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Three score years and ten ( 30x 20 +10 = 70)
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∙ 12y ago
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How do you convert 2 coordinate points into slope intercept form y equals mx plus b?
Suppose the points are P = (r,s) and Q = (t,u) Then gradient of PQ = (u - s)/(t - r) = m, say. Let Z = (x,y) be any point on the line. Then gradient ZP also = m that is (y - s)/(x - r) = m Equating the two expressions for m, (y - s)/(x - r) = (u - s)/(t - r) or y - s = (x - r)*(u - s)/(t - r) y = (x - r)*(u - s)/(t - r) + s = x*m - r*m - s This is of the form y = mx + b, where m is as defined above and b = -r*m + s
Factor 4x-4y-ty plus xt?
4x-4y-ty+xt = 4x+tx-4y-ty = x(4+t) - y(4+t) = (x-y)*(4+t)
When the sum of the two angle is 180 their difference is 40 What is the angles?
Let x represent the larger angle and y the smaller. Solve the following: x+y=180 x-y=40 The second equation tells us that x=y+40. Plugging this into the first equation, we get, (y+40)+y = 180 2y+40=180 2y=180-40=140 y=140/2=70 So, y=70 and x=70+40=110.
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How do you write the equation of a line with only the x-intercept and coordinates pair?
You can only do so if the coordinate pair is other than the x intercept. Suppose the x intercept is A = (a, 0) and suppose the coordinate pair is R = (s, t) Then gradient = (t-0)/(s-a) = t/(s-a) Suppose P = (x, y) are the coordinates of any point on the line. Then gradient of PA = (y-0)/(x-a) = y/(x-a) The two gradients must be the same so t/(s-a) = y/(x-a) or y*(s-a) = t*(x-a) or y = t/(s-a)*x - ta/(s-a) which is of the form y = mx + c with m = t/(s-a) and c = -ta(/(s-a)
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