CPhill

x(-x+6)=0

Setting both factors  to 0 , we have

x = 0         and      -x + 6 = 0

The first is obvious

For the second, we have

-x + 6 = 0     add x to both sides

6 = x

These are the two x intercepts  at x = 0 and x = 6

It's asking you to show that g(x) is it's own inverse

Fof g(x), write  y   and we have

y = ( x + 1) /(x - 1)     the object is to first get a single x by itself......multiply both sides by  x - 1

y ( x - 1)   =  x + 1

yx - y = x + 1       rearrange as

xy - x = y + 1       factor out x on the left

x(y - 1)  = y + 1    divide both sides by y - 1

x = [ y + 1 ] / [ y - 1]       "exchange" x with y

y = [x + 1 ] / [x - 1]     and for y, write g^-1(x)

g^-1(x) =  [x + 1 ] / [x - 1]     notice that this is the same as g(x)

y = ∛x    

To find the x intercept, let y = 0

0 = ∛x     cube each side

0 = x     so, wehave the point (0, 0)

To find the y intercept,let x = 0

y = ∛0   = 0

So........again we have the point (0,0).....thus the x and y intercepts are the same, i.e., the origin

Imaginary.......example......

a * bi  =  

3 * 5i  =

15i

|(-3)3| – (19 – |-10|)  =

l -9 l  - ( 19 - 10 ) =

9 - 9   =

0

See my explanation under the original post....

Think  about the problem in this way:

If we square any positive real number greater or equal to 1, we will always get a result that is greater or equal to 1

Likewise....{and this is tougher to see} ......, squaring any negative real number that  less than or equal to -1 will also give a result that is greater or equal to 1......example   (-1.1)^2 =  1.21

This is what geno is trying to say.......that x cannot come from either of the intervals  x ≥ 1 or   x ≤ -1

Thus.....x   must come from the interval  (-1, 1)

There are 30 days in June and we want to choose any two of them....this is a binomial probability problem

"At most"   means it either rains 0 days, 1 day or 2 days

0 days  = (9/10)^30  = 0.042391158275216203514294433201

1 day = C(30,1) (1/10)(9/10)^29  = 0.14130386091738734504764811067

2 days =  C(30,2) * (1/10)^2 * (9/10)^28 =  0.227656220366901833687877511635

The sum of these =  0.411351239559505382249820055506  ≈  0.411

The possible primes are ;

2, 3, 5, 7, 11

Probability of a 2  = 1/36

Probability of a 3 = 2/36

Probability of a 5 = 4/36

Probability of a 7  = 6/36

Probability of an 11  = 2/36

The sum of these probabilities = [1 + 2 + 4 + 6 + 2] / 36 =  15/36 =  5/12

Sorry...I mis-read the question....

10.2^3 =  1061.208

10.3^3  = 1092.727

So.........there are  [ 1092 - 1061 ]   = 31  integers between these two values