Answer:
1. 15
2. 8
Step-by-step explanation:
The two sequence are geometric progression GP, because they follow a constant multiple (common ratio)
The nth term of a GP is;
Tn = ar^(n-1)
Where;
a = first term
r = common ratio
For the first sequence;
The common ratio r is
r = T3/T2 = 540/90 = 6
r = 6
T2 = ar^(2-1) = ar
T2 = 90 = ar
Substituting the values of r;
90 = a × 6
a = 90/6
a = 15
First term = 15
2. The sam method applies here.
Common ratio r = T3/T2 = 128/32 = 4
r = 4
T2 = ar^(2-1) = ar
T2 = 32 = ar
Substituting the values of r;
32 = a × 4
a = 32/4
a = 8
First term = 8
Answer:
X= 8
Y= 130
(8, 130)
Step-by-step explanation:
I used a graphing calculator to see where the two equations intersected.
Hope this helped :)
(6y - 11)(6y + 11) = ay
² - b |use (a - b)(a + b) = a² - b²
(6y)² - 11² =ay² - b
36y² - 121 = ay² - b |add b to both sides
ay² = 36y² - 121 + b |divide both sides by y² ≠ 0
a = (36y² - 121 + b)/y²