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What is 18 fives tens?

10


How many tens are in eighteen fives?

9


How much does 100000 dollars weigh?

it all depends how do you want type of bills you use like ones, fives, tens and so on


Is two fives greater than five tens?

5/10 is greater


Is rolling the dice 10 times and counting the number of fives binomial?

Yes.


What might you be counting by?

You might be counting by pairs, threes, fours, or even fives. Sometimes, the numbers of the crowds of people are estimated by the busloads.


How many twenties in 56 fives?

Oh, dude, let me grab my calculator... just kidding. So, there are 11 twenties in 56 fives. It's like basic math, but hey, who's counting, right?


How many pennies nickels dimes pennies tens fives and ones do I have to make 200?

If you have 20 tens you have 200. You don't need the rest.


How many fives have the same value as the sum of 2 tens and 10 twos?

To solve this problem, we first calculate the sum of 2 tens and 10 twos: 2 tens = 20, 10 twos = 20, so the sum is 20 + 20 = 40. Next, we determine how many fives are equal to 40 by dividing 40 by 5, which equals 8. Therefore, there are 8 fives that have the same value as the sum of 2 tens and 10 twos.


What is the relationship between the values of the two fives in 55?

Counting from the left, the first is ten times the second.


How much money is 3 tens and 3 fives Maths?

Well, isn't that a happy little question! If you have 3 tens, that's 30, and if you add 3 fives, that's another 15. So in total, you have 45. Just imagine all the wonderful things you could buy with that!


When would you use skip-counting to find the number in an array?

Skip-counting can be useful when you want to quickly determine the total number of items in an array that are evenly spaced or follow a specific pattern, such as counting by twos, fives, or tens. For example, if you have an array of even numbers or multiples of a certain number, skip-counting allows you to efficiently calculate the total by counting only the significant intervals rather than each individual item. This method enhances efficiency and can simplify calculations in various mathematical contexts.