How many four-letter codes can be made if no letter can be used twice?
2 answers:
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Answer:
The number is 12
Step-by-step explanation:
[] First, let's turn all these words into something mathematical. N will equal "my number"
-><u> If you add 12 to my number</u> and then multiply the result by 3, you will get 64 more than two-thirds of my number.
-> n + 12 <u>and then multiply the result by 3</u>, you will get 64 more than two-thirds of my number.
-> 3(n + 12), you will get 64 more than <u>two-thirds of my number</u>.
-> 3(n + 12) = <u>64 more than</u>
-> 3(n + 12) = 64 +
[] Phew, okay. Now it is something we can solve and less scary;
[Given]
3(n + 12) = 64 +
[Distribute]
3n + 36 = 64 +
[Multiply both sides by 3]
9n + 108 = 192 + 2n
[Subtract 108 and 2n from both sides]
7n = 84
[Divide both sides by 4]
n = 12
Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)
- Heather