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.
C because if it was constant then for day 3 the distance would be 65.4 instead of 64.8
Answer:
The percentage change from July to November is 67.27 %
Step-by-step explanation:
Given as :
The number of tourists at the beach per weekend in the month of July = 
= 55,000
The number of tourists at the beach per weekend in the month of November = 
= 18,000
Let the percentage change from July to November = A %
Or, % decrease change = 
× 100
So , A % = ![\dfrac{\tetxtrm [tex]x_2](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctetxtrm%20%5Btex%5Dx_2)
- \textrm }{\textrm }[/tex] × 100
or, A % = 
× 100
Or, A % = 
× 100
Or, A = 67.27 %
So percentage change between two months = 67.27 %
Hence The percentage change from July to November is 67.27 % Answer
Answer:
2
Step-by-step explanation:
68+48=116
116÷16=7.24 (wrong)
116÷12=9.7 (wrong)
116÷4=29 (lowest even)
116÷2=58 (highest even)
If I'm wrong, let me know
Equation 1 ==> y - x = -13
Equation 2 ==> -4x + 3y = -51
3(y - x) = 3(-13)
Equation 3 ==> 3y - 3x = -39
Equation 2 - 3
= (3y - 3y) + ( -4x - (-3x) ) = -51 - (-39)
-x = -12
x = 12
Substitude x into equation 1
y - 12 = -13
y = -1
