39
Exactly 18 of them
To determine how many 2s go into 30, you divide 30 by 2. The calculation is 30 ÷ 2 = 15. Therefore, there are 15 twos in 30.
It has 17 of them with a remainder of 1
214 ÷ 2 = 107
1
39
96 ÷ 2 = 48
300 sir
There are 202 2s from 1 to 512.
Two will go into forty two twenty one times.
Exactly 18 of them do, with nothing hanging out and no room to spare.
A = (s, 2s), B = (3s, 8s) The midpoint of AB is C = [(s + 3s)/2, (2s + 8s)/2] = [4s/2, 10s/2] = (2s, 5s) Gradient of AB = (8s - 2s)/(3s - s) = 6s/2s = 3 Gradient of perpendicular to AB = -1/(slope AB) = -1/3 Now, line through C = (2s, 5s) with gradient -1/3 is y - 5s = -1/3*(x - 2s) = 1/3*(2s - x) or 3y - 15s = 2s - x or x + 3y = 17s
There is only one 2s orbital in an atom.
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
Points: (s, 2s) and (3s, 8s) Slope: (8s-2s)/(3s-s) = 6s/2s = 3 Perpendicular slope: -1/3 Midpoint: (s+3s)/2 and (2s+8s)/2 = (2s, 5s) Equation: y-5s = -1/3(x-2s) => 3y-15s = -1(x-2s) => 3y = -x+17x Perpendicular bisector equation in its general form: x+3y-17s = 0
2s + 17 = 2s + 17 1) First, you want to start on the left side of the equation and subtract 17 from both sides. 2s = 2s 2) Then, you take the 2 on the left side and divide it on both sides. s = s 3) You are left with s (Or 1s) on both sides, so s = 1.