Answer:
$4.50
Step-by-step explanation:
Please give brainliest!!
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.
At the end of the 6 months the balance would be 360.54
Answer:
-7/4
Step-by-step explanation:
You are looking for the composite g(f(2)). The simplest way to solve this is to evaluate f(2) and enter the solution in to your g function.
g(f(2))=g(-(2)^2-2(2)+4)=g(-4-4+4)=g(-4)
g(-4)=4/(-4(-4)-2)=4/(16-2)=4/14=2/7
Therfor, g(f(2))=2/7 **I'm assuming the -4x-2 is all in the denominator of the g(x) function. If -2 is not in the denominator you would have
g(f(2))=4/(-4(-4)) -2=4/16 -2=1/4 -2=1/4-8/4= -7/4
10 total pieces of the pie
-5
-2
And only 3 remained so your answer would be
3/10
