MaxWong

\((2xy^2)^3\div(3x^2y^3)\times (9x^3y^2)\\ =(2^3\div 3\times 9)\times (x^3\div x^2 \times x^3)\times ((y^2)^3\div y^3\times y^2)\\ =24x^4y^5\)

\(\dfrac{5}{6}\div \dfrac{3}{6}\times \dfrac{3}{15}\\ =\dfrac{5}{6}\div \dfrac{1}{2}\times \dfrac{1}{5}\\ =\dfrac{5}{6}\times 2 \times \dfrac{1}{5}\\ =\dfrac{5\times 2}{6\times 5}\\ =\dfrac{10}{30}\\ =\dfrac{1}{3}\)

1/8 = 0.125, so we have 3 1/8 = 3 + 1/8 = 3 + 0.125 = 3.125.

Do you mean partial fractions?

\(\dfrac{1}{x^2 - 1}=\dfrac{a}{x-1}+\dfrac{b}{x+1}\\ 1=a(x+1)+b(x-1)=(a+b)x+a-b \\a=1/2\text{ and }b=-1/2\\ \therefore \dfrac{1}{x^2-1}=\dfrac{1}{2(x-1)}-\dfrac{1}{2(x+1)}\)

We made 1/(x^2-1) seperate into 2 fractions: 1/(2(x-1)) and -1/(2(x+1)).

Let x be the length of the 3rd side.

From triangle inequality, we know:

8 + 15 > x

x < 23

and 

8 + x > 15

x > 7

combine the 2 results, we get:

7 < x < 23.

The question now becomes: 

Find the number of positive integers that satisfies 7 < x < 23.

We can count the answer, which is 15.

Therefore there are 15 possible lengths for the 3rd side, if it is a +ve int.

\(\text{LCM}(9a^2bc,8abc,6ab^3)\\ =\text{LCM}(9,8,6)\cdot a^2b^3c\\ =72a^2b^3c\)

\(\text{LCM}(xy,x^2,xy^3)\\ \text{You find the highest power of x and y among the 3 and multiply together.}\\ =x^2y^3\)

Use the quadratic formula:

\(x^2+by-4=0\\ x=\dfrac{-b\pm\sqrt{b^2+16}}{2}\)

Yes...

I have been less careful these days...

I love math challenges :)

I did math challenges long time ago too... https://web2.0calc.com/questions/i-dont-rly-need-help-some-challenging-questions