7p - 4 + 12p = -3(5 + p)
7p - 4 + 12p = -15 - 3p
+ 4 + 4
7p + 12p = -11 - 3p
+ 3p + 3p
22p = -11
p = -1/2 Answer
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.
Answer:
36
Step-by-step explanation:
positive version
5. The graph of g(x) is narrower. Both graphs open upward. The vertex of g(x), (0,10), is translated 10 units up from the vertex of f(x) at (0,0)
6. The graph of g(x) is wider. Both graphs open upward. The vertex of g(x), (0,-3) is translated 3 units down from the vertex of f(x) at (0,0)
7.The graph of g(x) is narrower. g(x) opens downward and f(x) opens upward. The vertex of g(x), (0,8) is translated 8 units up from the vertex of f(x) at (0,0).
8. The graph of g(x) is wider. g(x) opens downward and f(x) opens upward. The vertex of g(x), (0,1/4) is translated 1/4 units up from the vertex of f(x) at (0,0).
9. A. h1(t)=-16t^2+400 h2(t)= -16t^2+1600
9. B The graph of h2 is a vertical translation of the graph of h1 : 1200 units up.
9. C sandbag dropped from 400 ft: 5 s
sandbag dropped from 1600 ft: 10 s
Basically get two slopes -50(1)+100 will get you 1,50 (1 is x and 50 is y since its the answer)
-50(0)+100 (0,100) Y₂-Y₁/X₂-X₁ 50-100/1-0
Rate of change per month = -$50
