If there are n coins, then the smallest (minimum) amount of coins we need to flip is 1. The reason why is because the best case scenario would be flipping all heads except the last coin which is tails. This scenario is rare if n is large, but still possible.
Answer:
Exactly One Solution
Step-by-step explanation:
4x + 2 = x + 8
3x = 6 Simplify.
x = 2 Divide by 2.
To complete this algebra problem, just like any other, we have to complete the proper steps in order; here as follows.
1. <span><span>−<span>3<span>x^2 </span></span></span>+ <span>20x </span></span>− <span>12
-------------
x - 6
2. </span><span><span>(<span><span>−<span>3x </span></span>+ 2</span>) </span><span>(<span>x − 6</span><span>)
</span></span></span> -----------
x - 6
And finally..
You come up with the answer of:
-3x + 2
I hope this cleared up the little confusion you may have been having. If you need more information on the steps, just ask.
Answer:
hello your question is incomplete below is the missing parts
(a) A\ (A\B) = B\(B\A)
(b) A\ (BA) = B\(A\B)
answer : A\ (A\B) = B\(B\A) = always true
A\ (BA) = B\(A\B) = sometimes true and sometimes false
Step-by-step explanation:
(a) A\ (A\B) = B\(B\A). = ALWAYS TRUE
using de Morgan's law to prove this
A\ (A\B) = A\ ( A ∩ B^c )
= A ∩ ( A^C ∪ B )
= ( A ∩ A^C ) ∪ ( A ∩ B )
= Ф ∪ ( A ∩ B )
= ( A ∩ B )
ALSO : B\(B\A) = attached below is the remaining parts of the solution
B) A\ (BA) = B\(A\B) = Sometimes true and sometimes false
attached below is the prove using De Morgan's law