I know I’m late but
WOOOOOOOOOOOOOOOOO JESUS!!!
Can’t wait for next year Haha!!!
Y1 is the simplest parabola. Its vertex is at (0,0) and it passes thru (2,4). This is enough info to conclude that y1 = x^2.
y4, the lower red graph, is a bit more of a challenge. We can easily identify its vertex, which is (-4,0), and several points on the grah, such as (2,-3).
Let's try this: assume that the general equation for a parabola is
y-k = a(x-h)^2, where (h,k) is the vertex. Subst. the known values,
-3-(-4) = a(2-0)^2. Then 1 = a(2)^2, or 1 = 4a, or a = 1/4.
The equation of parabola y4 is y+4 = (1/4)x^2
Or you could elim. the fraction and write the eqn as 4y+16=x^2, or
4y = x^2-16, or y = (1/4)x - 4. Take your pick! Hope this helps you find "a" for the other parabolas.
It could be? the 7 could be 7 to literally any number soooo
Answer:
- (4x -4)° +x° +(6x-3)° = 180°
- J = 99°
- K = 64°
- L = 17°
Step-by-step explanation:
The relation that helps you write an equation for x is, "the sum of angles in a triangle is 180°."
__
<h3>equation</h3>
(4x -4)° +x° +(6x -3)° = 180° . . . . . sum of angles in this triangle
<h3>solution for x</h3>
11x -7 = 180 . . . . . . . divide by °, collect terms
11x = 187 . . . . . . . . add 7
x = 17 . . . . . . . . . divide by 11
<h3>angle values</h3>
m∠J = (6x -3)° = (6(17) -3)° = 99°
m∠K = (4x -4)° = (4(17) -4)° = 64°
m∠L = x° = 17°
