## The Bargain Bin

Forum for discussing where to pick up cheap deals

# Xbox One Sale Roundup: July 11th, 2017

• BonnieTheBigVig247,256
• Cornerscout561,732
I gotta say, as sad as it is that I'm not interested in anything on sale, I sure don't mind saving money several weeks in a row. Among the best in my state and the world. I blog about it, and I tweet about things I take interest in: @Cornerscout
• sanityends339,278
I don't get it every week its nothing but whining. If you are not going to buy anything then just do not. There does not need to even be a sale they owe you nothing. Do you go to the store and complain shampoo is not on sale? Of course not. The bitching and complaining is unbelievable.
• ThreeHaddock62545,732
sanityends said:
I don't get it every week its nothing but whining. If you are not going to buy anything then just do not. There does not need to even be a sale they owe you nothing. Do you go to the store and complain shampoo is not on sale? Of course not. The bitching and complaining is unbelievable.
Especially after a huge sale....Those post add nothing of value to this site- I do appreciate the ones w/ recommendations though.
Hola
• Noodles Jr1,062,983
Posted on 11 July 17 at 03:16, Edited on 11 July 17 at 03:23 by Noodles Jr Twobby said:
thisisgman said:
M+E+H = MEH
You failed algebra didn't you?
Well if we assume M, E and H are all equal to an arbitrary value A, then the equation can be rewritten as a 3A=A^3. Divide out an A, so 3=A^2, therefore A=±root3. We also know intuitively that M, E and H can be 0 for the equation to check out, but now we also know they can equal root3 and -root3.

We know that if two of the variables are equal to 0, then the equation cannot check out as the MEH would always equal to 0, whereas M+E+H would equal a positive or negative integer. However, if one variable is equal to 0, then there are an infinite number of possibilities, if you let those two variables be the same number, but with separate signs.

If M, E and H are not equal to one another, then we can generalize by saying one of the three variable is larger than the other 2 (we'll say H). Therefore M≤E≤H. With this, we know that 3H≥MEH, and therefore 3≥ME. If 3≥ME, M≠0 and E≠0 then (M,E) can only be (±1,±1), (±1,±2), (±1,±3) (M and E values are interchangeable of course).

By substituting our points into M+E+H=MEH, then we can find that there are no possible values of H that satisfy (1,1), (-1,-1), but there are H values that satisfy (1,2), (1,-2), (-1,-2), (-1,2), (1,3), (1,-3), (-1,-3) and (-1,3), which are 3, 1/3, 3/4, -1/3, 1/2, -2, -1/2 respectively.

These are just a few solutions, and technically, thisisgman is infinitely right. He's infinitely wrong too... but still.
• Dannyboi511141,736 • billbillbill721,516
LV 1 Blue Slime said:
As a note for Dex... I would say the combat gets quite tedious despite how short of a completion it is so a mere 33% off would not be worth it, in my eyes.
I disagree. I thought DEX was sweet-as.
Old-school RPG with a (for once!) decent cyberpunk story & setting. I'm not gonna talk-up the combat, but then it's not a necessary component by any means. Most all of the game can be completed with little-to-no combat, with multiple routes to the objective.

Basically: It's a simplified, side-scrolling female-led 'Deus Ex' with an open-hub, numerous side-quests and several choice-based, divergent outcomes.

I felt it was decent-enough to grab at full-price. Would recommend to anyone interested in side-scroller 'Metroid-vania's, sci-fi stories or the artwork.
• sanityends339,278
sanityends said:
I don't get it every week its nothing but whining. If you are not going to buy anything then just do not. There does not need to even be a sale they owe you nothing. Do you go to the store and complain shampoo is not on sale? Of course not. The bitching and complaining is unbelievable.
Especially after a huge sale....Those post add nothing of value to this site- I do appreciate the ones w/ recommendations though.
As do I ty all for your recommendations of games. Other than that I have no idea why you post anything.
• BigWiIIieStyIe393,224
Noodles Jr said:
Twobby said:
thisisgman said:
M+E+H = MEH
You failed algebra didn't you?
Well if we assume M, E and H are all equal to an arbitrary value A, then the equation can be rewritten as a 3A=A^3. Divide out an A, so 3=A^2, therefore A=±root3. We also know intuitively that M, E and H can be 0 for the equation to check out, but now we also know they can equal root3 and -root3.

We know that if two of the variables are equal to 0, then the equation cannot check out as the MEH would always equal to 0, whereas M+E+H would equal a positive or negative integer. However, if one variable is equal to 0, then there are an infinite number of possibilities, if you let those two variables be the same number, but with separate signs.

If M, E and H are not equal to one another, then we can generalize by saying one of the three variable is larger than the other 2 (we'll say H). Therefore M≤E≤H. With this, we know that 3H≥MEH, and therefore 3≥ME. If 3≥ME, M≠0 and E≠0 then (M,E) can only be (±1,±1), (±1,±2), (±1,±3) (M and E values are interchangeable of course).

