i think its pretty much the same thing because matrix X1 X2 IS ACTUALLY X1 X2
This question can only be answered if the probability distribution functions of X1, X2 and X3 are known. They are not and so the question cannot be answered.
x1:y1 = x2:y2 4:-2 = x2:5 x2 = (4*5)/-2 x2 = -10
Let P(x1, y1), Q(x2, y2), and M(x3, y3).If M is the midpoint of PQ, then,(x3, y3) = [(x1 + x2)/2, (y1 + y2)/2]We need to verify that,√[[(x1 + x2)/2 - x1]^2 + [(y1 + y2)/2 - y1]^2] = √[[x2 - (x1 + x2)/2]^2 + [y2 - (y1 + y2)/2]^2]]Let's work separately in both sides. Left side:√[[(x1 + x2)/2 - x1]^2 + [(y1 + y2)/2 - y1]^2]= √[[(x1/2 + x2/2)]^2 - (2)(x1)[(x1/2 + x2/2)) + x1^2] + [(y1/2 + y2/2)]^2 - (2)(y1)[(y1/2 + y2/2)] + y1^2]]= √[[(x1)^2]/4 + [(x1)(x2)]/2 + [(x2)^2]/4 - (x1)^2 - (x1)(x2) + (x1)^2 +[(y1)^2]/4 + [(y1)(y2)]/2 + [(y2)^2]/4 - (y1)^2 - (y1)(y2) + (y1)^2]]= √[[(x1)^2]/4 - [(x1)(x2)]/2 + [(x2)^2]/4 + [(y1)^2]/4 - [(y1)(y2)]/2 + [(y2)^2]/4]]Right side:√[[x2 - (x1 + x2)/2]^2 + [y2 - (y1 + y2)/2]^2]]= √[[(x2)^2 - (2)(x2)[(x1/2 + x2/2)] + [(x1/2 + x2/2)]^2 + [(y2)^2 - (2)(y2)[(y1/2 + y2/2)] + [(y1/2 + y2/2)]^2]]= √[[(x2)^2 - (x1)(x2) - (x2)^2 + [(x1)^2]/4 + [(x1)(x2)]/2 + [(x2)^2]/4 + (y2)^2 - (y1)[(y2) - (y2)^2 + [(y1)^2]/4) + [(y1)(y2)]/2 + [(y2)^2]/4]]= √[[(x1)^2]/4 - [(x1)(x2)]/2 + [(x2)^2]/4 + [(y1)^2]/4 - [(y1)(y2)]/2 + [(y2)^2]/4]]Since the left and right sides are equals, the identity is true. Thus, the length of PM equals the length of MQ. As the result, M is the midpoint of PQ
The equation for the slope between the points A = (x1, y1) and B = (x2, y2) = (y2 - y1)/(x2 - x1), provided x1 is different from x2. If x1 and x2 are the same then the slope is not defined.
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i think its pretty much the same thing because matrix X1 X2 IS ACTUALLY X1 X2
f(x1) = (-5)2 + 3*(-5) + 5.1 = 25 - 15 + 5.1 = 15.1 f(x2) = (6)2 + 3*(6) + 5.1 = 36 + 18 + 5.1 = 59.1 f(x2)-f(x1) = 59.1 - 15.1 = 44 x2 - x1 = 6 - (-5) = 11 So average rate of change = 44/11 = 4
x²+6x+9=49 x²+6x-40=0 x1=-6/2 - Square root of ((6/2)²+40) x1=-3 - 7 x1= -10 x2=-6/2 + Square root of ((6/2)²+40) x2=-3 + 7 x2= 4
x2 + 10x = 0 x2 + 10x + 25 = 25 (x + 5)2 = 25 x + 5 = +-5 x1 = 0 x2 =10
x²+2x-15=0 x1=-2/2 - Square root of ((2/2)²+15) x1=-1-4 x1=-5 x2=-2/2 + Square root of ((2/2)²+15) x2=-1+4 x2=3
x² = x+110 x²-x-110 = 0 x1= -(-1/2) - Square root of ((1/2)²+110) x1 = 0.5 - 10.5 x1 = - 10 x2 = -(-1/2) + Square root of ((1/2)²+110) x2 = 0.5 + 10.5 x2 = 11
It's m = y2 - y1/ x2- x1 It's m equals y2 minus y1 over x2 minus x1
if you mean X²+8X-5=0: X1=-8/2 - Square root of ((8/2)²+5) X1=-4 - Square root of 21 X1 is about -8.58 X2=-8/2 + Square root of ((8/2)²+5) X2=-4 + Square root of 21 X2 is about 0.58
Given the limited information in the question, Z is maximised when x1 or x2 (or both) are maximised. There is no trade-off between x1 and x2 to worry about.
This question can only be answered if the probability distribution functions of X1, X2 and X3 are known. They are not and so the question cannot be answered.
it equals x1 it equals x1