ABC is a equilateral triangle .
Proof :-
Let's assume both circles as C1 and C2 [ as shown in the figure ]
AC is the radius of circle C1.
BC is the radius of circle C2 .
AB and AC both are radius of circle C1 so both are equal ie AB = AC .
AB and BC both are radius of circle C 2 so both are equal ie AB = BC .
Hence we conclude that .
AB = BC = AC.
So the triangle is equilateral triangle.
Answer:
.
Step-by-step explanation:
Start by separating this decimal number into its integer part and its fraction part:
The most challenging task here is to express as a proper fraction. Once that fraction is found, expressing the original number
will be as simple as rewriting a mixed number as an improper fraction.
Let .
would then represent the original number.
Note that the repeating digits appear in groups of two. Therefore, if the digits in are shifted to the left by two places, the repeating part will continue to match:
.
Note, that this "shifting" is as simple as multiplying the initial number by (same as
raised to the power of the number of digits that needs to be shifted.)
Subtract the original number from the shifted number to eliminate the fraction part completely:
.
In other words:
.
.
Therefore, the original number would be:
.