tertre

1. I'll just try this problem...I don't know if I'll get it !

Let \(\sqrt{49-20\sqrt{6}}=a-b\).

We can square both sides to get, \(49-20\sqrt{6}=(a-b)^2\).

Expanding this, gives us \(49-20\sqrt{6}=a^2+b^2-2ab\).

Now, by a bit of matching, we can see that \(a^2-b^2=49\) and \(-2ab=-20\sqrt{6}\) (Same thing as \(2ab=20\sqrt{6}\))

This might be a bit of tedious work, but by a bit of inspection \(a=5\) and \(b=2\sqrt{6}\) work.

Thus, the answer is \(a-b=\boxed{5-2\sqrt{6}}.\)

Row 0 has one number

Row 1 has two numbers

.

.

.

and so on

Therefore, row 42 must have 43 numbers.

The first number starts \(\binom{n}{0}\)with, the second number,  \(\binom{n}{1}\) , and so on..

Thus, the answer is \(\binom{42}{1}=\boxed{42}.\)

Hope that helped!

Well, the median is half the hypotenuse in a right triangle. And, just divide that length by 2.

Yex, \(5^{x+1}=5^x*5^1=30 \longrightarrow 5^x=6.\) Breaking. \(5^{3x+1}=5^{3x}*5^1, (5^x)^3*5^1=(6)^3*5=\boxed{1080}.\)

Hint: If it's easier for you, split the hexagon into six equilateral triangles, and calculate the area of each one using \(\frac{\sqrt{3}}{4}s^2\) . Then, multiply that by six.

To generalize. the area of a hexagon is \(\frac{3\sqrt{3}}{2}s^2\) .

1. 28^2+21^2=c^2,\(c=35.\)

2. 6^2+7^2=c^2, c^2=85, \(c=\sqrt{85}.\)

3. \(c^2+(\sqrt{3})^2=(2\sqrt{3})^2, c^2=9, c=3.\)

That's what I get too, is it right, ant101?

Erstens ist 9765 durch 2 geteilt durch 4882 und damit verbleibt ein Rest von 1. Wenn 4882 durch 2 dividiert wird, ergibt sich 2441 mit einem Rest von 0. Dann werden Dividieren und Teilen 10011000100101.

1. You can easily see a 5-12-13 Pythagorean triplet...so the answer is 12 !

2. Again, a 9-12-15 Pythagorean Triplet, and the answer is 15.

3. Pythagorean Triplet, 12-16-20, so (C), 16.

Break 1000 into its prime factorization: 2^3*5^2, so 4*4=16 factors. (add one to each power, then multiply)