Answer:
Tn = 2Tn-1 - Tn-2
Step-by-step explanation:
Before we can generate the recursive sequence, we need to find the nth term of the given sequence.
nth term of an AP is given as:
Tn = a+(n-1)d
If a17 = -40
T17 = a+(17-1)d = -40
a+16d = -40 ...(1)
If a28 = -73
T28 = a+(28-1)d = -73
a+27d = -73 ...(2)
Solving both equations simultaneously using elimination method.
Subtracting 1 from 2 we have:
27d - 16d = -73-(-40)
11d = -73+40
11d = -33
d = -3
Substituting d = -3 into 1
a+16(-3) = -40
a - 48 = -40
a = -40+48
a = 8
Given a = 8, d = -3, the nth term of the sequence will be
Tn = 8+(n-1) (-3)
Tn = 8+(-3n+3)
Tn = 8-3n+3
Tn = 11-3n
Given Tn = 11-3n and d = -3
Tn-1 = Tn - d... (3)
Tn-1 = 11-3n +3
Tn-1 = 14-3n
Tn-2 = Tn-2d...(4)
Tn-2 = 11-3n-2(-3)
Tn-2 = 11-3n+6
Tn-2 = 17-3n
From 3, d = Tn - Tn-1
From 4, d = (Tn - Tn-2)/2
Equating both common difference
(Tn - Tn-2)/2 = Tn - Tn-1
Tn - Tn-2 = 2(Tn - Tn-1)
Tn - Tn-2 = 2Tn-2Tn-1
2Tn-Tn = 2Tn-1 - Tn-2
Tn = 2Tn-1 - Tn-2
The recursive formula will be
Tn = 2Tn-1 - Tn-2
Answer:
see below
Step-by-step explanation:
<h3>Given</h3>
- Distance is 142.2 m, correct to 1 decimal place
- Time is 7 seconds, correct to nearest second
<h3>To find:</h3>
- Upper bound for the speed
<h3>Solution </h3>
<em>Upper bound for the speed = upper bound for distance/lower bound for time</em>
- Upper bound for distance = 142.25 m (added 0.1/5 = 0.05)
- Lower bound for time = 6.5 seconds (subtracted 1/2 = 0.5)
<u>Then, the speed is:</u>
- 142.25/6.5 = 21.88 m/s
- 21.88 = 21.9 m/s correct to 1 decimal place
- 21.88 = 22 m/s correct to nearest m/s
Answer: Both A and E have a slope of 5/2
Step-by-step explanation:
Answer:
since output value is paried with input value teh relationship IS a function
Step-by-step explanation:
