geno3141

When I try graphing your function using my TI-83+, I keep getting an error ... I think because may calculator can't handle 365!

I believe that your formula is correct; it's just that the numbers are larger than your calculator's ability to handle them.

Since  81  =  3^4

then:  1/81  =  3^-4

So:     3^x  =  3^-4     --->     x  =  -4

-1/2  =  sin( πx/11 )

Find the invsin of both sides:  invsin(-1/2)  =   πx/11

Multiply both sides by 11 and divide by π   --->   (11/π)invsin(-1/2)  =  x

(When you finish this, remember that there are two answers, one in Quadrant III and the other in Quadrant IV.)

10^(-2)   =   1 / (10^2)   =   1 / 100

Yep! That's good.

When you wrote log4x, did you mean:  log (to the base 4) of x  or log (to the base to) of 4x?

First:  log (to the base 4) of x:  I'll skip the '4' when I type:

   (1/3)(logx + logz)  =  (1/3)(log(xy)      (because adding logs means that the numbers were multiplied)

    =  log[ (xy)^(1/3) ]       (because a multiplied goes into a log as an exponent) 

Second:  log (to the base 10 )

   (1/3)[log(4x) + log(4y)]  =  (1/3)[log(4x·4y)]  =  (1/3)[log(16xy)]  =  log[ (16xy)^(1/3) ]

Oops, that wasn't you question; sorry!

First, subtract 6 from 3Q784 and then sum those digits to get a multiple of 9.

A number is divisible by 9 if the sum of its digits is divisible by 9.

Can you find a digit Q such that  3 + Q + 7 + 8 + 4  is a multiple of 9? 

Theorem of the medians of a triangle:  they are concurrent at a point that is 2/3rds of the distance from a vertex to the middle of the opposite side.

Can you take it from here?