Y = mx + b
m is your slope AKA rate of change
b is the y intercept.
from the graph we can see that the y intercept (b) is 15. (look at the y axis and see what point the line crosses on the y, here it is 15, in-between 10 and 20)
slope is rate of change. to figure this out remember slope = rise over run:
so pick two points. how much does one point rise after the other point you picked? In the graph it rises 1 square so rise = 1
Now looking at the same two points, how far away from each other are they on the x axis? Looking at the graph they are 2 squares away from each other. so run = 2
Now plug these numbers into the rise over run formula:
Now we have everything we need to put into our formula for a line:
For number 16 you should get:
after you do the same steps again yourself
Choose the one on the very top.
<span><span><span>1. An altitude of a triangle is a line segment from a vertex perpendicular to the opposite side. Find the equations of the altitudes of the triangle with vertices (4, 5),(-4, 1) and (2, -5). Do this by solving a system of two of two of the altitude equations and showing that the intersection point also belongs to the third line. </span>
(Scroll Down for Answer!)</span><span>Answer by </span>jim_thompson5910(34047) (Show Source):You can put this solution on YOUR website!
<span>If we plot the points and connect them, we get this triangle:
Let point
A=(xA,yA)
B=(xB,yB)
C=(xC,yC)
-------------------------------
Let's find the equation of the segment AB
Start with the general formula
Plug in the given points
Simplify and combine like terms
So the equation of the line through AB is
-------------------------------
Let's find the equation of the segment BC
Start with the general formula
Plug in the given points
Simplify and combine like terms
So the equation of the line through BC is
-------------------------------
Let's find the equation of the segment CA
Start with the general formula
Plug in the given points
Simplify and combine like terms
So the equation of the line through CA is
So we have these equations of the lines that make up the triangle
So to find the equation of the line that is perpendicular to that goes through the point C(2,-5), simply negate and invert the slope to get
Now plug the slope and the point (2,-5) into
Solve for y and simplify
So the altitude for vertex C is
Now to find the equation of the line that is perpendicular to that goes through the point A(4,5), simply negate and invert the slope to get
Now plug the slope and the point (2,-5) into
Solve for y and simplify
So the altitude for vertex A is
Now to find the equation of the line that is perpendicular to that goes through the point B(-4,1), simply negate and invert the slope to get
Now plug the slope and the point (-4,1) into
Solve for y and simplify
So the altitude for vertex B is
------------------------------------------------------------
Now let's solve the system
Plug in into the first equation
Add 2x to both sides and subtract 2 from both sides
Divide both sides by 3 to isolate x
Now plug this into
So the orthocenter is (-2/3,1/3)
So if we plug in into the third equation , we get
So the orthocenter lies on the third altitude
</span><span>
</span></span>
If you take the original amount 45 and subtract the admission fee 17 you get what is left over which would be 28 dollars. Good Luck and I hope that this helps!!
