CPhill

Oops, sorry...ignore that statement.....I got sidetracked !!!   I meant to say that 0 was in the excluded interval!!!

The domain is (-∞, -1/√2) U ( 1/√2, ∞ )

And 0 obviously isn't in that interval !!!!!

I assume that we have √[(-2-3)^2 + (4-0)^2 ] = √[(-5)^2 + (4)^2] = √[25 + 16] = √41 = about 6.403

Let x  be the number for the first night.....so we have

(x) + (x + 25) + (x + 25 + 25) + (x + 25 + 25 + 25) = 1050    simplify

4x + 150 = 1050    subtract 150 from each side

4x = 900     divide both sides by 4

x = 225

So....the number for each night are

225..250..275..300

Notice xvxvxv, that x cannot be 0  in the denominator

For the function iside the root....it must be > 0

So we have

40 = ab⁸  and 8 = ab⁴⁰

Using the first, we can solve for a  =   40/b⁸ ........and substituting this into the second, we have

8 =(40/b⁸)b⁴⁰  

8 = 40b³²        divide both sides by 8

(1/5) = .20 = b³²            take the (positive) 32nd root of both sides

b = about .951

And substituting this into 40 = ab⁸  to find a, we have

40 = a(.951)⁸        divide both sides by (.951)⁸ 

a = 40 / (.951)⁸  = about 59.79

So....our function is

w(x) = 59.79(.951)^x

Hey, h7.....you left out Melody !!!

I think we have

x * (x + 1) = x +(x+1) + 155   simplify

x^2 + x  = 2x + 156    

x^2 - x - 156 = 0    factor

(x -13)(x +12) = 0

Setting each to 0, we have that x = 13 and x = -12

So....since the answer is positive....x = 13  and (x +1) = 14

And 13*14 = 182   and 13 + 14 = 27

And 182 is 155 more than 27

And there you go....

Thanks, h7.....I like my avatar too much to change it........maybe I can just be an "honorary" member.......!!!