Answer:
b1 = 2 ; r = 3
Step-by-step explanation:
Given that :
if b3 −b1 = 16 and b5 −b3 = 144.
For a geometric series :
Ist term = a
Second term = ar
3rd term = ar^2
4th term = ar^3
5th term = ar^4 ;...
If b3 - b1 = 16;
ar^2 - a = 16
a(r^2 - 1) = 16 - - - (1)
b5 - b3 = 144
ar^4 - ar^2 = 144
ar^2(r^2 - 1) = 144 - - - - (2)
Divide (1) by (2)
a(r^2 - 1) / ar^2(r^2 - 1) = 16 /144
a / ar^2 = 1 / 9
ar^2 = 9a
Substitute for a in ar^2 - a = 16
9a - a = 16
8a = 16
a = 2
From ar^2 - a = 16
2r^2 - 2 = 16
2r^2 = 16 + 2
2r^2 = 18
r^2 = 18 / 2
r^2 = 9
r = √9
r = 3
Hence ;
a = b1 = 2 ; r = 3
440,000 is rounded to the nearest ten thousandth
Answer:
No, a regular pentagon does not tessellate.
In a tessellation, all the angles at a point have to add to 360 degrees, as this means there is no overlap, nor are there gaps. To find the interior angle sum of a pentagon, we use the following formula:
(n-2)*180 (where n is the number of sides)
We plug in the number of sides (5) and get:
Angle sum = (5–2)*180
Angle sum = 3*180
Angle sum = 540
Regular pentagons have equal sides and equal angles, so to find the size of the interior angle of a pentagon, we divide the angle sum by 5 and get 108 degrees for every angle.
As I said before, the angles at a point need to add up to 360, so we need to know if 108 divides evenly into 360. If it does, the shape tessellates, and, if it doesn’t, the shape does not.
360/108 = 3.33333…
This means that a regular pentagon does not tessellate.
Hope this helps!
Answer:
d amos annabell_aaaa
Step-by-step explanation:
