The point slope form of the equation of the line that passes through (-4 -3) and (12, 1) is y - 1 = 1/4 (x - 12). what is the standard form of the equation for this line?

Standard Form is ...

ax + by = c

where a ≥ 0, b > 0 (if a=0), and {a, b, c} are mutually prime integers (unless irrational numbers are involved).

You can get there from where you're starting in a couple of steps. Multiply by the least common denominator of any fractions. (Here, you multiply by 4.) Then put the variable terms on the opposite side from the constant term. Make sure the leading coefficient is positive.

... 4y -4 = x -12

... x -4y = 8 . . . . add 12-4y

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752

y=sec x is the correct answer

~batmans wife dun dun