Being a random number "x"
1 + x²
Answer: x=4 y=14 z=4
Step-by-step explanation:
On edge it is that
Answer:
Step-by-step explanation:
(6e-3f-3/4) contains two terms which do not involve fractions and one fractional term (3/4).
We can safely remove the parentheses. Then:
(6e-3f-3/4) => 6e - 3f - 3/4
That is "an equivalent expression."
We could go further and create one equivalent fraction. Multiply the first two terms by 4/4, obtaining:
24e 12f 3 24e - 12f - 3 3(8e - 4f - 1
------ - ------ - ----- => ---------------------- => -------------------
4 4 4 4 4
Other equivalent expressions exist here.
Answer:
Step-by-step explanation:
One of the rules of logarithms is ...
log(a^b) = b·log(a)
So ...
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.
