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.
Answer:
D
Step-by-step explanation:
Using the Cosine rule to find AC
AC² = BC² + AB² - (2 × BC × AB × cosB )
= 18² + 12² - ( 2 × 18 × 12 × cos75° )
= 324 + 144 - 432cos75°
= 468 - 111.8
= 356.2 ( take the square root of both sides )
AC = 
≈ 18.9
-----------------------------------------
Using the Sine rule to find ∠ A
= 
( cross- multiply )
18.9 sinA = 18 sin75° ( divide both sides by 18.9 )
sinA = 
, then
∠ A = 
( ) ≈ 66.9°
Y - 190 = 55(x - 2)....distribute thru the parenthesis
y - 190 = 55x - 110....now add 190 to both sides
y = 55x - 110 + 190..simplify
y = 55x + 80 <== slope intercept form (y = mx + b)
