Answer:8 I think, sorry it it is wrong
Step-by-step explanation:
3x + 4y = 16 Write original equation
3x + 4y - 4y = -4y + 16 Subtract 4y from each side
3x = -4y +16 Simplify
3x/3 = -4y/3 + 16/3 Divide each side by three
x = -4y/3 +16/3 Simplify
I hope this helps!
Perimeter = a + b + c = 30
Area = 1/2 x a x b = 30
Multiples of 30: 2, 3, 5, 6, 10, 12, 15
For perimeter c = 30- (a+b)
C= sqrt( a^2 + b^2)
Using the possible combinations of the above:
5 and 12:
C = sqrt(5^2 + 12^2) = 13
5 + 12 + 13 = 30 for the perimeter
Area = 1/2 x 5 x 12 = 30
The sides are 5, 12 and 13 cm
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
