What you want to do here is take this information and plug it into point-slope form. any time you're given a point and a slope, you generally want to plug it into this equation: y - y1 = m(x - x1).
in this equation, m is your slope and (x1, y1) is a given point. plug in your info--slope of -3 and (-5, 2).
y - 2 = -3(x + 5)
that is the equation of your line. however, if you want to graph it, this doesn't really make much sense to you. convert it to slope-intercept form, y = mx + b, by solving for y.
y - 2 = -3(x + 5) ... distribute -3
y - 2 = -3x - 15 ... add 2
y = -3x - 13 is your equation.
to graph this, and any other y = mx + b equation, you want to start with your y-intercept if it's present. your y intercept here is -13, which means the line you wasn't to graph crosses the y-axis at y = -13, or (0, -13). put a point there.
after you've plotted that point, you use your slope to graph more. remember that your slope is "rise over run"--you rise up/go down however many units, you run left/right however many units. if your slope is -3, you want to go down 3 units, then go to the right 1 unit. remember that whole numbers have a 1 beneath them as a fraction. -3/1 is your rise over 1.
The answer is
5mn(KUASA DUA) -6gh(KUASA DUA) +2
Let x= months
So 50x is the total amount he pays after x months. You went to add 100 to this because there's a fixed cost of 100 for the phone
So you have the expression 100+50x which is the amount he pays after x months
Since he has a limit of 300,
100+50x < 300
