CPhill

The slope of this line is   ( 7 - -3) / ( 1 - -2)   =  10/ 3

Using either point to write the equation, we have

y =  (10/3) ( x - 1) + 7      multiply both sides by 3

3y  = 10 (x - 1)  + 21

3y  = 10x - 10 + 21

3y =  10x + 11     rearrange as

10x - 3y  =  -11

N(O(N(O(N(O(3))))))

O(3)  =  9

N(O(3)) = N(9)  = 6

O(6)  =  36

N(O(6))  = N(36) =   12

O (12)  =  144

N(O(12)))  =  N(144)  =   36

( a^2/5 b² )⁵    =

(a^2/5)⁵  * (b²)⁵  =

Using  (a^m)ⁿ   =   a^m*n      we have

a^10/5  *  b¹⁰   =     a²b¹⁰

[ Note, Manuel.....we don't have any fractional exponents as a result... ]

{ 2, 3, 4 ,9 }

Possible outcomes  =  [ P(4,2) + 4 ]    = [12  +  4]   = 16

It may be easier to  look at the possibilities that don't work

9 paired with itself, 4 or 3 are not favorable outcomes

So....this is  5 unfavorable outcomes  (9,9), (9,4), (4, 9), (9,3), (3,9)

3 paired with itself is also unfavorable

So....this is 6 unfavorable outcomes

So....there are 10 favorable outcomes

(2,2) (2,3) (3,2) (2,4) (4,2) (2,9, (9,2), (3,4) (4,3) (4,4)

³ √ [27x - 81]    -  5    =

³ √ [ 27 (x - 3) ]  - 5                     [Note    ³√27  =   3  ]

3 ³√ (x - 3)  -  5

This is the cube root function vertically stretched by a factor of 3

Shifted to the right 3  units

And shifted down 5 units

Here's the graph of both :

https://www.desmos.com/calculator/man8dgxb7u

The above answers are identical...

-4 + x/3  =

-12/3 + x/3   =

[ x - 12 ]  / 3

Average rate of change  =

[ f(9)  - f(5) ]  /  [ 9 - 5 ]  =

[ 14 - (-2) ]  /  [ 4 ]  =

[ 16 ]  /  [4 ]  =

4

1.

Consider  "UA"  to be a single entity

Can  put these in any one of 5 positions    12, 23, 34, 45, 56

And for each of these, there are 4!  ways to arrange the other letters

But....we can also write UA  as AU

So.....the number of total "words"  is

2 ways to arrange UA  x  5 positons  x 4! arrangements of the other letters

2   x  5  x  4!   =

2 x  5  x  24  =

240  "words"

2.  

Similarly, for   UAR,  consider this as a single entity

It can appear in  4 different positions    123    234    345    456

And for each of these, the other letters can be arranged in 3! ways

But....UAR  can also be arranged in 3!  ways

So....the total possible "words"   =   

4  x  3!  x 3!  =    

4 x 6 x 6  =

144  "words"

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

a_n  =  23 - 3 (n - 1)      where  n  is an integer from 1 to 6 inclusive

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

a₆  =  23 - 3(6 - 1)  =     23 - 3(5)   =  23  - 15  =  8 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

[ 23 + 8 ] * [6/2]   =

[31] * [3]  =

93 boxes in total

Mmmmm....OK.......