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Q: How many two-digit numbers are there in which the digit on the right is larger than the digit on the left?

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From right to left.

you right the numbers and divide

the right answer is 10 cause there is o,1,2,3,4,5,6,7,8,9 so there 10 numbers in total which have 1 digit

93 and 42

This is impossible to answer ! According to your rules - starting from the right-ham digit... Say the right-hand digit is 1 - the value of the figure to the left of that would be 7, the next value would be 28, and the fourth would be 196 !

To the right of the digit in the unit's column.

Compare one digit at a time, from left to right, until you find a digit that is different. The number with the greater digit in this position is the larger number.

Think about it as moving the decimal point to the right until you come to a digit that is different in the two numbers. In this case it becomes 90 and 92 so 0.92755 is larger because 92 is larger than 90

All decimal numbers are simply a way of representing numbers in such a way that the place value of each digit is ten times that of the digit to its right.

10 times larger

The greatest place value will be whatever non-zero digit is farthest to the left. Look at the digit immediately to the right of it. If that digit is four or less, zero it and everything to the right of it out. If that digit is five or higher, increase the greatest digit by one and zero everything to the right of it out.

To round any number, look at the digit to the right of the place you are rounding to. If the digit is 5 or more, change the digit in the place you are rounding to to the next higher digit. If the digit to the right of the place you are rounding to is less than 5, leave the digit in the place you are rounding to as it is. Change all digits to the right of the place you rounded to to zeros.

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