By substituting our points into M+E+H=MEH, then we can find that there are no possible values of H that satisfy (1,1), (-1,-1), but there are H values that satisfy (1,2), (1,-2), (-1,-2), (-1,2), (1,3), (1,-3), (-1,-3) and (-1,3), which are 3, 1/3, 3/4, -1/3, 1/2, -2, -1/2 respectively.

These are just a few solutions, and technically, thisisgman is infinitely right. He's infinitely wrong too... but still.
You just broke the internet lol, I read all that and almost made sense of it lol
• Rhhe82295,844
K4rn4ge said:
If there's only ONE game you buy on sale this week, definitely make it Stardust Galaxy Warriors - only TWO bucks right now & such a great game! Out of the hundreds of indie games I own on Xbox One, Stardust Galaxy Warriors is actually in my personal top 10. Such a steal - dont miss it guys. Also, most of the gamerscore is easy enough but there are a few tricky achs, but nothing too crazy.
I had never heard of this one until now. Took a quick peek to a gameplay video of this and yeah, looks like it's time to add to my everlasting backlog.
• GV Loc472,454
• Saint G Man 93195,317
• matdan748,072
33% off ain't a sale. I only see 4 games on sale and none I would touch, very happy with that. I need a few weeks to clean out my back-log anyway.
Out of the abyss peer the eyes of a demon, Behold the Razgriz, its wings of black sheath!
• Saint G Man 93195,317
Noodles Jr said:
Twobby said:
thisisgman said:
M+E+H = MEH
You failed algebra didn't you?
Well if we assume M, E and H are all equal to an arbitrary value A, then the equation can be rewritten as a 3A=A^3. Divide out an A, so 3=A^2, therefore A=±root3. We also know intuitively that M, E and H can be 0 for the equation to check out, but now we also know they can equal root3 and -root3.

We know that if two of the variables are equal to 0, then the equation cannot check out as the MEH would always equal to 0, whereas M+E+H would equal a positive or negative integer. However, if one variable is equal to 0, then there are an infinite number of possibilities, if you let those two variables be the same number, but with separate signs.

If M, E and H are not equal to one another, then we can generalize by saying one of the three variable is larger than the other 2 (we'll say H). Therefore M≤E≤H. With this, we know that 3H≥MEH, and therefore 3≥ME. If 3≥ME, M≠0 and E≠0 then (M,E) can only be (±1,±1), (±1,±2), (±1,±3) (M and E values are interchangeable of course).

By substituting our points into M+E+H=MEH, then we can find that there are no possible values of H that satisfy (1,1), (-1,-1), but there are H values that satisfy (1,2), (1,-2), (-1,-2), (-1,2), (1,3), (1,-3), (-1,-3) and (-1,3), which are 3, 1/3, 3/4, -1/3, 1/2, -2, -1/2 respectively.

These are just a few solutions, and technically, thisisgman is infinitely right. He's infinitely wrong too... but still.
W0W
thisisgman
• Infamous1,100,288
• matdan748,072
thisisgman said:
Noodles Jr said:
Twobby said:
thisisgman said:
M+E+H = MEH
You failed algebra didn't you?
Well if we assume M, E and H are all equal to an arbitrary value A, then the equation can be rewritten as a 3A=A^3. Divide out an A, so 3=A^2, therefore A=±root3. We also know intuitively that M, E and H can be 0 for the equation to check out, but now we also know they can equal root3 and -root3.

We know that if two of the variables are equal to 0, then the equation cannot check out as the MEH would always equal to 0, whereas M+E+H would equal a positive or negative integer. However, if one variable is equal to 0, then there are an infinite number of possibilities, if you let those two variables be the same number, but with separate signs.

If M, E and H are not equal to one another, then we can generalize by saying one of the three variable is larger than the other 2 (we'll say H). Therefore M≤E≤H. With this, we know that 3H≥MEH, and therefore 3≥ME. If 3≥ME, M≠0 and E≠0 then (M,E) can only be (±1,±1), (±1,±2), (±1,±3) (M and E values are interchangeable of course).

By substituting our points into M+E+H=MEH, then we can find that there are no possible values of H that satisfy (1,1), (-1,-1), but there are H values that satisfy (1,2), (1,-2), (-1,-2), (-1,2), (1,3), (1,-3), (-1,-3) and (-1,3), which are 3, 1/3, 3/4, -1/3, 1/2, -2, -1/2 respectively.

These are just a few solutions, and technically, thisisgman is infinitely right. He's infinitely wrong too... but still.
W0W
He is the best math cook in town, only writes in blue ink usually find him in back alleys dealing math equations. 100% pure math equations.
Out of the abyss peer the eyes of a demon, Behold the Razgriz, its wings of black sheath!
• MattQuin89343,882
• WildWhiteNoise1,103,422
• XvalgurX183,280
• MattQuin89343,882