Math 1

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Problem 1

Problem 2

Problem 3

[ctapobep]

Problem 1

In this comparison the 3d number is greater because its "missing" part is less than the "missing" part of the 1st number:

$$ 1 - \frac{1110}{1111} = \frac{1}{1111} = \frac{3}{3333} > 1 - \frac{3331}{3334} = \frac{3}{3334} $$

Now let's try with the 2nd pair of numbers:

$$ 1 - \frac{2221}{2223} = \frac{2}{2223} = \frac{6}{6669} > 1 - \frac{3331}{3334} = \frac{3}{3334} = \frac{6}{6668} $$

The "missing" part of the 2nd number is less than the 3d, this means that 2nd number is the greatest here: $\frac{2221}{2223}$

Problem 2

Because the distance between AC and CB is the same, and the distance between CB and BD is also the same, then all the segments are equal. Since the length of AC is 60, then we can get the length of each small segment $60/3 = 20$: