You can get 70 sweets for 1.40 if they are 2 each.
3p = 2p + 12 subtract 2p from both sides 3p - 2p = 2p - 2p + 12 1p = 12 p = 12 this is how you solve this problem.
25
20p, 2p, 2p, 2p, 1p.
there are 36 sweets in a jar
78p in British coins can be made in many ways.Here are a few, 78 x 1p = 78p, 39 x 2p = 78p, 20 x 2p + 39 x 1p = 78p, 50p + 20p + 5p + 2p + 1p = 78p and a great many more combinations.
You can buy 36 sweets because: 74 divided by 2 is 36.
I do not know but my teacher said about 1 1/2 p for small sweets, 2p for medium swees and 5p for large sweets.
There are six 2p electrons in Iron
The two factors of 2p are 2 and p.
There are 6 2p electrons in argon.
There are 50 2p coins in 1.00.
The GCF is 2p.
There are 31 ways:1p × 211p × 19 + 2p × 11p × 17 + 2p × 21p × 16 + 5p × 11p × 15 + 2p × 31p × 14 + 2p × 1 + 5p × 11p × 13 + 2p × 41p × 12 + 2p × 2 + 5p × 11p × 11 + 2p × 51p × 11 + 5p × 21p × 10 + 2p × 3 + 5p × 11p × 9 + 2p × 61p × 9 + 2p × 1 + 5p × 21p × 8 + 2p × 4 + 5p1p × 7 + 2p × 71p × 7 + 2p × 2 + 5p × 21p × 6 + 2p × 5 + 5p1p × 6 + 5p × 31p × 5 + 2p × 81p × 5 + 2p × 3 + 5p × 21p × 4 + 2p × 6 + 5p × 11p × 4 + 2p × 1 + 5p × 31p × 3 + 2p × 91p × 3 + 2p × 4 + 5p × 21p × 2 + 2p × 7 + 5p × 11p × 2 + 2p × 2 + 5p × 31p × 1 + 2p × 101p × 1 + 2p × 5 + 5p × 21p × 1 + 5p × 42p × 8 + 5p × 12p × 3 + 5p × 3
It has four factors: 1, 2, p, and 2p.
80
Nitrogen has 2s^3 2p^3 valence electrons so the answer would be 3
there are so many sweets in England you can never be sure how many there are.