what are two ways to skip count to find 2x 5?
is 115 a multiple of 5? i also need to explain how i would know...
This site shows how to skip count.
http://www.mathsisfun.com/numbers/skip-counting.html
You could divide 115 by 5.
Or you could remember that any number that ends in 0 or 5 is divisible by 5.
o ok thank u 4 the site i didn't know they had stuff like that :)
It's amazing what is on the internet. That site is especially useful.
my mom showed me what to do on the site thank u very much ms sue!
You're very welcome, Shy.
To skip count to find 2x 5, you can use two different methods:
1. Addition method:
Start at zero and keep adding 5 repeatedly until you reach the desired number. In this case, you would start with 0, add 5, and continue adding 5 until you have done it twice. So, 0 + 5 + 5 = 10. Therefore, 2x 5 is equal to 10.
2. Multiplication method:
You can also multiply 5 directly by 2 to find the answer. Multiply the number you want to skip count by (in this case, 2) with the number you are skip counting (in this case, 5), so 2 x 5 = 10.
To determine if 115 is a multiple of 5, you need to check if 115 can be evenly divided by 5 without leaving a remainder. There are a couple of ways to determine this:
1. Division method: Divide 115 by 5 and see if there is any remainder. If the division results in an even number, then 115 is a multiple of 5. If there is any remainder, then it is not a multiple.
When you divide 115 by 5, you get 23 with no remainder. So, 115 is divisible by 5.
2. Digit method: One way to determine if a number is divisible by 5 is by checking the last digit. If the last digit of the number ends with 0 or 5, then the number is divisible by 5. In this case, the last digit of 115 is 5, so it is divisible by 5.
Therefore, using either method, we can conclude that 115 is indeed a multiple of 5.