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.
Here is how you find the answer.
Do remember that the term Bisect means to cut it in half.
So here it goes.
<span>5x = 3x + 10
5x - 3x = 10
2x = 10
x = 5
Then, substitute the values, so 5*5 or 3*5+10
Then, the answer for each smaller angle is 25.
</span>Remember bisect? so 25 x 2 so the final answer is 50.
Hope this is the answer that you are looking for. Thanks for posting your question!
Answer:
Step-by-step explanation:
First problem:
blank 2) on the left: PR = PQ + QR
blank 3) on the right: Replace PR in statement 1) with what we just completed in statement 2. Therefore:
2PQ = PR
2PQ = PQ + QR
blank 4) on the right: Subtract PQ from both sides, therefore
2PQ = PQ + QR
2PQ - PQ = PQ + QR - PQ
PQ = QR
Jose has 3 more shirts than he has pants.
