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 electron configuration of Fe2+ is 1s2 2s2 2p6 3s2 3p6 3d6. Fe2+ has lost two electrons from its neutral Fe atom configuration.
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
2s2+7s-15 = (2s-3)(s+5) when factored
1s2 2s2 2p3 or otherwise denoted [He]2s2 2p3
The noble gas notation of oxygen is [He] 2s2 2p4.
The electron configuration of 1s2 2s2 2p6 corresponds to a neutral atom of Neon (atomic number 10). A 3+ ion would have lost all its valence electrons, leaving behind a stable 1s2 2s2 2p6 configuration, giving it the electron configuration of a Neon ion.
The full configuration of He 2s2 2p6 is 1s2 2s2 2p6. This configuration represents a neutral oxygen atom with its electrons distributed in the 1s, 2s, and 2p orbitals.
The electron configuration of beryllium is written as [He] 2s2. This means that it has 2s2 electrons above the configuration of Helium.
2s2-s-15 = (2s+5)(s-3)