Answer:
see explanation
Step-by-step explanation:
The n th term ( explicit formula ) of an arithmetic sequence is
= a₁ + (n - 1)d
where a₁ is the first term and d the common difference
Given a₁₂ = - 95 and a₃₇ = - 270 , then
a₁ + 11d = - 95 → (1)
a₁ + 36d = - 270 → (2)
Subtract (1) from (2) term by term to eliminate a₁
25d = - 175 ( divide both sides by 25 )
d = - 7
Substitute d = - 7 into (1) and solve for a₁
a₁ + 11(- 7) = - 95
a₁ - 77 = - 95 ( add 77 to both sides )
a₁ = - 18 , thus
= - 18 - 7(n - 1) = - 18 - 7n + 7 = - 7n - 11
= - 7n - 11 ← explicit formula
--------------------------------------------------------------
The recursive formula allows a term in the sequence to be found by adding the common difference d to the previous term, thus
= 
- 7 with a₁ = - 18 ← recursive formula
Or mine means that itself and 1 can only go into it. Conposit means that multiple numbers can go into it like 24. 12,6,8,2 can all go into it
2/6, 3/9, 4/12, 5/15, 6/18 ect.
Hope this helped!
Answer:
10 and 15
Step-by-step explanation:
Let 'x' and 'y' are the numbers we need to find.
x + y = 25 (two numbers whose sum is 25)
(1/x) + (1/y) = 1/6 (the sum of whose reciprocals is 1/6)
The solutions of the this system of equations are the numbers we need to find.
x = 25 - y
1/(25 - y) + 1/y = 1/6 multiply both sides by 6(25-y)y
6y + 6(25-y) = (25-y)y
6y + 150 - 6y = 25y - (y^2)
y^2 - 25y + 150 = 0 quadratic equation has 2 solutions
y1 = 15
y2 = 10
Thus we have
:
First solution: for y = 15, x = 25 - 15 = 10
Second solution: for y = 10, x = 25 - 10 = 15
The first and the second solution are in fact the same one solution we are looking for: the two numbers are 10 and 15 (since the combination 10 and 15 is the same as 15 and 10).
7.62 a^2+b^2=c^2 have a great day
