Nine tens = 90 Ten ones = 10 90 + 10 = 100
(9*10)+(15*1) = 105
The number that has 9 tens and 7 fewer ones than tens is 83. This is because 9 tens equal 90, and if we have 7 fewer ones than tens, we subtract 7 from the tens place, giving us 90 - 7 = 83. Therefore, the number with 9 tens and 7 fewer ones than tens is 83.
4 tens + 9 ones 3 tens + 19 ones 2 tens + 29 ones 1 ten + 39 ones 49 ones.
9 tens = 90. 4 fewer ones than tens = 9 - 4 = 5. Therefore the answer is 95.
Nine tens = 90 Ten ones = 10 90 + 10 = 100
9 tens + 12 ones = 9 × 10 + 12 × 1 = 90 + 12 = 102.
100 = (9x10) + (10x1) = 90 + 10
5 thousands = 5000. 10 tens = 100. 3 hundreds=300. 9 ones = 9. That's 5000 + 400 + 9, which is 5409
9 tens = 90. 4 fewer ones than tens = 9 - 4 = 5. Therefore the answer is 95.
What number has 9 tens and 4 fewer ones than tens
(9 × 10) + 9 = 99 is the number.
(9*10)+(15*1) = 105
The number is 16*10 plus 9*1 = 169
The general function is:1. y = a*x+bb is irrelevant and we can be removed2. y = a*xlets split x into ones and tens3. x = tens*10 + ones /e.g. 23 = 2*10 + 34. p1 = Multiplier of the onesp2 = Multiplier of the tens5. y = tens*10*p2 + ones*p1 /according to the question6. x*a = tens*10*p2 + ones*p1 /according to 2.7. (tens*10 + ones)*a = tens*10*p2 + ones*p1 /according to 3.8. tens*10*a + ones*a = tens*10*p2 + ones*p1 /regroup9. tens*10*a - tens*10*p2 + ones*a - ones*p1 = 0 /regroup10. tens*10*(a-p2) + ones*(a-p1) = 0 /regroup11. assuming "tens" and "ones" are not 0 then (a-p2) and (a-p1) must be 012. a-p2 = 0a-p1 = 013. a = p2a = p114. a = p1 = p2the answer is: when the Multipliers of ones and tens are equal then the product is called a.
(3*10)+(9*1)=39
The number that has 9 tens and 7 fewer ones than tens is 83. This is because 9 tens equal 90, and if we have 7 fewer ones than tens, we subtract 7 from the tens place, giving us 90 - 7 = 83. Therefore, the number with 9 tens and 7 fewer ones than tens is 83.