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.
x=4 y=5
if we put two pharases together
the answer is as follows pic :
x=4 y=5
Answer:
espero sirva qwq
Step-by-step explanation:
4= 0.56666666666
5=1.66666666667
6=1.375
7=1.425
8=0.92857142857
9=1.666666667
10=1.44444444444
11=0.825
Answer:
x = -1
Step-by-step explanation:
8-2x=-8x+14
8-14=-8x+2x
-6/-6=x
x=-1
