Math Is Fun Forum

Uh, oh, Zachs back!

And how are you my tunnel-running friend?

Example: Square Root of 2 (but works generally)

Guess: 1.5

2/1.5 = 1.333
Average the guess and the answer: (1.5+1.333)/2 = 1.417

Again:
2/1.417 = 1.411
Average the guess and the answer: (1.417+1.411)/2 = 1.414    DONE!   

So, the more times you do this, the more accurate it becomes, but you can see that we have quickly arrived at three decimal places of accuracy. Stuff like this is called Successive Approximation.

Let's try another one. Square root of 10

Guess: 4
10/4=2.5
(4+2.5)/2 = 3.25
10/3.25=3.077
(3.25+3.077)/2 = 3.163
Test: 3.163*3.163 = 10.004 Amazing!

This can be useful if you are stuck without a calculator and need to find the diagonal of a square. I used it once for placing survey pegs in the ground when I was far from home.

And that is a good point. I guess it depends on how you look at it.

Taking your argument, they will always be one decimal place out, but that difference is at infinity.

So the difference is "infinitesimal", which means infinitely small, or "approaching zero as a limit".

So, 0.999... = (1 - infinitesimal),  perhaps? Is that the same as saying 0.999... = 1   ?

There is a Times Table Test which you can do online, and there are also Multiplication Worksheets which you can print out and practice on.

Print a few worksheets, and also print each one's answer sheet (put these aside to check your work later on), and then practice every day.

One thing to make it easier is to look at the Times Table Chart and find patterns: for example the last digit of the 5-times table always goes 5,0,5,0, and the last digit of 9-times goes 9,8,7,6, ... (AND if you add the digits of each 9-times, you get 9. What I mean is that 9x5=45 and 4+5=9, or 9x9=81 and 8+1=9. But what about 9x11=99?)

Because the circle sits at eye-level, and also has the 40° line grazing its upper surface, then a line at 20° (half of 40°) would go thru the circles center.

This makes a triangle ABC with A being the eye, B being the base of the circle and C being the center of the circle. We know the circle has an angle at A of 20°, and line BC is 3m (half of the 6m diameter).

Solving this triangle: tan 20° = 3m / line length (AB), so AB = 3m / tan 20° = 8.24 m

The question asks "the distance  the boy is from the nearest point of the tower.", well we have just figured out that the boy is 8.24m from the centre line of the tower, the tower is 6m in diameter, so he is 8.24 - 3  = 5.24m away from the outer wall

Now all that is left is that spike !

Hahahaha. Smart duck

Q2. A boy walks a certain distance at 9km/h. He finds that if he increases his speed by 1/3, he would have saved 4/3 hrs. Find the distance he walks.

I agree, it is hard to get your head around it.

Let's rephrase a little: 9km/h times (1+1/3) saves 4/3h, or an extra 3km/h saves 4/3h, so his time "t" is reduced.

OK, I think I can work this ... it has TWO things we don't know, time AND distance.

Let's call time "t" (in hours), and distance "x" (in km)

x = 9t (he went at 9km/h)
x = 9(1+1/3)(t-4/3) (increasing speed by 1/3 reduces time by 4/3)

Put the two together:

9t = 9(1+1/3)(t-4/3)
9t = 12(t-4/3)
9t = 12t-48/3
9t = 12t-16
-3t = -16
t = 16/3

Check this, too

Oh, cool. London. Do anything interesting?

Good one, WTF !

Egads and Gazooks!  Give me a coupla days!

This is how I would start:

Forget all the 3 dimensional stuff, it is really a 2d puzzle, so just write it down on a sheet of paper:

                               | (spike)
        .-'             (circle 6m diameter)
     .-'                      ^         
eye------------        7.5m                 
   1.5m               
---+-----------------------------------

Draw a circle 6m in diameter with its centre 7.5m up. Draw a line (eye height) 1.5m up. Now draw a line at 40° that just grazes the circle.

You could now measure off the paper how far away the boy is standing (by where the 40° and the "eye-line" intersect), and the length of the spike (the distance from the top of the circle upwards to the 40° line), but we should really do all the math!

The puzzler has made it easy for us, I think, by having the circle's diameter 6m, which is exactly the same as the difference in heights (7.5m-1.5m=6m), or else it is just coincidence

Beep !

We have some pages on this: Regular 2-D Shapes , and Quadrilaterals

What kind of help would you like?