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/Step-by-step explanation:
The equation of the line that passes through the two points would be correct if each point, when substituted into the equation, satisfy the equation.
This is what I mean:
Given the equation of the line, y = 2x - 5, and the two points (-2, -9) and (3, 1):
For the first point, substitute x = -2, and y = -9 into y = 2x - 5.
Thus:
-9 = 2(-2) - 5
-9 = -4 - 5
-9 = -9 (this is true). It means the line runs through the point (-2, -9)
For the second point, substitute x = 3, and y = 1 into y = 2x - 5
This:
1 = 2(3) - 5
1 = 6 - 5
1 = 1 (this is true). This also means the point, (3, 1) is also a point that the equation runs across.
Answer:
16Km due east of school P
Step-by-step explanation:
Given
A school P is 16km due west of a school Q
Thus, we can say that distance PQ = 16 km.
________________________________
now we have to find the bearing of Q from P
As distance is same
distance PQ = distance QP
Thus,
Distance will remain same of 16 km.
For direction,
If Q is west of P, then P will be east of Q
as shown in figure P is west of Q,
now from point P , Q is west P.
Thus,
Bearing of School Q from P is 16Km due east of school P
All you have to do is divide the left side of the ratio by 5 and the right side of the ratio by 7. The one that comes out even on both sides is the correct one.
