Answer:
sorry i dont know
Step-by-step explanation:
but do the ones in the () first always
∠1 and ∠2 are RIGHT ANGLES, VERTICAL ANGLES, and SUPPLEMENTARY ANGLES.
Hope this helps!
Answer:
I am 90% sure that the answer is M'N'O' is 90*
Step-by-step explanation:
A line segment MN is shown as M = -1,1 and N= -5,4 this makes a diagonal line across the grid. Another line segment NO is shown with O as ordered pair negative 1, 4. Angle MNO is rotated 90 degrees counterclockwise about the origin to form angle M′N′O′. this is where the angles meet, Rotating the shape (or the angles) does not change the degree of the angle of the shape. I am 99% sure that an 90* angle would be formed. but im just in middle school so what do i know?
Notice that
So as
you have
. Clearly
must converge.
The second sequence requires a bit more work.
The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then
will converge.
Monotonicity is often easier to establish IMO. You can do so by induction. When
, you have
Assume
, i.e. that
. Then for
, you have
which suggests that for all
, you have
, so the sequence is increasing monotonically.
Next, based on the fact that both
and
, a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.
We have
and so on. We're getting an inkling that the explicit closed form for the sequence may be
, but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.
Clearly,
. Let's assume this is the case for
, i.e. that
. Now for
, we have
and so by induction, it follows that
for all
.
Therefore the second sequence must also converge (to 2).
