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(a + b)  = 69          square both sides    a^2 + 2ab + b^2  =  4761  ⇒  2ab = 4761 - a^2 - b^2  ⇒

2ab = 4761 - (a^2 + b^2) ⇒

2ab = 4761  - ( 1944)

2ab =  2817

ab = 2817/2

Note that  a^3 +b^3    =    (a + b) ( a^2  - ab   + b^2)

So  we have 

a^3 + b^3  =  ( a + b ) ( [ a^2 + b^2 ] - ab )   =

(69) [ 1944 - 2817/2 ]  =  

73899 / 2

I just got that before I saw your post.  That is correct!!  Can you please try and answer this question? https://web2.0calc.com/questions/pls-halp_27

Can somebody pls try and answer this one?

300 factors as     5^2 * 2^2 * 3

I see the possible identifiable sequences as

5 * 5 * 2 * 2 * 3 =    5!/(2! * 2!)  = 30

5 * 5 * 4 * 3 * 1 =    5! / 2! =  60

5 * 5  * 6 * 2 * 1 =  5! / 2! =  60

I get  150

Anyone else ???

I have gotten that answer before.  It is incorrect.  Can you show how you got it?

6 sided.

3 - 5 - 7  by the triangle inequality works

It's super hard.  I've tried that multiple times, but I just get so stuck.  Pls help me!!  

Looks good, qjin  !!!

That is true, 

You seem to get these counting and prob questions with ease. These type of questions always mess up my brain!!!

-Vinculum

You are very good at this  counting stuff......I'm usually stuck on many  of these  !!!

Rewrite the equation using the known values of x and y in terms of k as: \(k+1 = 4k - 14\)

Solving for k, we find \(k = 5\)

Substituting the known value of k into the point, we find it passes through the point \(\color{brown}\boxed{(5,6)}\)

There are 2 options for 1 question

There are another 2 options for the other question

The remaining 8 must all be false.

There are \(10 \choose 8 \) ways to choose which questions are guaranteed to be false

So, in total, there is: \(2 \times 2 \times {10 \choose 8} = \color{brown}\boxed{180}\) ways to answer. 

1) \(\color{brown}\boxed{f(x)= 63-7(x-1)}\), where \(1 \leq x \leq 6\)

2) There are 6 rows, the bottom row with 63 and the top row with 28. 

Writing as factors of 7, we have: \((7 \times 4) + (7 \times 5) + \space ... \space + (7 \times 9)\)

We can rewrite this as: \(7 \times (4+5 + \space ... + \space 9)\)

This is equal to \(7 \times 39 = \color{brown}\boxed{273}\)

a)  To find the number of boxes in any row...we can use

an  =  63 - 7 (n - 1)      where  n  is an integer from 1 to 6 inclusive

So...in the 6th row, we have

a6  =  63 - 7(6 - 1)  =  28 boxes

b)  To find the total number of boxes, we can use

Sum  = 

[ Number of boxes in the first row + Number of boxes in last row ] * [ No. of rows / 2 ]

So we have

[ 63 + 28 ] * [6/2]   =

273 boxes in total

Used answer by CPhill for framework: https://web2.0calc.com/questions/anyone_11

-Vinculum