To find the number represented by 7 tens and 9 ones, you would multiply 7 by 10 (since there are 10 ones in a ten) to get 70, and then add the 9 ones to get the final number. Therefore, 7 tens and 9 ones is equal to 79.
(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.
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
What number has 9 tens and 4 fewer ones than tens
(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
There is no four digit number where the ones is twice the tens, the hundreds is five less than the ones, and the thousands is the sum of the tens and hundreds. int ones, tens, hundreds, thousands; for (thousands=1; thousands<10; thousands++) { /**/ for (hundreds=0; hundreds<10; hundreds++) { /**/ /**/ for (tens=0; tens<10; tens++) { /**/ /**/ /**/ for (ones=0; ones<10; ones++) { /**/ /**/ /**/ /**/ if (ones != 2 * tens) break; /**/ /**/ /**/ /**/ if (hundreds != ones - 5) break; /**/ /**/ /**/ /**/ if (thousands != tens + hundreds) break; /**/ /**/ /**/ /**/ printf ("dd\n", thousands, hundreds, tens, ones); /**/ /**/ /**/ } /**/ /**/ } /**/ } }
To express the number "2 hundreds, 5 ones, 9 tens, and 3 tenths" in standard form, you would write it as 259.3. This is because the number 2 represents the hundreds place, the number 5 represents the ones place, the number 9 represents the tens place, and the number 3 represents the tenths place. Therefore, the number can be written as 259.3.
99
8 tens.