Answer:
yes
Step-by-step explanation:
Each day Donna and Mary toss a coin to see who buys the other person coffee ($2.34 a cup). One tosses and the other calls the outcome. If the person who calls the outcome is correct, the other buys the coffee; otherwise the caller pays. Assume that an honest coin is used, that Mary tosses the coin, and that Donna calls the outcome.
Answer:
Step-by-step explanation:
we can write -9 instead of x and y=1 instead of 1
so we write solution again
-9a+1b=-31
-9a-1b=-41
-18a=-72
a=4
we should write 4 instead of a
-9(4)+1b=-31
-36+b=-31
b=5
a=4
Answer: (b)
Step-by-step explanation:
Given
The letter "BHUTAN" has to be arranged
For filing the first place of 6 letter word, we have 6 choices
Similarly for filling the second place of the word, we have 5 choices as 1 letter is already filled.
In this way, the remaining places can be filled in 4, 3, 2, and 1 way
On combining the above, 6 places can be filled in

Thus, option (b) is correct.
Recall Euler's theorem: if
, then

where
is Euler's totient function.
We have
- in fact,
for any
since
and
share no common divisors - as well as
.
Now,

where the
are positive integer coefficients from the binomial expansion. By Euler's theorem,

so that
