I believe it would be h/3+2.5=w
(75/3)+6 = X 25 + 6 = X 31 = X
Yes. The remainder cannot be more that the divisor but there is no issue with it being greater than the quotient. For example, if you divide 5 by 3, 5/3 = 1 and remainder 2 (out of 3) So you get quotient = 1, remainder = 2.
10+c
3
probably...................(n+3) -------- 4
25 is.
10+c 3
That depends upon to what you are rounding, but can always be solved by using the following algorithm: 1) The nearest is defined to be the number whose multiple you want the number to be rounded to, eg when rounding to the nearest ten, the nearest is 10; then: 2) Divide 25 by the nearest to get a whole quotient and remainder. 3) If the remainder is half or more of the nearest, then add one to the whole quotient. 4) Multiply the whole number quotient by the nearest to get the answer. eg round 25 to the nearest 5: 1) nearest is 5 2) 25 / 5 = 5 r 0 3) 0 is less than half 5, so quotient stays the same 4) answers is 5 x 5 = 25 → 25 to the nearest 5 is 25 eg round 25 to the nearest 7: 1) nearest is 7 2) 25 / 7 = 3 r 4 3) 4 is more than half 7 → quotient becomes 3+1 = 4 4) answer is 4 x 7 = 28 → 25 to the nearest 7 is 28. eg round 25 to the nearest 10 1) nearest is 10 2) 25/10 = 2 r 5 3) 5 is half 10 → quotient becomes 2 + 1 = 3 4) answer is 3 x 10 = 30 → 25 to the nearest 10 is 30. eg round 25 to the nearest 100 1) nearest is 100 2) 25 / 100 = 0 r 25 3) 25 is less than half 100, so quotient stays the same 4) answer is 0 x 100 = 0 → 25 to the nearest 100 is 0.
(n/-2) + 10 = 3
n/10+8
2 divided by 3 + x.
2.5 more than the quotient of h and 3 is w