Q: What is the numeral for 11 tens and 5 ones?

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6 tens and 5 ones is 65 (which is sixty five).

6 ones and 5 tens

45 and the nearest ten to 45.23 is 45.20

16 tens= 16 * 10= 160 5 ones= 5 * 1= 5 160 + 5= 165

254 - has 2 hundreds, 5 tens, and 4 ones

There are 5 'tens' in the number 55 & 5 'ones'

7 tens and 4 ones 5 thousandths = 74.005

9 tens = 90. 4 fewer ones than tens = 9 - 4 = 5. Therefore the answer is 95.

It has 2 ones and 5 tens

35 tens 35 ones

5, with 9 ones left over.

You can't subtract 15 ones from 15 tens. The bills are all tens, so you'd have to get change first. Then you'd be subtracting 15 ones from 13 tens and 20 ones, or from 13 tens, one 5, and 15 ones.

49

51

105 is.

50

the tens place is on the left for example the # 15 1 is the tens and 5 is the ones

11131.15

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.

2 hundred = 2 × 100 = 200 5 ones = 5 × 1 = 5 9 tens = 9 × 10 = 90 3 tenths = 3 × 0.1 = 0.3 → 2 hundreds 5 ones 9 tens 3 tenths = 200 + 5 + 90 + 0.3 = 295.3

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); /**/ /**/ /**/ } /**/ /**/ } /**/ } }

43 tens and 5 ones.

165

95

Yes

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