Rewriting each number in base 3:

3^1998 + (3^2)^999 + (3^3)^n = 3^1999

3^1998 + 3^1998 + 3^(3n) = 3^1999

It takes 3 3^1998s to make 3^1999 as 3 x 3^1998 = 3^1999

Therefore 3^(3n) = 3^1998

so 3n = 1998 and n = 666

Q; For what value of n does 3^1998 + 9^999 + 27^n = 3^1999?

(^ = to the power of)

A: 666

Rewriting each number in base 3:

3^1998 + (3^2)^999 + (3^3)^n = 3^1999

3^1998 + 3^1998 + 3^(3n) = 3^1999

Rewriting each number in base 3:

3^1998 + (3^2)^999 + (3^3)^n = 3^1999

3^1998 + 3^1998 + 3^(3n) = 3^1999

It takes 3 3^1998s to make 3^1999 as 3 x 3^1998 = 3^1999

Therefore 3^(3n) = 3^1998

so 3n = 1998 and n = 666

Q: With one straight cut a cake can be sliced into two pieces. A second cut that crosses the first one will make four pieces, and a third cut can produce as many as seven pieces. What is the largest number of pieces that you can get with six straight cuts?

A: 22

With 0 cuts, you get 1 big cake

With 1 cut, 2 pieces

With 2 cuts, 4 pieces

For the 3rd cut, if you cut cleverly you can get 7 pieces

With 0 cuts, you get 1 big cake

With 1 cut, 2 pieces

With 2 cuts, 4 pieces

For the 3rd cut, if you cut cleverly you can get 7 pieces

click here for more info and nice graphics thanks to the Guardian

http://bit.ly/CuttingthePieGardner

http://bit.ly/CuttingthePieGardner

Definition of a function:

Every x maps to a unique y

Notice Jan has 31 days and that’s it

March has 31 days and April has 30 days

But the number of days in February depends on leap year vs non leap year

So which month maps to two outputs?

(https://www.timeanddate.com/calendar/months/

Using substitution, we can rewrite the second equation as

3(-2x + 2x + 8) = 12

3(8) = 12

24 = 12

huh? NO solution

If we were to solve the second equation for Y=, we would get:

by distributing the 3

-6x + 3y = 12

add 6x to each side

3y = 6x + 12

Divide both sides by 3

y = 2x + 4 this is parallel to the first line y = 2x + 8 and therefore the lines will never meet

making the answer “No solution”

First look at the graph — looks super nonlinear and pretty exponential

Use the point (0,4).

We can eliminate answers (2) and (3), as if we substitute h=0, the output would be 6/5 and 4.2 as opposed to the 4 for the y value we are looking for.

Now we can use the point (1,8) to see if y = 8 when x = 1

For answer (4), if we substitute in x = 1, we get 2/3(1)^3 – 1^2 + 3(1) + 4 = 2/3 – 1 + 3 + 4≠ 8

Try answer (1): using order of operations: 4(2)^1 = 4(2) = 8

4(2)^2 = 16 and 4(2)^3 = 32. That’s it –2 points

You can also use Y= and check the table for a match: