H + 88 - 4 which can be simplified to H + 84.
60 degrees
4 : Each coin toss multiplies the generated outcome by two. The outcome for two tosses can be Heads (H) Tails (T), H H, T T or T H. As such, with three tosses the outcome can be H H H, H H T, H T H, T H H, T T H, T T T, T H T, H T T. We can define the number of outcomes as 2^n, where n is the amount of tosses.
2(h + 3) = 2h + 6
6
h = 4 and 8 +4=12
H + 88 - 4 which can be simplified to H + 84.
So this statement is an equation whose left side contains the sum of (5) and (twice g), and the right side contains only 23. 2g + 5 = 23
the quotient of 12 and a number h
18 different combinations. When a coin is tossed twice there are four possible outcomes, (H,H), (H,T), (T,H) and (T,T) considering the order in which they appear (first or second). But if we are talking of combinations of the two individual events, then the order in which they come out is not considered. So for this case the number of combinations is three: (H,H), (H,T) and (T,T). For the case of tossing a die once there are six possible events. The number of different combinations when tossing a coin twice and a die once is: 3x6 = 18 different combinations.
-13/12 is a rational number
H. M. A. El-Sum has written: 'Reconstructed wave-front microscopy'
The oxidation number for H in H3PO2 is +1, as it is in compounds with nonmetals. The oxidation number for P in H3PO2 is +1, as the sum of oxidation numbers in a compound is zero.
60 degrees
4 : Each coin toss multiplies the generated outcome by two. The outcome for two tosses can be Heads (H) Tails (T), H H, T T or T H. As such, with three tosses the outcome can be H H H, H H T, H T H, T H H, T T H, T T T, T H T, H T T. We can define the number of outcomes as 2^n, where n is the amount of tosses.
2(h + 3) = 2h + 6
The oxidation number of H in H2SO4 is +1, the oxidation number of S is +6, and the oxidation number of O is -2. This is because the sum of the oxidation numbers in a neutral compound like H2SO4 must equal zero.