The orthocenter is the point where the three altitudes meet.
sketch the graph and you will see that AB is a horizontal line, the altitude is a vertical line through the point (1,3), so the equation of this altitude is x=1
next, find another altitude. I'll use the altitude of BC.
the slope of BC is (6-3)/(4-1)=1, so the slope of the altitude, which is perpendicular to BC going through the point A (0,6), is -1, the equation of the altitude of BC is y=-x+6
the system of equation : x=1
y=-x+6
has a solution (1, 5)
the solution is where the two lines meet, the meeting point is the orthocenter.
Double check by find the equation for other altitude:
slope of AC: (3-6)/(1-0)=-3
slope of altitude of AC: 1/3
equation of altitude of AC: y=(1/3)x+b
the altitude of AC goes through point B (4,6), so we can find out b by plug x=4, y=6 in the equation: 6=(1/3)*4+b, b=14/3
y=(1/3)x+14/3
Is (1,5) also a solution to this equation? Plug x=1 in the equation, we get y=5, so yes, (1,5) is a point on the third altitude.
Answer:
1 x 48 = 48
2 x 24 = 48
3 x 16 = 48
4 x 12 = 48
6 x 8 = 48
8 x 6 = 48
12 x 4 = 48
16 x 3 = 48
24 x 2 = 48
48 x 1 = 48
Step-by-step explanation:
Sorry for the poor graph. The solution to the inequality is
Yaaaaaayyy hurry let’s gooooo!!!!!
I thinks it C, this a difficult question but idk, that my closest guess...
