's see with 11.
C = 11!/(6!*5!) =  462
So you need at least 11 members in the club.
Then the minimum number of members such we have more than 365 combinations is 11 members

5324 , 5624 , 5724 , 5824 , 5924 , 6324 , 6524 , 6724 , 6824 , 6924 , 7324 , 7524 , 7624 , 7824 , 7924 , 8324 , 8524 , 8624 , 8724 , 8924 , 9324 , 9524 , 9624 , 9724 , 9824 , Total =  25

5326 , 5426 , 5726 , 5826 , 5926 , 7326 , 7426 , 7526 , 7826 , 7926 , 8326 , 8426 , 8526 , 8726 , 8926 , 9326 , 9426 , 9526 , 9726 , 9826 , Total =  20

5328 , 5428 , 5628 , 5728 , 5928 , 6328 , 6428 , 6528 , 6728 , 6928 , 7328 , 7428 , 7528 , 7628 , 7928 , 9328 , 9428 , 9528 , 9628 , 9728 , Total =  20

25 + 20 + 20 ==65 - This is the 2nd solution where they all end in 4, 6, 8. If you look at them carefully, they are very easy to understand. My problem with their solution is that I cannot get the denominator of 840!

You should always use Rad.

The & functions cancel, and you are left with 0.

To find the volume of pyramid CFAH, we need to first find the height of the pyramid. Since CFAH is a pyramid with a rectangular base, the height of the pyramid is the perpendicular distance from point F to plane ABCD.

Let's first draw a diagram of the rectangular prism:

We know that EF = 4, EH = 5, and EA = 6. Let's use the Pythagorean theorem to find the length of segment FH:

FH² = EF² + EH² FH² = 4² + 5² FH² = 41 FH = sqrt(41)

Now, let's draw a line segment from point F perpendicular to plane ABCD:

Let's call the point where this line segment intersects plane ABCD point X. We want to find the length of segment FX, which is the height of the pyramid.

Since EF is perpendicular to plane ABCD, we know that segment FX is perpendicular to segment EF. Therefore, triangle FEX is a right triangle, with legs of length FX and EF, and hypotenuse of length FH.

Using the Pythagorean theorem again, we can solve for FX:

FH² = FX² + EF² 41 = FX² + 4² FX² = 41 - 16 FX = sqrt(25) FX = 5

So the height of the pyramid is 5.

Now, to find the volume of the pyramid, we need to multiply the area of the base by the height and divide by 3 (since it's a pyramid). The area of the base is the area of rectangle AFCH, which is the product of the length and width:

AF = CH = EA = 6 CF = AH = EH - EF = 5 Area of rectangle AFCH = 6 * 5 = 30

So the volume of pyramid CFAH is:

(1/3) * 30 * 5 = 50

Therefore, the volume of pyramid CFAH is 50 cubic units.

Hi Tiggsy,

Finally somebody responds...thank you. I get what you say about how you calculate things...it really makes a lot of sense. Thank you so much. I will go ahead and try to do the last one....I will let you know..Thanks

Hi Juriemagic,

I used a method like the one you are asking about, but my numbers were different.

I split the counting into four separate groups, first two digits odd - odd, odd - even, even - even or even - odd.

For odd - odd, there are three possibles for the first digit, 5, 7 or 9.

For the second digit there are again three possibles. There are four odd numbers 3, 5, 7, 9 but one of them has already been chosen, so three left to choose from.

The third digit is fixed and there are three possibles for the fourth digit.

That means that the total number of odd-odds is (3*3*1*3) = 27.

For the odd-even group it would be (3*3*1*2) = 18. The 2 at the end because one of the even numbers has been knocked out by the even second digit.

The other two, even-even and even-odd I'll leave you to work out, but the total of all four is 65.

As to the answer that you've posted, I think that the first digit in each bracket relates to the choices, odd, 6, 8, so 3, 1, 1.

If that's the case though, I don't understand where the 5 in the second position comes from.

Let's label the six children A, B, C, D, E, and F. We can assume that the three pairs of siblings are AB, CD, and EF (without loss of generality).

First, let's consider the ways in which the first row can be filled. We have three choices for the leftmost chair: A, C, or E. Without loss of generality, let's say we choose A. Then, there are two choices for the middle chair: either D or F. Finally, there is only one choice for the rightmost chair, which is either C or E (whichever one was not chosen for the leftmost chair).

So there are 3 × 2 × 1 = 6 ways to fill the first row.

Now, let's consider the second row. We can't put any siblings directly in front of each other, so we have to be a bit more careful. Without loss of generality, let's say we start by putting B in the leftmost chair of the second row. Then, we can't put A in the middle chair directly in front of B, so we have to choose either C or E. Let's say we choose C. Then, we can't put D in the rightmost chair directly in front of C, so we have to put F there.

So the second row must be filled with the children B, C, and F, in that order. There are 3! = 6 ways to do this.

Putting it all together, there are 6 ways to fill the first row and 6 ways to fill the second row, for a total of 6 × 6 = 36 ways to seat the six children as described.

It is too hard. I think you should study harder to make up for being smart. or you can play some puzzle game to train your brain.

We can begin by drawing a diagram of triangle XYZ with legs XY = 12 and YZ = 6. Let's label the right angle as angle Z, and the hypotenuse as XZ.

At station A

There were 45 passengers on the bus when it left station A

At station B, some passengers boarded the bus but no passenger alighted from it.

Let , number of passengers boarded at B = x

at B no of passenger = 45 + x

At station C, half of the passengers in the bus alighted and 20 passengers boarded it. 

at C = \( {45 + x \over 2} + 20\)

Finally at station D, another half of the passengers in the bus alighted and 30 passengers boarded it (45 + x)/2 + 20)/2 + 30

There were 80 passengers on the bus as it left station D for station E.

(45 + x)/2 + 20)/2 + 30 = 80
(45 + x)/2 + 20)/2 = 50

(x + 45)/2 + 20 = 100

(x + 45)/2 = 80

x + 45 = 160

x = 115

number of passengers boarded at B = x = 115

There are 34 ways.

1.25 grams will be left after 48 hours.

The answer is 12.

26 passengers boarded the bus at station B.

The constant term is 27.

why, could you provide a proof or what kind of shape the locus is?

Well, I believe the answer is m=6 and m=11.

Yes, that's true.

Answer is here:

First we use Stars and bars to find all possible cases:

Stars and bars tells us that there are (9+7-1) choose (7-1) = 5005 nonnegative integer solutions to this.

Now we assume Dhruv (assume that he is x7) gets 2 chocolates. We can find the number of cases which Dhruv gets 2 chocolates. 

Now we use stars and bars again to find that there are (7+6-1) choose (6-1) = 792 nonnegative integer solutions.

So the probability that Dhruv gets 2 chocolates is (792/5005) = (72/455)

(There might be a calculation error if there is tell me)

Credit to hairyberry for the solution!

The answer is 4/3.

The answer is a^2 - 4a + 4b^2 - 12 = 0.

n=18;r=5;k=1; a=n * (n - (r*k) - 1) nPr (r - 1)==213,840 - number of ways of seating 5 people.

/* where k stands for minmum chair separation between any 2 persons */
/* and k<=[n - r] / r and n=number of chairs, r=number of people to be seated */
/* and k=1 or number of chairs 2 people must be separated by.*/

0.3n + 0.4n + 0.7n = 4/9, solve for n

n = 20 / 63