Yes
I'm not positive, but they look like 2s or 0s. Maybe 00s. Don't hold me on that though.
2r + 2s = 50 2r - s = 17 therefore 4r - 2s = 34 Add so that you can eliminate one of the variables: 2r + 2s = 50 4r - 2s = 34 ---------------- 6r + 0s = 84 Solve for r: 6r = 84 r = 14 Substitute r into one of the original equations: 2(14) + 2s = 50 28 + 2s = 50 2s = 22 s = 11 Doublecheck with the other original equation: 2(14) - 11 = 28 - 11 = 17
Yes, although in conversation, this is not usually confusing.
To get the 2s complement, find the 1s complement (by inverting the bits) and add 1. Assuming that number is [4-bit] binary it would be 1000. If it is preceded by 0s, as in, for example, 0000 1000, then it would be 1111 1000.
4s2 - 9 can be expressed by using the identity: a2 - b2 = (a-b)(a+b) Therefore, 4s2 - 9 = (2s)2 - 32 = (2s-3)(2s+3)
The equivalent is 7s+2s = 9s
2s - 12 + 2s = 4s - 124s - 12 = 4s - 124s = 4ss = s==========this is an identity and any number can be s
2s + 16 = 4s - 6 Subtract 2s from both sides: 16 = 2s - 6 Add 6 to both sides: 22 = 2s divide both sides by 2: s = 11
There is only one 2s orbital in an atom.
It simplifies to: 2s+4R
9r2-4s2/9r+6sIt looks like you can factors the numerator(3r + 2s)(3r - 2s) [This is the factored form of 9r2-4s2]Put this back into the equation(3r+2s)(3r-2s)/9r+6sYou can also factor the denominator3(3r+2s)Put this back into the equation(3r+2s)(3r-2s)/3(3r+2s)You can cancel out the 3r+2s on top and bottom because they are the same they equal 1. Therefore your final answer is3r-2s over 3You could go further and say this is...r-(2/3)seither one is correct