T(1) = 3, t(n) = -2t(n-1) + 1
So t(2) = -2(t(1)) + 1 = -2(3) + 1 = -5
t(3) = -2(t(2)) +1 = -2(-5) +1 = 11
t(4) = -2(t(3)) +1 = -2(11) +1 = -21
t(5) = -2(t(4)) +1 = -2(-21) +1 = 43
I've seen this question on Brainly before, and I always shake my head.
Please think about this for a few seconds. Maybe even make some
scribbles on a piece of paper.
-- A triangle has 3 sides and 3 angles.
-- A square, rectangle, rhombus or parallelogram has 4 sides and 4 angles.
-- Draw anything with 5 sides. It doesn't have to be pretty, and they don't
all have to be the same length or anything special. Just draw any shape
with 5 sides. Count the angles, and you'll find that there are five of them.
By now you should be starting to get the creepy hunch that maybe a
polygon always has the SAME number of sides and angles. I hope so.
That's the correct creepy hunch.
You can get all kinds of hunches, and even work most of them out,
just by using your thinker for a while.
Do you mean the phythagorus theorum: a squared + b squared= c squared
Answer:
No real root.
Complex roots:
Step-by-step explanation:
There are no two integers whose product is 5 and whose sum is 2, so this trinomial is not factorable. We can use the quadratic formula.
Since we have a square root of a negative number, there are no real roots. If you have learned complex numbers, then we can continue.