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( 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

all whole numbers is the answer

The above answers are identical...

-4 + x/3  =

-12/3 + x/3   =

[ x - 12 ]  / 3

sqrt(2x+10)  -6  = 2     Add 6 to both sides of the equation

sqrt(2x+10) = 8            Square both sides

2x+10 = 64                  Subtract 10 to both sides

2x= 54

x = 27

Is "find (f * g)(4)" meant to be written in that way, or is it suppose to be (f * g)(x)?

I think that the answer is x=27.

f(x) - g(x)

(-2x + 4) - (-6x - 7) = 4x + 11 

Answer = 4x + 11

Repace 'y' with 'x'    and 'x' with 'y'  then sove for 'y'

x= 3y+12

x-12 = 3y

y = (x-12)/3    is the inverse

I believe that the inverse is f^-1(x) = -4 + x / 3

Human egg cell : Approx  0.1 mm dia

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

A). First replace y in the first equation by the right hand side of the second equation:

4x - 1 = 2x + 3

subtract 2x from both sides: 2x - 1 = 3

add 1 to both sides: 2x = 4

divide both sides by 2:  x = 2

put this into the first equation: y = 2*2 + 3 or y = 7

hence the point of intersection is (2, 7)

B) You have only supplied one equation

C) Try using the same approach I used in A) above.

Must it be "UA", or is "AU" acceptable as well ???

The same for "UAR", or are other perumations allowed such as "AUR", "RUA"......etc.??

A triangle with vertices A(5,3), B(1,-4) and C(-3,1) is rotated 90 degrees counterclockwise about vertex B.

Find the co-ordinates of the image of vertex A.

\(\begin{array}{llcll} \text{Formula Rotation:} & \boxed{\vec{A}' = (\vec{A}-\vec{B_{rotation\ axis}})\cdot D +\vec{B_{rotation\ axis}} } \\ & \vec{A} \text{ before rotation} \\ & \vec{A}' \text{ after rotation} \\ & \text{$\vec{B_{rotation\ axis}}$ at $\binom{1}{-4} $ } \\ & \overset{\curvearrowleft}{D}_{\varphi} = \begin{pmatrix} \cos(\varphi) & \sin(\varphi) \\ -\sin(\varphi) & \cos(\varphi) \end{pmatrix} \ \text{ Matrix of rotation counterclockwise}\\ & \overset{\curvearrowleft}{D}_{90^{\circ}} =\begin{pmatrix} \cos(90^{\circ}) & \sin(90^{\circ}) \\-\sin(90^{\circ}) & -\cos(90^{\circ}) \end{pmatrix} =\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \\ \end{array} \)

\(\begin{array}{llcll} \text{Formula Rotation:} & \boxed{\vec{A}' = (\vec{A}-\vec{B_{rotation\ axis}})\cdot D +\vec{B_{rotation\ axis}} } \\ & \begin{array}{|rcll|} \hline \vec{A}' &=& \left(\dbinom{5}{3}-\dbinom{1}{-4} \right)\cdot \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} +\dbinom{1}{-4} \\ &=& \dbinom{4}{7} \cdot \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} +\dbinom{1}{-4} \\ &=& \dbinom{-7}{4}+\dbinom{1}{-4} \\ &=& \dbinom{-6}{0} \\ \hline \end{array} \end{array} \)

ok thanks, I will think about it some more.

I did it as simultaneous, diophantine equation. :/

Melody: I wanted a number that was less than the LCM of 360, and that was divisible EVENLY by 8 and that ended in 2. Also, quickly realized that 352 - 1 =351 is divisible EVENLY by 9 since 3+5+1 =9. And finally, realized that 351 - 1=350 which is also divisible by 10 EVENLY !!.

As follows:

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idk598 53. I need help 😛😛

Could you please explain your logic?

Why did you subtract 8 from 360?