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I bound 2P together , there are three rooms _B_B_ , the 2p can be placed in them , and the 2p can also switch position themself, so 3x2=6 arrangements
what'd you mean by particular 2 books, dont get it , 6 arrangements perhaps?
I was just wandering around equation and stuff,
After some calculation I get this
√(b^2 - 4c) = |(b^2-2c)/b|
It's so wierd , by it , I can calcultion a square root approximately.
As the number becomes greater , the result become more accurate
when I assume b^2-4c = 99999, I calculate the sqr99999 =316.22857
which by calculator it's 316.2261
when I assume b^2-4c=999999 , I get 999.9995 which is the same as the result come out of the calculator
Wierd man,.~
you're welcome. I just happen to have an inspiration,hehe
There only exist two situation , L precdeds the N , or N precedes the L ,that means there are only half of the arrangement satisfy the need , the total arrangements 7! , then the satifying arrangement is 7!/2 =2520
denote the unknown side as y , perimeter 77=2*(x/3)+18+y
we get 2x/3+y=59
since 18 is the base , so y<18
y=59-2x/3<18
then we get 20.5<x/3
Now we have all the length of this trapezoid , 18, 20.5<x/3 ,y<18
assume y is close to 0 , then x/3<29.5
so 32 foot is greater
"there is a boy at one end and a girl at the other end" . I assume they dont switch end ~lol
the first seat can have one of the 4 boys , the last one can have one of the 3 girls , so there are 4x3=12 ways , after chosing one boy and one girl ,there are 5 people left , so it has 5!=120 combination. In total there are 120x12=1440 arrangment
yeah , you are right mathsyperson
since everyone at least gets 1 dollar , the total way to distribute the money is 3x3x3x3=81 . when each of them gets 2 dollars there is one dollar left , that means there will always be a people who has 3 dollars or more .so the number of ways of cola is equal to the total ways Then there are 81 ways to distribute the money.
First of all , you should know that the y are the same , and so are the x
we shall eliminate as many unknow factor as possible when solving a group of equation
So we shall eliminate the y or x , since y=3x-1=2x+5 , 3x-1=2x+5 then x=6
and we also get y=17
18.ak-a(k-1)=4 , that means it's a sequence , ak=a1+(k-1)d = 3+(k-1)*4=4k-1=399
dont understand the 11th question ,lol
The value of 0 dg , 30dg , 45dg , 90dg must be remembered , or you can draw a coordinate graph of triangle function or an acute triangle to find these values.
hmmm, non-zero integer
Hmmm, I like Ensiferum the Tale of Revenge song , any viking folk metal fans here ?
well follow the equality
10^2=8^2+11^2 - 2x8x11cosx
cosx=85/176
11^2=8^2+10^2-8x10x2cosy
cosy=43/160
That's my solution , Don't know if I am correct
Is the length of the third side 11? is it given?
but to find out exactly how many letters are repeated and how many pair of letters , that's kinda hard ~
Let's suppose it's an n letters words ( don't know if some of the letters are repeated, I assume it n different letters)
n!=180
n!=180 , no solution
so there must be some letters which are repeated , then n will be greater than 5
So according to the options you provided it's a 6 letters word
Ricky ,you are such a clever guy~ , Thanks for all the helping
I think the p should be 10 , since the length p shoudn't be the adjacent to angle x
6. Denote boy as B , girl as G, when there are fewer boys than girls , the satisfying condition is 2B4G, 1B5G, 0B6G
Every birth can be boy or girl , so there are 2^6 combinations
Consider the 2B4G condition , due to the differents order of the birth , there are C6(2) combination
Consider the 1B5G condition , there are 6 combination
Consider the 0B6G condition , there is only one
the probabilty p= [C6(2)+6+1] /[2^6]
5. statisticians A,B,C,D , Hotel W,X, Y ,Z
Assume A picks W , we have [BX,CY,DZ] [BY,CX,DZ] ......... 6 combination
Since A can choose between W,X,Y,Z , ,so there will be 6x4 =24 combinations
That's the situation when everyone chooses different hotel.
Consider all the situations , 4 can choose one hotel , maybe 3 , 2 ,1 .. so A may choose W,X ,Y,Z, since each of them has 4 choices , the sum of the situation is 4x4x4x4 =256
the probability p=24/256 =0.09375
4.everytime the coin is tossed the probability is 1/2 , the fourth head is fixed , So there are 3 heads in the early toss,that's like 10 place for 3 heads ,there are C10(3) kinds of arragnement , then 4 heads , 7 tails ,we get (1/2)^4*(1/2)^7 . I think the probability is C10(3) *(1/2)^4*(1/2)^7 = 0.05859375 , Dont know if I am correct