Math Is Fun Forum

Thanks, you're a lifesaver. I had forgotten that you need at least as many equations as unknowns, and didn't think to use the derivatives.

All of yours look right to me.

B.4 is 2n - 1, so the missing number is 9.
B.5 alternates between adding 2 and subtracting 1, so the missing number is 5.

Thanks kylekaturn. After reading it about 10 times, I think I understand.

MathsIsFun, a page on dx would be great. I would say that, instead of following odd rules, it just sort of breaks them sometimes. Though offensive to mathematicians, it's accurate.

speed = distance / time
Average speed also = distance / time

Assume two chunks of time: t1 spent driving at 97 km/hr and t2=26.4min spend driving at 0 km/hr

Convert break to hours: 26.4/60 = 0.44 hr
Average the speeds:  (s1 + s2) / (t1 + 0.44)

(97 + 0) / (t1 + 0.44) = 64
t1 = 97/64 - 0.44 = 1.075625

Or, total time = 1.515625 hr and total distance = 64 km/hr * 1.075625 hr = 68.84 km.

But, I may have done it wrong. It's late and I haven't checked for mistakes...

Looks good.

A few questions remain, however. Like, how do you translate a square from english to spanish? You move it, of course! s->c q->u u->a a->d r->r e->o. Repitan juntos conmigo. Cuadro!

I'm not sure if I understand what you're asking, but here goes.

I think here:

You meant to say: (1+a1)(1+a2)
...as that would make a lot more sense.

n >=2, so let's start with the minimum case: n=2.

(1+a1)(1+a2) = 1 + a1 + a2 + a1a2
1+(a1+a2) = 1 + a1 + a2

The first expression > the second, because there's that extra a1a2 term in there.

This will be true for any n >= 2! This is because the expanded multiplication will always have all the terms of the simple addition, plus some additionals that are combinations of the a's.

That proof probably wouldn't satisfy a mathematician, but it sure works for me.

Hey, wait, I was wrong--radians will not always be positive in the important physical sense of direction of motion...

Hey ganesh, I was trying to learn from you , and I found a calculation error:

You forgot to square the 5/12.

Also, dh/dt should have units of ft/min. I have no idea what, if anything, went wrong.

The formula for the volume of a cone is V = (1/3) * (pi * r²) * h.

The volume of water in the tank is a function of time:
V(t) = 10 (ft³ / min) * t (min).

As the water flows into the tank, it fills a smaller cone in the tip of the larger one, the volume of which is the volume of water that has entered the tank so far. The radius also varies with time, and this will give us trouble unless we do something about it now.

You can think of a cone as a right triangle that has been spun about its height. This triangle will let us find r as a function of h, so we can eliminate r altogether in our original formula.

For the cone given in this problem, r = h * tan[sin-¹(5/12)]. You should probably draw the triangle to convince yourself that this is true.

Ok, now plug that in for radius in our original formula:
V = (1/3) pi * h * {h * tan[sin-¹(5/12)] }²

We need h as a function of time, so start by solving for h:
h = (3V / (pi * tan[sin-¹(5/12)]² ))^(1/3)
Oof.

Remember that V is a function of time, so we can substitute that to get h as a function of time:
h(t) = (30t / (pi * tan[sin-¹(5/12)]² ))^(1/3)

Differentiate this function you must. I don't want to .

Too bad they didn't just give you a time. Instead, you have to find the rate of change of h when h = 8. You can find the time at which this happens by setting h = 8 and solving for t. Plug that t into h'(t), and you should be set.

Oh, I just read the whole discussion. Your imprecision is probably coming from the conversion to radians, since you only use 5 decimal places of pi. And, it looks like your imprecision starts where your precision of pi ends, which makes sense if you've had the misfortune of having to work with significant figures.

Oh! Why didn't you say so? If you know that your triangle is exactly half of a screen pixel, then you're talking about a very specific thing: a 45-45-90 triangle.

Two of these make a square. Now, let me ask a stupid question. Why are you even bothering with triangles, when squares are so much easier to work with given a square texture tile?

Ok, I was writing the sequences out wrong. Let's try a few and see what happens:

1  2  3
    2

1  2  3  4
    2     4
           4

1  2  3  4  5
    2     4
    2

1  2  3  4  5  6
    2      4     6
            4

1  2  3  4  5  6  7
    2      4     6
    2             6
                   6

1  2  3  4  5  6  7  8
    2      4     6      8
            4             8
                          8

1  2  3  4  5  6  7  8  9
    2      4     6      8
    2             6
    2

1  2  3  4  5  6  7  8  9  10
    2      4     6      8      10
            4             8
            4

1  2  3  4  5  6  7  8  9  10  11
    2      4     6      8      10
    2             6              10
                   6

1  2  3  4  5  6  7  8  9  10  11  12
    2      4     6      8      10       12
            4             8                12
                          8

1  2  3  4  5  6  7  8  9  10  11  12  13
    2      4     6      8      10       12
    2             6             10
    2                            10
                                  10

1  2  3  4  5  6  7  8  9  10  11  12  13  14
    2      4     6      8      10       12        14
            4             8                12
            4                               12
                                             12

The pattern I see in the last cards is that they increase by 2 cyclically with each successively greater number of cards. In other numbers,

3    (2)
4    (4)
5    (2)
6    (4)
7    (6)
8    (8)
9    (2)
10  (4)
11  (6)
12  (8)
13  (10)
14  (12)

If this pattern holds, we should be able to predict the next numbers:

15  (14)
16  (16)
17  (2)
18  (4)
19  (6)
20  (8)
21  (10)
22  (12)
23  (14)
24  (16)
25  (18)
26  (20)
27  (22)
28  (24)
29  (26)
30  (28)
31  (30)
32  (32)
33  (2)

etc. The cycle "resets" when the next even in line is greater than the total number of cards in the stack.

Fortunately, it seems that the resetting follows another pattern; the cycle resets at every power of two! We now have enough info gleaned from patterns to write a general relationship between the number of cards and the last card standing!

First, some definitions to improve readability:
Last card standing = L
Number of cards to start with = n

L = 2 * [ n - (Largest power of 2 < n) ]

This just tells us how many steps we are above the last point the pattern was reset, and multiplies by 2 to convert from that to the last card standing.

Does it work? Let's try it on 6:
L = 2 * [6 - 4] = 4

Well, it worked for six. Whew! *pants*

I have no idea if this formula, which is just based off of observed patterns, actually works all the time, or is provable. I'll let somebody else figure that one out...;)