Answer:
Step-by-step explanation:
we have:
we also have:
from (1)(2) => proven
You would assume that in this figure, the number of colored sections with which are not colored with respect to a " touching " colored section, would be half of the total colored sections. However that is not the case, the sections are not alternating as they still meet at a common point. After all, it notes no two touching sections, not adjacent sections. Their is no equation to calculate this requirement with respect to the total number of sections.
Let's say that we take one triangle as the starting. This triangle will be the start of a chain of other triangles that have no two touching sections, specifically 7 triangles. If a square were to be this starting shape, there are 5 shapes that have no touching sections, 3 being a square, the other two triangles. This is presumably a lower value as a square occupies two times as much space, but it also depends on the positioning. Therefore, the least number of colored sections you can color in the sections meeting the given requirement, is 5 sections for this first figure.
Respectively the solution for this second figure is 5 sections as well.
The exact answer would be <span>392.78
An estimated value could be either 390 or 393</span>
Ooh, fun
geometric sequences can be represented as
so the first 3 terms are
the sum is -7/10
and their product is -1/125
from the 2nd equation we can take the cube root of both sides to get
note that a=ar/r and ar²=(ar)r
so now rewrite 1st equation as
subsituting -1/5 for ar
which simplifies to
multiply both sides by 10r
-7r=-2-2r-2r²
add (2r²+2r+2) to both sides
2r²-5r+2=0
solve using quadratic formula
for
so
for 2r²-5r+2=0
a=2
b=-5
c=2
so
or
use them to solve for the value of a
try for r=2 and 1/2
or
test each
for a=-1/10 and r=2
a+ar+ar²=
it works
for a=-2/5 and r=1/2
a+ar+ar²=
it works
both have the same terms but one is simplified
the 3 numbers are
,
, and
Answer:
2.8828283e+16
Step-by-step explanation: