Answer: 14,515,200
Note: this is a single number (not an ordered triple or a collection of three different numbers) roughly equal to about 14.5 million if you round to the nearest hundred thousand.
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Explanation:
There are 13 people. Let's call them person A, person B, person C, ... all the way up to person M. The first four people are given who we'll call A through D. The rest (E through M) aren't really important since they aren't named.
A = Monsier Thenardier
B = Madame Thenardier
C = Cosette
D = Marius
Peron's E through M = remaining 9 people
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A and B must stick together. Because of this, we can consider "AB" as one "person".
So we go from 13 people to 13-2+1 = 12 "people".
Likewise, C and D must stick together. We can consider "CD" as one "person". So we go from 12 "people" to 12-2+1 = 11 "people"
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The question is now: how many ways can we arrange these 11 "people" around a circular table? The answer is (n-1)! ways where n = 11 in this case
So, (n-1)! = (11-1)! = 10! = 10*9*8*7*6*5*4*3*2*1 = 3,628,800
We're almost at the answer. We need to do two adjustments.
First off, for any single permutation, there are two ways to arrange "AB". The first is "AB" itself and the second is the reverse of that "BA". So we will multiply 3,628,800 by 2 to get 2*3,628,800 = 7,257,600
Using similar logic for "CD", we double 7,257,600 to get 2*7,257,600 = 14,515,200
The final answer is 14,515,200
Answer:
84.8 in³
Step-by-step explanation:
The formula for the volume of a cylinder is ...
V = πr²h
The volume Mary will be filling will be 3/4 of the 9-inch height of the vase, so is ...
V = π(2 in)²(3/4·9 in) = 27π in³ ≈ 84.8 in³
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<em>Comment on the answer</em>
In more conventional units of measure, that is very nearly 3 pints of water.
Answer:
Step-by-step explanation:
far away
That would depend on the car and the amount of miles the car can go on one tank of gas. Divide that amount from 1048 to get the answer.
Yes, we can obtain a diagonal matrix by multiplying two non diagonal matrix.
Consider the matrix multiplication below
![\left[\begin{array}{cc}a&b\\c&d\end{array}\right] \left[\begin{array}{cc}e&f\\g&h\end{array}\right] = \left[\begin{array}{cc}a e+b g&a f+b h\\c e+d g&c f+d h\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7De%26f%5C%5Cg%26h%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%20e%2Bb%20g%26a%20f%2Bb%20h%5C%5Cc%20e%2Bd%20g%26c%20f%2Bd%20h%5Cend%7Barray%7D%5Cright%5D%20)
For the product to be a diagonal matrix,
a f + b h = 0 ⇒ a f = -b h
and c e + d g = 0 ⇒ c e = -d g
Consider the following sets of values

The the matrix product becomes:
![\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}\frac{1}{3}&-1\\-\frac{1}{4}&\frac{1}{2}\end{array}\right] = \left[\begin{array}{cc}\frac{1}{3}-\frac{1}{2}&-1+1\\1-1&-3+2\end{array}\right]= \left[\begin{array}{cc}-\frac{1}{6}&0\\0&-1\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%262%5C%5C3%264%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Cfrac%7B1%7D%7B3%7D%26-1%5C%5C-%5Cfrac%7B1%7D%7B4%7D%26%5Cfrac%7B1%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Cfrac%7B1%7D%7B3%7D-%5Cfrac%7B1%7D%7B2%7D%26-1%2B1%5C%5C1-1%26-3%2B2%5Cend%7Barray%7D%5Cright%5D%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-%5Cfrac%7B1%7D%7B6%7D%260%5C%5C0%26-1%5Cend%7Barray%7D%5Cright%5D)
Thus, as can be seen we can obtain a diagonal matrix that is a product of non diagonal matrices.