380
There are twenty numbers between 19 through 39, including 19 and 39. So the problem is 20 times 20 times 20 times 20 times 20. 320,0000
The expression equivalent to (19 \times 8) can be rewritten using the distributive property. For example, it can be expressed as ((20 - 1) \times 8), which simplifies to (20 \times 8 - 1 \times 8), resulting in (160 - 8 = 152). Thus, (19 \times 8) is equivalent to (152).
54 goes into 1099 20 times with remainder 19.
Deducting 5 from 5 times 20 gives 100 - 5 ie 95.
495 ÷ 25 = 19 with remainder 20.
There are twenty numbers between 19 through 39, including 19 and 39. So the problem is 20 times 20 times 20 times 20 times 20. 320,0000
359 ÷ 20 = 17.95 or 17 times with a remainder of 19.
19/20 x 5 = 95/20 or 4 and 15/20 or 4 and 3/4 19/20 = 0.95 x 5 = 4.75
20 times 19 iz 380
19 with remainder 19.
(133/160) / (19/20) First. edit the expression above to be multiplication. You can do this by finding the reciprocal of 19/20 and multiplying it by 133/160. so 133/160 * 20/19 19 goes into 133 seven times, and 20 goes into 160 eight times. This leaves us with a 7 in the numerator and an 8 in the denominator. so 7/8
The expression equivalent to (19 \times 8) can be rewritten using the distributive property. For example, it can be expressed as ((20 - 1) \times 8), which simplifies to (20 \times 8 - 1 \times 8), resulting in (160 - 8 = 152). Thus, (19 \times 8) is equivalent to (152).
9 times but there is a remainder.
54 goes into 1099 20 times with remainder 19.
It is: 19/20 times 100 = 95%
6y+1=19 6y=20 y=20/6=10/3~=3.333
Deducting 5 from 5 times 20 gives 100 - 5 ie 95.