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Well, we know we can split the balls into four groups: 4-0-0-0, 3-1-0-0, 2-2-0-0, 2-1-1-0, 1-1-1-1.

Case 1: 4-0-0-0 has only 1 way we can assign the balls to different boxes.

Case 2: 3-1-0-0 has (I think) 4 ways we can assign the balls to different boxes. The one ball can go anywhere, and the three other balls must go in a different box

Case 3: 2-2-0-0 has 4C2 = 6 ways to choose two balls to go in one box, then the other two balls must go in separate boxes. But since order doesn't matter, (there is no first box or second box), we divide by 2.

Case 4: 2-1-1-0 has (again) 4C2 = 6 ways to choose two balls to go in one box, then the other balls go in seperate boxes.

Case 5: 1-1-1-1 has only 1 way.

1 + 4 + 6/2 + 6 + 1 = 15.

What is an ellipse?

hi Guest!

I think your solution was just CPhill's but copied and pasted... make sure you give him credit next time :)

That isn't right.  Can you please show me how you got that?

Formula

proj _v u    =  dot product (u, v)

                  _______________   *      b        =

                 (  magnitude of v)^2

-6*1   +  -7*1

___________ *   ( 1 , 1)     =

  1^2 + 1^2

-13

___  * (1 , 1)     =

  2

(13/2, 13/2)

1/a  +  1/b   =  (a + b)   / ab

The  sum  of  the roots  =  a  + b   =   (7/2)

The product of  the  roots  =  ab  =   (13/2 )

So

1/a  + 1/b   =     (7/2)   /(13/2)    =   7  / 13

xy  =  3x + 2y +  z

xy - 3x   =  2y  + z

x(y - 3)   =  2y + z

            (2y + z)

x   =    _______

            (y -  3 )   

If this is true.....we should be able  to  plug  in values  for  x, y  and see if  we get the same result

Let  x =  2       Let y  = 3

(2 + 3)^2  =  (2)^2  + (3)^2     ???

(5)^2  =  4  + 9    ???

25  =  13

So....not true

This question has been answered before, search it up.

If you are going to post your questions 10 times an hour no one will answer it.

√[ 7  + 2√10 ]

Let  us suppose   that we can write  this  in a final form  of √a  +  √b

Square both sides

7 + 2√10   =   a + 2√[ab]   +  b

Equating coefficients  we  have  that

a + b   =  7

2√10  = 2√[ab]

Then   ab  =  10   ⇒   b  =  10 / a

And

a + 10/a  =  7         multiply through by a

a^2  + 10  =  7a       rearrange as

a^2  - 7a  + 10   =    0        factor

(a  - 5) ( a - 2)  =   0

Setting  each factor  to 0  and solving for  a we  get  that

a = 5          a   =   2

When  a  =  5,  b =  10/5    = 2

or

When a  = 2, b  =  10/2  = 5

So.....depending on your preference for a , b.......we have

√[ 7  + 2√10 ]  =  √2  +  √5

"Two circles of radius r are placed in a square with side length 2, without overlap  What is the largest possible value of r?"

The right answer is 12. 12 choose 4=495

495>365

f(n)  =  3f(n - 2)  - 2f(n - 1)

f(2)  =  3

f(1)  =  -1

f(3)  =  3f(3 - 2)  - 2f(3 - 1)   = 3f(1) - 2f(2)  =   3(-1)  - 2(3)  =  -3 - 6  =  - 9

f(4)  =  3f(4 - 2)  - 2f(4 - 1)  =  3f(2)  - 2f(3)  =  3(3)  - 2(-9)  =  9 + 18  =  27

f(5)  =  3f(5 - 2)   - 2f(5 - 1)  =  3f(3)  - 2f(4)  =  3(-9)  - 2(27)  =  -27 - 54  =  -81

First one

a) Look at the chart  for  the  blcck  that  represents the  intersection of Wrexham and Corwen.....notice that this distance  =  21  miles

b)  He travels  from  Mold to Wrexham   =  12 miles  and then from Wrexham  to Ostwestry = 15 miles

So...one  way  =  (12 + 15)  = 27 miles

Round trip  =  2(27)  =  54 miles

This is for one day

For 5 days we have  5 (54)  =  270  miles

看样子学中文有用了！

第一道题：看样子从 wrexham 到 corwen 的距离 是 21 英里，因为两个城市的交叉点是 21.

Mold 和 Oswestry 距离是 27 英里，每星期的走 5*2 = 10 次，所以他的走 270 英里。

第二道题：到达时间是 14:19， 完 8 分钟就会 是 19+8 = 27， 所以到达时间是 14:27

12:30 到 14:27 分总共有1 小时 （60-3）分钟。如果是14:30 的话，时间是2 小时， 但是他早了3分钟所以他就花了1:57 分。

Judy 正时到达Shrewsbery火车站。时间就是 1:57 - 8 = 1:49 分

她花1:50 所以 1:49 + 1:50 = 3：39分。

12:30 + 3:39 分 就会等于 15:69 = 16:09 分。

我说中文，没用翻译器，但是我在美国住，所以我的中文有可能出错。

如果不懂什么的我欢迎你的问题。

Note: I'm hoping Guest uses chinese since I'm not proficient in chinese and these may include errors(it was a struggle lol)

This is manderin chinese btw

Since he must divide the number of coins into six groups, the number of coins must be both a multiple of 2 and a multiple of three.

Since you know that there are an even number of coints, you already know that the number of coins is a multiple of two.

Therefore, for the number of coins to also be a multiple of three, there is a 1/3 probability that Jack survives.

Last one :

a)  A  and C  are congruent to E

b)  If A  and C  are congruent to E, they are also congruent to each other

Huh???...I agree with Hugo.....if you rotate  "X"  90° counterclockwise, you  get "D"

A, D and  E  are all the same.....they are just rotations of the same congruent right triangle...

i got it wrong dont know how i just tried it plz explain plz

thank you