The equation of the parabolas given will be found as follows:
a] general form of the parabolas is:
y=k(ax^2+bx+c)
taking to points form the first graph say (2,-2) (3,2), thus
y=k(x-2)(x-3)
y=k(x^2-5x+6)
taking another point (-1,5)
5=k((-1)^2-5(-1)+6)
5=k(1+5+6)
5=12k
k=5/12
thus the equation will be:
y=5/12(x^2-5x+6)
b] Using the vertex form of the quadratic equations:
y=a(x-h)^2+k
where (h,k) is the vertex
from the graph, the vertex is hence: (-2,1)
thus the equation will be:
y=a(x+2)^2+1
taking the point say (0,3) and solving for a
3=a(0+2)^2+1
3=4a+1
a=1/2
hence the equation will be:
y=1/2(x+2)^2+1
Answer:
The airplane would starts its descent from an altitude 500 feet lower.
Step-by-step explanation:
The x-axis is horizontal and the y-axis is vertical (altitude). It would be 500 feet lower because they took 500 away from the 30,000.
Answer:
He will be unable to tell whether a difference in exercise preference is related to a difference in age or to a difference in weight
Step-by-step explanation:
age could be a factor to exercise as well as weight so both affect the outcome so he has two samples
First of all, the modular inverse of n modulo k can only exist if GCD(n, k) = 1.
We have
130 = 2 • 5 • 13
231 = 3 • 7 • 11
so n must be free of 2, 3, 5, 7, 11, and 13, which are the first six primes. It follows that n = 17 must the least integer that satisfies the conditions.
To verify the claim, we try to solve the system of congruences

Use the Euclidean algorithm to express 1 as a linear combination of 130 and 17:
130 = 7 • 17 + 11
17 = 1 • 11 + 6
11 = 1 • 6 + 5
6 = 1 • 5 + 1
⇒ 1 = 23 • 17 - 3 • 130
Then
23 • 17 - 3 • 130 ≡ 23 • 17 ≡ 1 (mod 130)
so that x = 23.
Repeat for 231 and 17:
231 = 13 • 17 + 10
17 = 1 • 10 + 7
10 = 1 • 7 + 3
7 = 2 • 3 + 1
⇒ 1 = 68 • 17 - 5 • 231
Then
68 • 17 - 5 • 231 ≡ = 68 • 17 ≡ 1 (mod 231)
so that y = 68.