sqr.rtx/x= sqrt.x*sqr.rtx/sqr.rtx=x/x*sqrt.x=1/sqrt.x. x1/2 = x1/2 * x1/2 = x = 1 (x1/2) /x= 1/x1/2
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
2.
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 ============================================
A statement that two ratios are equal is called a proportion in math. An example of a proportion is 1/2 equals 2/4. In this proportion, if you cross multiply, you find that 4 x1 is equal to 2 x 2, which is a true statement or proportion.
x*x1/2= x3/2 Derivative = 3/2 * x1/2
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The square root of the sum of the squares of the differences of each coordinate. In space, the distance2=(x1-x2)2+(y1-y2)2+(z1-z2)2 On a plane, the distance2=(x1-x2)2+(y1-y2)2 On a line, the distance2=(x1-x2)2 or more simply distance=absolute value of (x1-x2)
It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]
Let's see this by an example. Find the slope m of the line that passes through the points (-3, 2) and (-2, 4). Definition of slope: The slope of the line that passes through distinct points (x1, y1) and (x2, y2) is Change in y/Change in x = Rise/ Run = (y2 - y1)/(x2 - x1) where x2 - x1 is different than 0. Solution: Let (x1, y1) = (-3, 2) and (x2, y2) = ( -2, 4). The slope is: m = (y2 - y1)/(x2 - x1) = (4 - 2)/[-2 - (-3)] = 2/1 = 2 The slope of the line is 2, indicating that there is a vertical change, a rise, of 2 units for each horizontal change, a run, of 1 unit. The slope is positive, and the line rises from left to right.
The distance between two points, (x1,y1) , (x2,y2) = squareroot[(x2-x1)2 + (y2-y1)2]
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.