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.
1 :D
2:a
3:c
I hope it hellp
Answer:
hai
Step-by-step explanation:
Pak Awi mau beli jam tangan Rolex hulk submarine dengan harga 300jt. Dia mau beli 2 tahun lagi
dengan asumsi harga jam naik 2% / tahun. Future Value
Uang pak awi sekarang hanya 100jt, berapa yang harus pak awi tabung tiap bulan nya agar 2 tahun lagi
dia bisa beli jam tangan tersebut, jika bunga tahunan adalah 15% / tahun (pajak = 10%) ??? Present
value. Future value. Present value annuity.
Bunga di terima per bulan, dan tetap di gulung.
Ya Berapa yang dia tabung tiap bulan
-pakai tabel
The area of an equilateral triangle of side "s" is s^2*sqrt(3)/4. So the volume of the slices in your problem is
(x - x^2)^2 * sqrt(3)/4.
Integrating from x = 0 to x = 1, we have
[(1/3)x^3 - (1/2)x^4 + (1/5)x^5]*sqrt(3)/4
= (1/30)*sqrt(3)/4 = sqrt(3)/120 = about 0.0144.
Since this seems quite small, it makes sense to ask what the base area might be...integral from 0 to 1 of (x - x^2) dx = (1/2) - (1/3) = 1/6. Yes, OK, the max height of the triangles occurs where x - x^2 = 1/4, and most of the triangles are quite a bit shorter...
Answer:
1) the signs of the numbers is not in standard form
2) always make sure that the signs of the two equations is standarised and then solve
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