The points L(10,9)L(10,9), M(10,-5)M(10,-5), N(-1,-5)N(-1,-5), and O(-1,9)O(-1,9) form rectangle LMNOLMNO. Which point is halfwa
Inessa [10]
You are trying to find the halfway point between OO and NN.
OO: (-1,9) NN: (-1,5)
The x-coordinate does not change, because in both instances it is -1. The y-coordinate is (9-5)/2 AWAY from each point. AKA the number that is equidistant from 5 and 9 (7).
The area of a trapezoid is basically the average width times the altitude, or as a formula:
Area = h ·
b 1 + b 2
2
where
b1, b2 are the lengths of each base
h is the altitude (height)
Recall that the bases are the two parallel sides of the trapezoid. The altitude (or height) of a trapezoid is the perpendicular distance between the two bases.
In the applet above, click on "freeze dimensions". As you drag any vertex, you will see that the trapezoid redraws itself keeping the height and bases constant. Notice how the area does not change in the displayed formula. The area depends only on the height and base lengths, so as you can see, there are many trapezoids with a given set of dimensions which all have the same area.
F(x) = x^2 + 24x + 138
f(x) = x^2 + 24x + (24/2)^2 + 138 - (24/2)^2
f(x) = (x+12)^2 - 6
Answer:
Overall answer: 10 1/9 - 1 5/18x (Most simplified answer)
Step-by-step explanation:
First you have to do all the Distributive property which the answer would be:
2 1/2 - 1/2x + 3 1/9 + 8/9x + 1/6x - 5/18x + 2 14/18 + 1 13/18
Then you have to find the common demoninator which would be 18
After that you have to put them in the correct order since there is two different terms which would be:
2 9/18 + 3 2/18 + 2 14/18 + 1 13/18 - 9/18x + 16/18x + 3/18x - 5/18x
After that you solve the problem which in the end after solving both terms it ends up as..:
10 2/18 - 1 5/18x
and I think that is as simplified as you can go term wise.
I hope this helps and I hope I didn't make a mistake :D
<h3>Answer:</h3>
y = -3(x -2)² -5
<h3>Explanation:</h3>
The equation of a parabola with vertex (h, k) is given by ...
... y = a(x -h)² +k . . . . . . . for some vertical scale factor "a".
We can find the value of "a" by substituting the coordinates of a point that is not the vertex. Here, that point is the y-intercept: (0, -17).
... y = a(x -2)² -5 . . . . . initial form of the equation for the parabola
... -17 = a(-2)² -5 . . . . . substitute y-intercept values
... -12 = 4a . . . . . . . . . . add 5, simplify
... -12/4 = a = -3
The desired equation is ... y = -3(x -2)² -5.
