180°=triangle
360°=quadrangle
180°*34+180=6300°
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
The equation which is equivalent to the equation as given in the task content is; 6x -8y = 36.
<h3>Which equation is equivalent to the equation as given in the task content?</h3>
According to the task content, it follows that the equation which is given in the task content is;
3x -4y = 18.
Hence, upon multiplication of the whole equation by 2; the resulting equation from the multiplication is;
6x -8y = 36
Ultimately, the equivalent equation is; 6x -8y = 36.
Read more on equivalent equations;
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Answer:
Step-by-step explanation:
Answer:
2 lockers have both
Step-by-step explanation:
Here, if we have stickers being placed on the 5th and 7th stickers, how many lockers have both stickers
What this question is simply asking us is to find the places where we have both stickers occurring
We can get this in places where there are multiples of both
We have this at 35th locker , 70th locker
Remember that a quadratic equation is a parabola. The equation is of the type y = Ax^2 + Bx + C
A linear equation is a straight line. The equation is of the type y = MX + N
The soluction of that system is Ax^2 + Bx + C = MX + N
=> Ax^2 + (B-M)x + (C-N) = 0
That is a quadratic equation.
A quadratic equation may have 0, 1 or 2 real solutions. Those are all the possibilitis.
So you must select 0, 1 and 2.
You can also get to that conclusion if you draw a parabola and figure out now many point of it you can intersect with a straight line.
You will realize that depending of the straight line position it can intersect the parabola in none point, or one point or two points.
