Answer:
vertex: (1, -49)
x-intercepts: (-6, 0) and (8, 0)
Step-by-step explanation:
I graphed it and looked for the needed information
here's the graph if you want it:
<h2>Answer: A trapezoid with bases of 6 mm and 14 mm and a height of 8 mm </h2>
The parallelogram in the figure has an area of 
, according to the following formula, which works for all rectangles and parallelograms:
(1)
Where 
is the base and 
is the height
The<u> area of a triangle</u> is given by the following formula:
(2)
So, for option A:
Now, the <u>area of a trapezoid </u>is:
(3)
For option B:
For option C:
>>>>This is the correct option!
For option D:
<h2>Therefore the correct option is C</h2>
Answer:the number of adults that attended the game is 135.
the number of children that attended the game is 31
Step-by-step explanation:
Let x represent the number of adults that attended the game.
Let y represent the number of children that attended the game.
There were 166 paid admissions to a game. This means that
x + y = 166
The price was $2 for adults and $.75 for children. The amount of data taken in was $293.25. This means that
2x + 0.75y = 293.25 - - - - - - - - - - 1
Substituting x = 166 - y into equation 1, it becomes
2(166 - y) + 0.75y = 293.25
332 - 2y + 0.75y = 293.25
- 2y + 0.75y = 293.25 - 332
- 1.25y = - 38.75
y = - 38.75/- 1.25
y = 31
x = 166 - y = x = 166 - 31
x = 135
<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>
