ordered pair rule (x,y) (-x,y)is a ___ over ____?
548747
(x,y) (-x,y)is a refelction over y axis. I think...
111021
If you are sloving for X then anything greater than 2 will work. basically it is 6 time what (X) will be greater than 12. Well 6x2=12 so that won't work.
1180199
To find an ordered pair on the line, you simply select one half of the pair, and then find what the other value would have to be for the equation to be true. For instance, (0,6) is a valid pair, because if m=0, then n must equal 6 for 2m+n=6. Other valid pairs are (1,4), (-1,8), (5, -4), (1/2, 5), and so forth.
863581
You have a couple of typos, I think.  As for the "proof", dividing by x - y is dividing by zero.  That is a no-no.  (In other words, a * 0 = b * 0 does not imply a = b.)
914223
difference of squares rule:\n\nx^2 - y^2 \n\n(x-y)(x+y)
848382
To figure this out, you need to know that an ordered pair is in the form: (x, y). Just take each ordered pair and stick it into the equation:\na) 3(1) = 2(-1) -1 ... 3 = -3 (Not correct.)\nb) 3(-1) = 2(-1) - 1 ... -3 = -3 (Correct.)\nc) 3(-2) = 2(-5/2) - 1 ... -6 = -6 (Correct.)\nd) 3(0) = 2(-1/2) - 1 ... 0 = -2 (Not correct.)\n\nSo, the correct answers are (b) and (c).
441742
Assume there is a pair a_i, b_j with a non zero product.\nWrite fg= (a_i x^i + other terms) (b_j x^j + other terms) =\na_i b_j x^(i+j) is non zero.\n\nContradiction.
406468
D = {x | x < -3 U x > 4 }
907364
Almost.  You would actually include the endpoints 0 and -3 (so you need to use a bracket instead of parentheses).\nD = [0, infinity)\nR = [-3, infinity)
511948
There are an infinite number of ordered pairs that would satisfy this equation. If you want to find a few of them, assign a value to x and then solve for y.
1125164
When the book asks for a rule, they usually want it in a formula so it can be used no matter what the value of x is. \n\n1) Since the value of 'y' has a steady decrease of 2, then we know if x=0, y=87.  \n2) Start your formula: "y = 87...". \n3) We know that we'll have to be subtracting something since the value is decreasing: "y = 87 - ..."  \n4) Plug in what you already know, using something other than x=1, to figure out what more needs to be done (I'll use x=3):  87-81=6. \n5) How does the value of 'x' relate to this new figure? 3*2=6. \n6) Check to see if multiplying the value of 'x' by 2 will make up the difference after 87 is subtracted from the value of 'y'. \n\nThus we know that in order to find the value of 'y' we must first start with the number 87 and then subtract whatever we get after multiplying x*2 (i.e. 2x). \n\nAnswer: y = 87 - 2x
917224
Simply use correct order of operations\n\nparentheses\nexponents\nmult/division (from left to right)\naddition/subtraction (from left to right)
202912
let o: X->Y be the bijection.  if alpha is bijection from u to v, then o(u) -> o(v) should also be a bijection from Y to Y.
909284
((1/x) + (1/y)) / ((1/x) - (1/y))\n\n((y + x)/xy) / ((y - x)/xy)\n\n((y + x)/xy) * ((xy)/(y - x))\n\n(xy(y + x))/(xy(y - x))\n\n(y + x)/(y - x) = (x + y)/(-x + y) = (x + y)/(-(x - y)) = -(x + y)/(x - y)
1106173
The direction (angle) of |u| and |v|?
518633
Call  Y of x like Y(x) \n\nY(x) = Y(x-1)+ (3*4^x)-1(1/9)x\nY(x) = Y(x-2)+ (3*4^(x-1))-1(1/9)(x-1) + (3*4^x)-1(1/9)x\nY(x) = Y(x-3)+ (3*4^(x-2))-1(1/9)(x-2) (3*4^(x-1))-1(1/9)(x-1) + (3*4^x)-1(1/9)x\n...\nY(x) = Y(1) + 3(4^x+...+4^2) - (1/9)(x+...+2)\nY(x) = Y(1) + 3*4[4^(x+1)-1]/[4-1] - [x(x+1)/2-1]/9\nY(x) = (sqrt3)/4 + 4^(x+2) - 4 - [x(x+1)/2-1]/9\nY(x) = 16 * 4^x - x(x+1)/18 - (37/9 - (sqrt3)/4)
882707
Independent on X, dependent on Y