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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 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
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
The compound (NH4)2S is ammonium sulfide.