super hard math problem!!?

InX OutY

1 9

2 16

3 25

y=(X+2)^2

[y=(X+2) Squared]

x=1: (1+2)^2 = 3x3 = 9

x=2: (2+2)^2 = 4x4 = 16

x=3: (3+2)^2 = 5x5 = 25

Well, I worked this out, then read the other answers and, to my surprise, no two answers, mine included, agree! I got 23877m^3. The method I used was as follows. 1) Draw the problem. I got a grid consisting of 7 north-south and 7 east-west passages, each 200m long (assuming 25m centres). There were 49 intersections. 2) Calculate the total volume of the passages, =14x200x3x3 =25200m^3. 3) Calculate the total volume of the intersections, =49x3x3x3 =1323m^3. 4) Subtract(3) from (2), =25200-1323 =23877 which is the answer in m^3. The reason for the subtraction is that, in calculating the total passage volume, the volume in each intersection was counted twice. (since each passage has 7 intersections so 14 passages means 98 intersections, twice the actual number). Interesting exercise. Why the different answers when we are all using similar logic? I think accurately determining the numbers for pasages and intersections is the key. Get this wrong and the answer has to be wrong, no mater how sound the arithmetic. Thanks for the question. Edit: From reading blue girl's link, I see I may have made a mistake. I think Linlyons (answer A) is closest, it comes down to what happens to the passages that don't end in 4-way intersections. I am a bit busy right now; if you keep the question open I will have another go later. Edit 2: I've had another look and have come up with the same pattern as Linlyons and PHILEMON; 64 passages and 25 intersections. As I said above, the problem is what happens at the end of the passages that don't end in 4-way intersections. There are 28 of these, 4 are dead ends but the other 24 each connect with 1 other, producing 12 more intersections, each of these overlapping by a volume of (3x1.5x1.5) 6.75m^3. So if the passage volume was calculated as above, 12 times 6.75 would need to be subtracted as before. But the rule for intersections is that each is a 3m square, of height also 3m, in other words a 3m cube. To bring each of these intersections to that shape, a volume of (3x1.5x1.5) 6.75m^3 needs to be added. This is the same as the amount in the previous paragraph so, in calculating the passage volume, these two cancel and the calculation is as Linlyons example A; 13725m^3. As I said, it depends on what happens at the ends of the passages, there are probably several reasonable interpretations of this, all of which could be considered correct. Indeed, my origional answer of 23877m^3 doesn't actually break any of the rules, so this could be considered correct too! Perhaps the question really does need a little more information...

y=(x+2)^2

y = (x + 2)^ (2)

If this deals with input and output. Then the output goes up by every odd number, if you see the pattern.

9+7 = 16 + 9= 25 + 11 = 36 will be the next number for 4. If this doesn't deal with input and output o.o then i have no idea

1 * 6 + 3 = 9

2 * 7 + 2 = 16

3 * 8 + 1 = 25

so your rule is ...

(x * (x+5)) + (4-x) = y

y=(x+2)^2

x=1: (1+2)^2 = 3x3 = 9

x=2: (2+2)^2 = 4x4 = 16

x=3: (3+2)^2 = 5x5 = 25

Try (x+2)^2

y = (x + 2)^2 (squared)

type the problem in google the answetr shoudl pull up. it usually does for me GL!