6h 8min
3h + (2h 20m) + 35m = (3+2)h + (20+35)m = 5h + 55m
It is [(2a+2h+5) - (2a+5)]/h = 2h/h = 2
Yes, provided p = perimeter, b= breadth and h = height.
In the reaction (2H^+ + SO_4^{2-} + Ca^{2+} + 2I^- \rightarrow CaSO_4 + 2H^+ + 2I^-), the spectator ions are those that do not change during the reaction. Here, the ( H^+ ) ions and ( I^- ) ions are present on both sides of the equation and do not participate in the formation of the precipitate ( CaSO_4 ). Therefore, the spectator ions are ( H^+ ) and ( I^- ).
It is: 11h
3h + (2h 20m) + 35m = (3+2)h + (20+35)m = 5h + 55m
2h + 2h + 2h = 6h
(t+h)(x+2)
3h-5h + 11 = 17 is ------2h + 11 = 17- 11 -11____________-2h = 6___ ___-2 -2h = -3 (This is the answer.)
It is [(2a+2h+5) - (2a+5)]/h = 2h/h = 2
2Si +2HF-2siF + 2H
2H+ + SO42- + Ca2+ + 2I- CaSO4 + 2H+ + 2I-
h = 0.538462 14 + 5h + 2h = 5h + 28h 14 + 7h = 33h 14 + 7h - 7h = 33h - 7h 14 = 26h 14/26 = 26/26h 0..538462 = h
No because it equals zero.
Yes, provided p = perimeter, b= breadth and h = height.
- 3h - 2h + 6h + 9 = h + 9
First set up the equation: 2h+10=40 Then get the variable alone by subtracting 10 from each side: 2h=30 And continue by dividing both sides by two 2h=30 ______ 2 2 h=15 So your answer is h=15