Sorry but, there’s nothing attached so I can not help you.
No because in 1/3 you get 2/6 so if you have 1/3 vs. 5/6 than you only have 1/3 containing 2/6 and we are looking for 5/6 so that is wrong. 5/6 is bigger that 1/3.
2/3 time 5
5×2 is 10 so the numerator is ten
1×3 is 3 so the denominator is three
the answer is 10/3
Answer:
27 Palindromes are possible in the given situation.
Step-by-step explanation:
As a palindrome is a number that reads the same forward and backward and in our case, it is a 5 digit palindrome.
Let's suppose the 5 digit palindrome is ABCBA:
As we know that the digits on 4th and 5th positions will be the same as the digits on 1st and 2nd positions.
- The number of possibilities that a digit can come on 1st position are 3
- The The number of possibilities that a digit can come on 2nd position are 3
- The number of possibilities that a digit can come on 3rd position are 3
So the total number of possibilities that we can get 5 digit palindromes from digits 1, 2 and 3 are:
= 3 x 3 x 3
Total Palindromes = 27
Answer:
Step-by-step explanation:
2x3−34x2+131x+11
2x3−34x2+131x+11
=(x−11)(2x2−12x−1)
Answer:
(x−11)(2x2−12x−1)
