Points equidistant from DE EF are in the bisector of angle DEF
points equidistant from EF DF are in the bisector of angle EFD
the sought after point is the intersection of bisectricess of triangle
200,000+ 70,000+ 7,000+ 100+ 80
Answer:
are you sure its b? b doesnt seem to be present in 2x-y=6
The answer is x < 9 because
x-4<5
+ 4 +4
x<9
If
N = a (mod 10)
N = b (mod 13)
gcd(10,13) = 1
then
N = 10 bx + 13 ay (mod 130)
Where
10x + 13y = 1
<span>-> </span>(10x + 13)
(mod 2) = 1 (mod 2)
<span>-> </span>y (mod 2) = 1
y = -3, x = 4
-> N = 40b – 39a
(mod 130)
It is given that ra + sb
should be non-negative:
N = 40b – 39a (mod 130)
N = 40b + (130 – 39)a (mod 130)
N = 40b + 91a (mod 130)
Therefore, N modulo 130, in terms of a and b is: <span>N = 40b + 91a
(mod 130).</span>
