It is second class sitting. It provides bench style seats for traveling.
June 1st
do cripples go to heaven and part 2s episode
The Bose QuietComfort 2s are the best I've tried...but they're pretty expensive.
And what is the question? If you want the average acceleration for that time, divide the change in velocity, by the time elapsed.
2, 4, 6 3, 6 The LCM of 2 and 3 is 6.
S = student A = adult 2S +3A = 1050 also S + A = 400 2S + 2A = 800 2S + 3A + 1050 subtract equations -A = - 250 A = 250 S = 150
If the question is regarding the pre-decimal British coin, the symbol is a lower case "s". For example, Two Shillings can be expressed as 2s, which is how it may have appeared on a train ticket, a restaurant menu or a theatre ticket. Alternatively, Two Shillings may have been written as 2/-. The slash is a seperater between the denominations, and the dash represents a value of zero (Pennies in this case).
The cheapest class of travel on the Indian Railways, Second Sitting has ordinary coaches which have seating arrangement on cushioned seats in two rows of 3 seats each.
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 of 7s plus 2s is 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
The compound (NH4)2S is ammonium sulfide.
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
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
5s - 6 = 2s, ie 5s - 2s = 6, ie 3s = 6, ie s = 2