Answer:
- digits used once: 12
- repeated digits: 128
Step-by-step explanation:
In order for a number to be divisible by 4, its last two digits must be divisible by 4. This will be the case if either of these conditions holds:
- the ones digit is an even multiple of 2, and the tens digit is even
- the ones digit is an odd multiple of 2, and the tens digit is odd.
We must count the ways these conditions can be met with the given digits.
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Since we only have even numbers to work with, the ones digit must be an even multiple of 2: 4 or 8. (The tens digit cannot be odd.) The digits 4 and 8 comprise half of the available digits, so half of all possible numbers made from these digits will be divisible by 4.
<h3>digits used once</h3>
If the numbers must use each digit exactly once, there will be 4! = 24 of them. 24/2 = 12 of these 4-digit numbers will be divisible by 4.
<h3>repeated digits</h3>
Each of the four digits can have any of four values, so there will be 4^4 = 256 possible 4-digit numbers. Of these, 256/2 = 128 will be divisible by 4.
y = ( 3 * x ) - 2.
This is because the card costs for 3 $ and should be multiplied by the number of total cards she buys in that particular day. So 3 * x.
Now as she has got a 2$ off coupon, whatever the total is she gets 2 dollars off on that price. So we will subtract 2 at that time. FInally all this summed up would be equal to y.
Hey there,
The answer is: 18a^3b2/2ab=9a^2*b
:)
Answer:
![n^{3} + 15n^{2} + 56n](https://tex.z-dn.net/?f=n%5E%7B3%7D%20%2B%2015n%5E%7B2%7D%20%2B%2056n)
Step-by-step explanation:
First FOIL (n + 7)(n + 8) and then distribute n
(n + 7)(n + 8) = ![n^{2} +15n + 56](https://tex.z-dn.net/?f=n%5E%7B2%7D%20%2B15n%20%2B%2056)
n (
)
= ![n^{3} + 15n^{2} + 56n](https://tex.z-dn.net/?f=n%5E%7B3%7D%20%2B%2015n%5E%7B2%7D%20%2B%2056n)
Answer:
The answer is 5184. Hope it helps luv