1. Approximate: log5 47 to the nearest ten thousandth. (I put 2.3922 but that is incorrect)
The 5 is tiny
2. Approximate: log12 5 to the nearest ten thousandth. (I put 0.6477 but that is incorrect)
The 12 is tiny

b. 18

step-by-step explanation:

so the equation for this is:

ab^2=ac⋅ad

ab being the top length of 6 and ac being the length of 2 and ad being the length of x. let's plug into the equation:

6^2 = 2x

now lets just solve for x:

36 = 2x

18 = x

i hope that !

b. f(x) = –√x + 3this is the correct answer