Answer:
domain: 2x+1 not equal to zero, then x not equal to -1/2
Set 2x+1>0
then x>-1/2
100^(3-4x)/(2x+1)<=0.01^(5x-8)/(2x+1)
100^(3-4x)<=0.01^(5x-8)
(10^2)^(3-4x)<=[10^(-2)]^(5x-8)
10^[2*(3-4x)]<=10^[(-2)*(5x-8)]
10^(6-8x)<=10^(16-10x)
6-8x<=16-10x
2x<=10
x<=5
combine x>-1/2 and x<=5
then - 1/2<x<=5
Set 2x+1<0
then x<-1/2
100^(3-4x)/(2x+1)<=0.01^(5x-8)/(2x+1)
100^(3-4x)>=0.01^(5x-8)
(10^2)^(3-4x)>=[10^(-2)]^(5x-8)
10^[2*(3-4x)]>=10^[(-2)*(5x-8)]
10^(6-8x)>=10^(16-10x)
6-8x>=16-10x
2x>=10
x>=5
combine x<-1/2 and x>=5
, then x does not exist while 2x+1<0
so answer is - 1/2<x<=5
This is the answer your looking for 525.84
Yes. B/c as long as it has the same variable, you can add them, subtract, or anything else to them. As long as they have the same variable in common.
Hope this helps (:
If your Main Street line has a couple of points that you can find the coordinates ( x1, y1) and (x2, y2) you could determine the slope of the Main Street line : m= (y2-y1)/(x2-x1). Or maybe you can determine what the slope is from the line being a particular line , horizontal, etc. The slope of your path line would be m1 = -1/m. Use some point on this path to write the equation of a line in point- slope from. Suppose ( x3,y3) would be a point , ( look for example for the entrance coordinates), write y= m1(x-x3)+ y3 and plug in then values of m1, x3, y3 to have your path line equation. ~Nesabug :3