There is no answer as expressed, since it is not a question.
Are you trying to simplify it?
9r2s + 4rs + 5s2 - 9s2 - 9r2 - 5rs - s2 + 2r2s + 4rs + 2s2
= (9 + 2)r2s + (4 - 5 + 4)rs + (5 - 9 - 1 + 2)s2
= 11r2s + 3rs -3s2
= s(11r2 + 3r -3s)
4 is the number of electrons in that sublevel.
They are letters. s, p, and d 2s2 2p6 3s2 3p6 3d10 A full n = 3 level.
The area is 72 square units. Explanation: the diagonal,d, is the hypotenuse of a rt. triangle whose legs are both s, the side of the square, so d2= s2 + s2 = 2s2 Therefore , the area of the square , s2 = d2/2 = (12)2/ 2 = 144/2 = 72.
Call the length of the shorter leg s. Then, from the Pythagorean theorem, s2 + (s + 17)2 = 252. Expanding the binomial results in: s2 + s2 + 34s + 172 = 252, or 2s2 + 34s + 289= 625, or s2 + 17s - 168 = 0. This can be factored into: (s + 24)(s - 7) = 0, which is true when s = -24 or 7. Only the latter answer makes sense for a physical triangle; therefore, the shorter leg of this triangle is 7 cm long.
first divide each side by 2 so you get... sine^2(X)=1/2 Then make sine ^2(X)=sine(x^2) SO you get... sine(X^2)=1/2 Then take the sine^-1 of each side it will look like this X^2=sine^-1(1/2) type the right side into a calc which will give you a gross decimal but it works (0.5235987756) so now you have X^2=0.5235987756 then take the square root of each side to make it linear and you will get X=.7236012546 and that is your answer!!!! make sure to check it on your calculator...I did and it worked * * * * * Not quite correct, I fear. Try this: Let s = sin θ. Then, 2s2 = 1; s2 = ½; and s = ±½√2. Therefore, θ = 45°, 135°, 225°, or 315°; or, if you prefer, θ = ¼π, ¾π, 1¼π, or 1¾π.
1s2 2s2 2p6 3s2 3p6
1s2 2s2 2p6 3s2 3p6
The Fe2 plus electron configuration is 1s2 2s2 2p6 3s2 3p6 3d6.
Al+3 aluminum
The electron configuration for a magnesium cation Mg2 plus is 1s2.2s2.2p6.
Normal Ca atom electron configuration: 1s2 2s2 2p6 3s2 3p6 4s2Ca+ (last electron is gone from the s orbital): 1s2 2s2 2p6 3s2 3p6 4s1
Electron Configuration: 1s2 2s2 2p4 Abbr: [He] 2s2 2p4
2s2+7s-15 = (2s-3)(s+5) when factored
[He] 2s2 2p2
1s2 2s2 2p6 and is N3- nitrate
1s2 2s2 2p3 or otherwise denoted [He]2s2 2p3
Beryllium- 1s2, 2s2 or [He] 2s2 Magnesium- 1s2, 2s2, 2p6, 3s2 or [Ne] 3s2 Calcium- 1s2, 2s2, 2p6, 3s2, 3p6,4s2 or [Ar] 4s2 Strontium- 1s2, 2s2, 2p6,3s2,3p6, 4s2, 3d10, 4p6,5s2 or [Kr] 5s2