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>
Answer:
54 students.
Step-by-step explanation:
If you have 600 and 9% the nearest rounding point is %10. %10 of 600 is 60, and 1% of 60 is 6, so 60-6=54.
Pretty sure it’s the first one
Answer:
A and D and E
Step-by-step explanation:
Substitute the variable(in this case) <em>t</em> for 3, then use the order of operations to simplify(PEMDAS), which will give you 33 for both expressions. To see if two expressions are equivalent, you need to substitute the variable for an odd number AND an even number, so next, substitute the variable <em>t </em>for the number 6, which will give you 51 for both expressions. That means the expressions are equivalent
Hope this helps :)
P.S. Mark brainliest pls
Answer:
Step-by-step explanation:20