20th line first number = 192 + 1 = 362
20th line last number = 20^2 = 400
20th line = 362 , 363, 364, 365 , ....... , 400
6.8 * 40
= 68 * 4
= 272
\(16^x = 2\\ x\ln 16 = \ln 2\\ x =\dfrac{\ln 2}{\ln 16} = \dfrac{\ln 2}{\ln(2^4)}=\dfrac{\ln2}{4\ln 2}=\dfrac{1}{4}\)
Whoops what am I doing
I should do:
\(2x^2 + 5x - 12 = 0\\ x = \dfrac{-5 \pm \sqrt{5^2-4(2)(-12)}}{2(2)}\\ \;\;=\dfrac{-5\pm 11}{4}\\ x = \dfrac{3}{2}\text{ OR }-4\\ \therefore 2x^2 + 5x - 12 = (2x - 3)(x + 4)\)
2x2 + 5x - 12
= 2 (x2 + 5x/2 - 6)
= 2 (x2 + 5x/2 + 25/8 - 73/8)
= 2(x + 5/4)2 - 73/4
= (2x + 2.5)2 - 73/4
= \((2x + \dfrac{5}{2}-\dfrac{1}{2}\sqrt{73})(2x+\dfrac{5}{2}+\dfrac{1}{2}\sqrt{73})\)
2/4 \(\div\) 3
= 1/2 x 1/3
= 1/6
N x 2 : 3 = 2 : 7
N x 2 : 3 x 3 x 7 = 2 : 7 x 3 x 7
N x 2 = 6
N = 3
Add them up!!
\(\text{Perimeter}\\ =\dfrac{3}{8} + \dfrac{1}{2}+\dfrac{3}{4}\\ =\dfrac{3}{8} + \dfrac{1\times 4}{2\times 4}+\dfrac{3\times 2}{4\times 2}\\ =\dfrac{3}{8} + \dfrac{4}{8}+\dfrac{6}{8}\\ =\dfrac{3+4+6}{8}\\ =\dfrac{13}{8}\\ =1 \dfrac{5}{8}\\ =1.625\)
Last 3 rows are just the same. Depends on which one do your teacher prefer you to use.
My definition:
The Pythagorean Theorem, also called Pythagoras's Theorem or Pythagoras Theorem, is the relations of lengths of the 3 sides of a right triangle. It says that the sum of the square of the lengths of the 2 legs is equal to the square of the hypotenuse.
Use properties of operations on the more complicated side, if it still not EXACTLY and OBVIOUSLY equivalent, use properties of operations on the other side. If the results are OBVIOUSLY the same (e.g. 10 = 10)<-- very obvious, then the expressions are equivalent. Vice Versa.
