Triangles have 3 vertices. The total is 21 vertices. 21 / 3 = 7. He drew 7 triangles.
Answer:
x = 3, −10
Step-by-step explanation:
The roots (zeros) are the x values where the graph intersects the x-axis. To find the roots (zeros), replace y with 0 and solve for x.
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.
Answer:
21 + 3a
Step-by-step explanation:
3 (7+a)
distribute 3 into parentheses:
21 + 3a
What are you trying to do here?
Solve the graph, or make it appear as something else?
First, we're going to take one sec (x) out so that we get:
sec (x) (2sec (x) -1 -1) = 0
sec (x) (2sec (x) -2) = 0
Then we're going to separate the two to find the zeros of each because anything time 0 is zero.
sec(x) = 0
2sec (x) - 2 = 0
Now, let's simplify the second one as the first one is already.
Add 2 to both sides:
2sec (x) = 2
Divide by 3 on both sides:
sec (x) = 1
I forgot my unit circle, so you'd have to do that by yourself. Hopefully, I helped a bit though!
