A^2 = b^2 + c^2 - 2bc cos a
= 11^2 + 5^2 - 2*5*11 cos 40
= 7.86 to 2 DP's
to find the remaining angles use the sine rule:-
a / sin A = b / sin B so
7.857/ sin 40 = 11 / sin B
sin B = 11 sin40 / 7.857 = 0.8999
<B = 64 degrees
so <C = 180 - 64-40 = 76 degrees
Answer:
Neon: 0
Oxide: -2
Copper: 2
Tin: 0
Step-by-step explanation: To solve you must add/subtract the numbers to get the charge, for example if its -30 charge from electrons and 10 charge from protons, -30 + 10 = -20 therefore the charge is -20
standard form for a linear equation means
• all coefficients must be integers, no fractions
• only the constant on the right-hand-side
• all variables on the left-hand-side, sorted
• "x" must not have a negative coefficient


1: 0.5 and 50%
2: .2 and 20%
3: .75 and 75%
10: (1/10) and 10%
11: (3/5) and 60%
12: (1/4) and 25%
Answer:
9
Step-by-step explanation:
Y ~ X
Y = KX
K = Y/X
K = 36/4
K = 9