Answer:
b.) 4.2
Step-by-step explanation:
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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.
Ok so <span>Interest = Principle * rate * time
P = 10,000
R = 3% = 0.03
T (we need to solve for t
I = 3600
now we sub..
3600 = (10,000)(0.03)(t)
3600 = 300t -- divide both sides by 300
3600/300 = t
12 = t
check..
I = prt
3600 = (10000)(0.03)(12)
3600 = (300)(12)
3600 = 3600 (correct)
It would take 12 years</span>
Found this on google to help you out but it states “ Find the first derivative of a function f(x) and find the critical numbers. Then, find the second derivative of a function f(x) and put the critical numbers. If the value is negative, the function has relative maxima at that point, if the value is positive, the function has relative maxima at that point.”
