QUESTION:
The code for a lock consists of 5 digits (0-9). The last number cannot be 0 or 1. How many different codes are possible.
ANSWER:
Since in this particular scenario, the order of the numbers matter, we can use the Permutation Formula:–
- P(n,r) = n!/(n−r)! where n is the number of numbers in the set and r is the subset.
Since there are 10 digits to choose from, we can assume that n = 10.
Similarly, since there are 5 numbers that need to be chosen out of the ten, we can assume that r = 5.
Now, plug these values into the formula and solve:
= 10!(10−5)!
= 10!5!
= 10⋅9⋅8⋅7⋅6
= 30240.
For the 1st one we need to solve x. x-y=2 step1: x-y+y=2+y answer: x=y+2
<span>2(3)× = 3×+1 is equal when f(x) = g(x).
f(x) is equal to g(x) when x = 0.
Therefore, the solution to the equation </span><span>2(3)×=3×+1 is x = 0.</span>
Answer:
{5, 7, 9, 11}
Step-by-step explanation:
Roster form just means list the elements of the set in the brackets.
It says the elements are greater than or equal to five and less than or equal to 11. So we start at 5 and end at 11.
We list {5, 7, 9, 11} because these elements follow the rule that they must be odd natural numbers. Natural numbers are the numbers you can count starting from 0 like {0, 1, 2, 3, ...}.
I think it’s B (0, 0) (5, -3), (4, 1).
