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:
diameter = m - c
Step-by-step explanation:
In ΔABC, let ∠C be the right angle. The length of the tangents from point C to the inscribed circle are "r", the radius. Then the lengths of tangents from point A are (b-r), and those from point B have length (a-r).
The sum of the lengths of the tangents from points A and B on side "c" is ...
(b-r) +(a-r) = c
(a+b) -2r = c
Now, the problem statement defines the sum of side lengths as ...
a+b = m
and, of course, the diameter (d) is 2r, so we can rewrite the above equation as ...
m -d = c
m - c = d . . . . add d-c
The diameter of the inscribed circle is the difference between the sum of leg lengths and the hypotenuse.
Rational number between 3 and 5 = 3.5
irrational number between 3 and 5 = (pi) ....3.14159265359.......
Answer:
(30,60)
90mph
Step-by-step explanation:
