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Hi Guest, this is what I meant :)

what is a

y=a(x-2)^2+3.5

\(y=a(x-2)^2+3.5\) Subtract 3.5,

                                        divide (x-2)² on both sides

\(\large a=\frac{y-3.5}{(x-2)^2}\)        !

Oh I see!!! Thank you for pointing it out :D

log base 3 of log base 2 of 8 equals what???

\(log_3(log_2 \ 8 )=what?\)

\(log_2 \ 8=3 \)

\(log_3 \ 3=1\)

\(log_3(log_2 \ 8 )=1\)

   !

log base 3 of log base 2 of 8 equals what???

\(log_3(log_28)\\ =log_3(3)\\ =1\)

1.5[3(14.5+7)-3]-7.4

\(1.5[3(14.5+7)-3]-7.4\)

\(=1.5[3*21.5-3]-7.4\)

\(=1.5*61.5-7.4\)

\(=84.85\)         !

Okay. I don't get what you are suposed to fill in for these cases. 

Case 1:

First pick multiple of 3

Second pick not multiple of 3.

Case 2:

First pick not multiple of 3

Second pick multiple of 3

Case 3:

Both picks multiples of 3

Isn't this like a math forum so.....

I am currently taking a course in Counting and Probability andi am posting the questions I don't understand onto here. I have also posted some Geometry questions too so.....

1.5 (3 (14.5 + 7) - 3) - 7.4 =

1.5 (3 (21.5) - 3) - 7.4 =

1.5 (64.5 - 3) - 7.4 =

1.5 (61.5) - 7.4 = 

92.25 - 7.4 = 84.85

It also SOMEtimes denotes concentration , such as when you are caculating pH

pH = - log [H30+]       ph is the negative log of the hydronium ion concentration....

There are TWO 'N' atoms in each molecule, so 

5.66 moles x 2 atoms/molecule = 11.32 moles of 'N'

(1/3)^3 =1/27

m^(2/3) = \(\sqrt[3]{m^2}\)

You can calculate this by simply doing the following:

(1/3)*(1/3)*(1/3)

or

(1/3)(1/3)(1/3)

Log[3, Log[2, 8]] = Log[Log[2, 8]] / Log[3] = 1

Solve for t:
120 = 5. t^2 + 10 t

5. t^2 + 10 t = 5 t^2 + 10 t:
120 = 5 t^2 + 10 t

120 = 5 t^2 + 10 t is equivalent to 5 t^2 + 10 t = 120:
5 t^2 + 10 t = 120

Divide both sides by 5:
t^2 + 2 t = 24

Add 1 to both sides:
t^2 + 2 t + 1 = 25

Write the left hand side as a square:
(t + 1)^2 = 25

Take the square root of both sides:
t + 1 = 5 or t + 1 = -5

Subtract 1 from both sides:
t = 4 or t + 1 = -5

Subtract 1 from both sides:
Answer: |t = 4                   or               t = -6

If you were trying to add 1 half and two quarters, you would input the following:

(1/2) + (2/4) =

A googol is 10^100, or a one with 100 zeros after it. This is 20 orders of magnitude higher than the estimated number of hydrogen atoms in the universe, which makes a googol difficult to use in real-world math. A googol to the googol power, a googolplex, is a one followed by 10^100 zeros.

CORRECTION: Sorry, I calculated the number of N atoms in 5.66 moles of Barium Nitrate.

If you want the number of moles of N, then just multiply:

10.719% x 5.66 =0.607 moles of N atoms in 5.66 moles of Barium Nitrate.

Percent composition by element

Element  Symbol  Atomic Mass  # of Atoms  Mass Percent

Barium    Ba         137.327              1              52.548%

Nitrogen  N           14.0067              2             10.719%

Oxygen   O           15.9994              6              36.733%

1 mole = 6.02 x 10^23 atoms.

10.719% x [6.02 x 10^23] =6.452838 x 10^22 atoms of Nitrogen in Barium Nitrate.

5.66 x [6.452838 x 10^22] =3.65 x 10^23 atoms of Nitrogen in 5.66 moles of Barium Nitrate.