CPhill

[Quad]rilateral = 4

[Hex]agon = 6

[Tri]angle = 3

[Pent]agon = 5

[Sept]agon ...also called a  [Hept]agon.....  = 7

12-2[5x-3(x-4)]-8x   work inside the brackets, first

12 - 2[ 5x - 3x + 12] - 8x      simplify inside the brackets

12 - 2 [ 2x + 12] - 8x      distribute the "2"  across the terms in the brackets

12 - 4x - 24 - 8x       combine like terms

-12 - 12x

11542 + l-922l  = 12464  ft   above sea level

1-(3/4)(v+2)= 5   subtract 1 from each side

-(3/4)(v + 2)  = 4     multiply both sides by -(4/3)

v + 2     = -16/3        subtract 2 from both sides

v = -16/3 - 2   =  -16/3  - 6/3   = -22/3

(3x-4)^3   .....we can use the Binomial Theorem, here......we have

C(3,0)*(3x)^3 - C(3,1)*(3x)^2*(4) + C(3,2)*(3x)*(4)^2 - C(3,3)*(4)^3  =

27x^3 - (3)(9x^2)(4) + 3(3x)(16) - 64 =

27x^3 - 108x^2 + 144x - 64

OK.....!!!!

Anonymous......I'd need to see a picture of what you have........you have to register to post pictures.....it's not hard to do........!!!!!

Seems as though I'm having to fill in a lot of "blanks" tonight  !!!

I assume we have :

32^{x - 3} = 64        we can write this as:

(2⁵)^{x - 3}  = 2⁶       simplify

(2)^{5x - 15}  = 2⁶      and since we have  equal bases, we can solve for the exponents...so we have...

5x - 15   = 6      add 15 to both sides

5x  = 21    divide both sides by 5

x = 21/5

I assume that this might be : b+(3/2)b+(b+45)+(2b-90)+90.....if so, we have:

4b + (3/2)b + 45 =

(8/2)b + (3/2)b +  45 =

(11/2)b + 45

We have.....

Pt = Pb^t-1     where P is the original population ....   Pt is the popoulation after "n" days  ... b is the growth rate... and t is the number of days ...[ t = 1 is the first day.....so, we're looking for the population 3 days after. the first day = the end of the 4th day ]....and we have

94 = 6b^{(4 - 1)}   =    6b⁽³⁾      divide both sides by 6

(94/6)  = b³        take the cube root of both sides

b ≈ 2.5022

So, our function is  :     Pn =   6(2.5022)^{(t - 1)}