Answer:
I believe that they are Alternate Exterior angles
Step-by-step explanation:
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.
Hey! Just in case you have to show your work.
89(54x-36)+2=-3/4(-40+16x)+90x
4806x-3204+2=-3/4×4(-10+4x)+90x
4806x-3202=30-12x+90x
4806x-78x=30+3202
4728x=3232
3232÷4728=
X=404/591
The area of a circle with radius of 13 millimeters is A≈530.93A=πr2=π·132≈530.92916
