1. Find the H.C.F. of 4x2y3 and 6xy2z.
Solution:
The H.C.F. of numerical coefficients = The H.C.F. of 4 and 6.
Since, 4 = 2 × 2 = 22 and 6 = 2 × 3 = 21 × 31
Therefore, the H.C.F. of 4 and 6 is 2
Answer:
it is C 10
Step-by-step explanation:
If the positions are distinct, as in executive offices, then P(9, 5).
P(9, 5) = 9!/(9 - 5)! = 15120
If the positions are equivalent, such as seats in a legislative body, then C(9, 5).
C(9, 5) = 9!/[(9 - 5)!(5!)] = 126
Assuming the five positions are unique in their duties and responsibilities (i.e. order matters): position 1 has 9 candidates to choose from, position 2 has 8, position 3 has 7, and so on. Otherwise, if you're talking about 5 distinct but duplicate positions - meaning their responsibilities are the same but 5 people are required to carry them out - you need to divide the previous total number of possibilities by the number of ways those possibilities could have been reordered.
Answer:
1. A, because 4(2)=8. 2 is the x.
2. B, because 2.5 x 4 = 10, which makes sense.
3. B, because 5x is the amount that you want to put into equal parts and the 3 is just being subtracted from the amount of the equal parts from the 5x.
4. C.
Step-by-step explanation:
13 and 2, I think that is the only one left.