Q: What is x-1 6?

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notation: natural numbers = 0,1, 2, 3, 4, 5, ....., (some define it without the zero, though) <= means smaller than or equal to, {} is set notation and means a set of numbers : (such that) then some condition. For example {x: x is not a duck} is the set of all things not a duck. Our goal is to prove that there are 21 different times. let x1 = hours, x2 = tens of minutes, x3 = minutes. We are going to prove the statement about the set {x1, x2,x3: 1<=x1 <= 12, 0<= x2<=5, 0<=x3 <= 59, x1 + x2 + x3 = 6}. It will be taken by assumption that this set is the set of digital clock combinations that add up to 6. So then, we must prove that there are unique 21 elements in the set {x1 + x2 + x3 : 1<= x1 <= 12, 0<= x2<=5, 0<=x3 <= 59, x1 + x2 + x3 = 6}. {x1 , x2 , x3 : 1<= x1 <= 12, 0<= x2<=5, 0<=x3 <= 59, x1 + x2 + x3 = 6} = {x1 , x2 , x3 : 1<= x1 <= 6, 0<= x2<=5, 0<=x3 <= 5, x1 + x2 + x3 = 6} because x3<=6, and because if x1 >=1, then x2 + x3 <=5, and x3, x2 >= 0 , so surely x3, x2 <= x5. Either x1 = 1, 2, 3, 4, 5, or 6. Next, x1 + x2 + x3 = 6, so x2 + x3 = 6 - x1. There are n+1 natural numbers between 0 and n (I'm being lazy and not proving this, but the proof would be so much longer if I proved it), and since 0 <= x2 <= 5 <= 6-x1, there are at most 6-x1+1 values of x2 for each value of x1. When x1 = 1, there are a maximum of 6, when x1 = 2, there are 6-2+1 = 5, when x1 = 3, there are 6-3+1 = 4, when x1 = 3, there are 3, then 2, and then 1. Summing this up gives us a maximum of 21. So it is at most 21 and at least 21, so exactly 21.

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

The answer depends on absolute deviation from what: the mean, median or some other measure. Suppose you have n observations, x1, x2, ... xn and you wish to calculate the sum of the absolute deviation of these observations from some fixed number c. The deviation of x1 from c is (x1 - c). The absolute deviation of x1 from c is |x1 - c|. This is the non-negative value of (x1 - c). That is, if (x1 - c) â‰¤ 0 then |x1 - c| = (x1 - c) while if (x1 - c) < 0 then |(x1 - c)| = - (x1 - c). Then the sum of absolute deviations is the above values, summed over x1, x2, ... xn.

The slope is (y2 - y1)/(x2 - x1) That is in this case (-6 + 6)/(4 +4) = 0/8 = 0

The midpoint is going to have an x and y value halfway between those of the two endpoints. The midpoint has an x value 6 higher than the first endpoint and a y value 4 lower. Just continue this pattern to get the other endpoint. (-2+6, 6-4)=(4, 2) The midpoint formula: [(x1 + x2)/2, (y1 + y2)/2] By substituting the given values into the formula we have: (x1 + -8)/2 = -2 and (y1 + 10)/2 = 6 x1 - 8 = -4 and y1 + 10 = 12 x1 -8 + 8 = -4 + 8 and y1 + 10 - 10 = 12 - 10 x1 = 4 and y1 = 2 Thus, the other endpoint is (4, 2).

Related questions

If the variables are x1 & x2 the solution is : 1) x1=x1+x2; 2) x2=x1-x2; 3) x1=x1-x2; EX: x1=1 , x2=6; 1) x1= 1+6 = 7 2) x2= 7-6 =1 3 x1=7-1 =6 ============================================

notation: natural numbers = 0,1, 2, 3, 4, 5, ....., (some define it without the zero, though) <= means smaller than or equal to, {} is set notation and means a set of numbers : (such that) then some condition. For example {x: x is not a duck} is the set of all things not a duck. Our goal is to prove that there are 21 different times. let x1 = hours, x2 = tens of minutes, x3 = minutes. We are going to prove the statement about the set {x1, x2,x3: 1<=x1 <= 12, 0<= x2<=5, 0<=x3 <= 59, x1 + x2 + x3 = 6}. It will be taken by assumption that this set is the set of digital clock combinations that add up to 6. So then, we must prove that there are unique 21 elements in the set {x1 + x2 + x3 : 1<= x1 <= 12, 0<= x2<=5, 0<=x3 <= 59, x1 + x2 + x3 = 6}. {x1 , x2 , x3 : 1<= x1 <= 12, 0<= x2<=5, 0<=x3 <= 59, x1 + x2 + x3 = 6} = {x1 , x2 , x3 : 1<= x1 <= 6, 0<= x2<=5, 0<=x3 <= 5, x1 + x2 + x3 = 6} because x3<=6, and because if x1 >=1, then x2 + x3 <=5, and x3, x2 >= 0 , so surely x3, x2 <= x5. Either x1 = 1, 2, 3, 4, 5, or 6. Next, x1 + x2 + x3 = 6, so x2 + x3 = 6 - x1. There are n+1 natural numbers between 0 and n (I'm being lazy and not proving this, but the proof would be so much longer if I proved it), and since 0 <= x2 <= 5 <= 6-x1, there are at most 6-x1+1 values of x2 for each value of x1. When x1 = 1, there are a maximum of 6, when x1 = 2, there are 6-2+1 = 5, when x1 = 3, there are 6-3+1 = 4, when x1 = 3, there are 3, then 2, and then 1. Summing this up gives us a maximum of 21. So it is at most 21 and at least 21, so exactly 21.

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

x1/6 = 6âˆšx

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1/6 x 1/6 x1/6 = 1/216

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

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we have the points (3,4) and (8,-6) let x1=3 y1=4 x2=8 y2=-6 slope=(y2-y1)/(x2-x1) =(-6-4)/(8-3) =-10/5 =-2

we have the points (3,4) and (8,-6) let x1=3 y1=4 x2=8 y2=-6 slope=(y2-y1)/(x2-x1) =(-6-4)/(8-3) =-10/5 =-2

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