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If \(m∠1 + m∠2 = 110, m∠3 = 70\). Since \(m∠3 = m∠18 = m∠13\), we have \(m∠11+m∠12 = 110\), so \(m∠16 = 180 - 110 = 70\).

- Daisy

Do you understand what is meant by this: (𝑎𝑏𝑐𝑑) = 𝜇(1234)𝜇^-1 ?? I don't !! Can you explain it please? Thank you.

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After extensive guessing and checking, I found out that the first column and third column both had a sum of \(15\), which means that the second column had a sum of \(15\), also. 

\(15 - (8+1) = 15 - 9 = 6\), so \(? = 6\).

- Daisy

Arithmetic sequence \(A\), starts at 30, and goes up 10 each step.

Arithmetic sequence \(B\), starts at 30, and goes down 10 each step.

To find the \(n\)th term of a sequenece, we have a formula: \(a + (n-1)d\), where \(a\) is the first number of the sequence and \(d\) is the difference between each term. Let's use this formula to find the \(51\)st term of sequence \(A\).

We have \(30 + (51-1)10= 30 + 50\times10 = 30+500=530\).

For sequence \(B\), we have \(30 + (51-1)(-10) = 30 + 50(-10) = 30 + -500 = -470\). 

So, the difference would be \(530 - (-470) = 530 + 470 = 1000\).

- Daisy

Oh wow! I never thought of it as an intersection question! That helps alot! oh Im acidentally using a guest...

At pt (-4, -7), y = 3x + 5 interect with y = kx - 7

Therefore Y = -7 when x = -4 for both equations

Sub the values in equation y = kx - 7 

-7 = k(-4) - 7

0 = k(-4)

k = 0

First die: any number

2nd die : 5/6 probability that it is different from the first

3rd die  : 4/6 probability that it is different from the first two

4th die  : 3/6 probability that it is different from the first three

5th die  : 2/6 probability that it is different from the first four

So, the probability that they are all different from each other is:

6/6 x 5/6 x 4/6 x 3/6 x 2/6 =720 / 6^5. Therefore:

The probability that at least two of them show the same face is:

1 - 720/ 7,776 =7,056 / 7,776 =49 / 54 =90.74%

Thanks Correct!

Thanks! Both are correct! Thanks for your hard work!

P(all white on first 5 draws)  =

5/11 * 4 / 10 * 3/ 9 * 2/ 8 * 1/7  =

5 / 11  * 2 / 5  * 1/3 * 1/4 * 1/7 =

(2 /11) *  (1/12) * (1/7)  =

2 / 924  =

1 / 462

(5ab)^(2/3)  =

(5^2 * a^2 * b^2) ^(1/3) =

³√ [ 25 a^2 b^2 ]

³√ [ 27 x^6 y^3 ]  =

( 3^3 * x^6 * y^3)^(1/3) =    [use the fact that   (a^m)^n  =  a^(m*n)

(3 * x^2 * y)

5√ [12xm^3 ]  =

5 √12 * √x * √m^3  =

5 * √4 * √3 * √x * √m^3 =

5 * 2 * √3 * √x * m√m  =

5 * 2 * m √ [ 3 * x * m ] =

10m √ [3xm ]

7√28  - 5√63   =

7√[4 * 7] - 5√[9 * 7 ]  =

7 * √4 *√7 -  5 *√9 * √7  =

7 * 2 * √7  - 5 * 3 * √7  =

14√7 - 15√7 =

-√7

2         √7                2√7

__  *   ___   =         ___

√7       √7                  7

LOL!!!!

Here's the way I interpret the second question

Let your truck be capable  of carrying  x  lbs

Let  the total number of pounds that your neighbor can carry in 4 identical trucks  =  y

So each of your neighbor's trucks can carry  y / 4   lbs

"My truck can carry at least 800 pounds more than each of her trucks"  translates to

x ≥  800 + y / 4

And 

"...but no more than one-third what all four of her trucks combined can carry"  translates to

x ≤  y/3

Look at the intersection point of these two inequalities :  https://www.desmos.com/calculator/xggshl0aos

Your truck can carry  3200 lbs  .....

Your neighbor's trucks can carry  9600 lbs

So  each of her trucks  can carry  9600/ 4  = 2400 lbs

And your truck carries at least 800 lbs more than each of her trucks [ 3200 - 2400 ]

Plus  your truck can carry no more than 1/3  of what all 4 of her trucks can carry.....9600/3  = 3200 lbs

Yes, sometimes it is just plain stupid.

inflammable and flamable is one  off the absolute stupidest examples of our language.

We both agree on that!

Another stupid word is 'pitted'.

If a date is pitted, does that mean its pit is taken out or does it mean that the pit is still in it.

OR

Can it mean either one?   Beats me!         

ok I will give it another shot

3 identical boxes and 4 different balls. Have I got it right this time?

possible numbers of balls in each of the identical boxes

4,0,0

3,1,0

2,2,0

2,1,1       

All four in a box,    1 way

how many ways can i choose 3 balls? well one must be left out so that is 4 ways

So there are 4 ways that 3 balls go into one box and 1 in another.

how many ways can I chose 2 balls.   mmm  4C2= 6 ways

so there are 6 ways that 2 can go in one box and 2 in another. 

BUT hang on, ive double counted here, there is really only 3 ways

There are also 6 ways that 2 can go in one box and one in each of the other boxes  6 ways

I get 14 ways.     

That is what CPhill got too :)

I think in this instant that Quora answerer is incorrect. They have double counted on the 2-2-0 posibility.

HAHAHA!!!.....that's like the words  "inflammable"  and "flammable"....they both mean the same thing......!!!

The English language is strange, isn't  it  ???

Oh okay XD i can't seem to find this stuff :P

First one answered here : https://web2.0calc.com/questions/help_9075

Yea sorry guest, I goofed up    

Sorry, Melody...but if the balls are distinguishable, we  can tell them apart....so Ball1  is different from Ball 2 ....and those are different from Ball 3...etc.

The boxes are "indistinguishable"....we cannot tell them apart....Box 1  is the same as Box 2....etc...

Here is the question on the Internet:

"In how many ways can 4 distinct balls be distributed into 3 identical boxes?"

Isn't the word "distinct" the same as "distinguishable", and the word "identical" the same as "indistinguishable"? If so, then it is exactly the same question.

LOL

Chris

"if the balls are distinguishable"     It is the balls that are the same.    

edit:

Woops you are right - I kept seeing indistinguishable.  

I hate that word, those words, distinguishable and indistinguishable The are totally confusing.

Why can't the question use the words different and identical  !!

I goofed up    Sorry.   

Here's my attempt at this....

Since the boxes are indistinguishable, it really just boils down to the selection of balls to put in the first box...also...I'm assuming that we can have empty boxes

[ I think Melody's assumption is slightly incorrect  ...the boxes are the same but the balls are different ]

4 balls in the first box  =  1 way

3 balls in the first box....the other ball in either two of the remaining boxes  = C(4,3)  =  4 ways

2 balls in the first box and 2 others in any remaining box  = C(4,2)  / 2  =

6 /2  = 3 ways

{  The reason for dividing by 2 is that  if we selected balls 1 and 3  for the first box, balls 2 and 4 would occupy either of the other two boxes....but...this is the same as  selecting balls 2 and 4 for the first box and balls 1 and 3 to occupy either of the other two boxes )

2 balls in the first box and  one in each of the other two  = C(4,2)  = 6 ways

So...by my count....the total ways  =  1 + 4 + 3 + 6  =  14 ways