MaxWong

\(13+\left(\dfrac{5}{8}-\dfrac{1}{4}\right)\times \dfrac{4}{5}\div \dfrac{3}{5}-8\\ =13-8+\left(\dfrac{5}{8}-\dfrac{1}{4}\right)\times \dfrac{4}{5}\div \dfrac{3}{5}\\ =13-8+\left(\dfrac{5}{8}-\dfrac{2}{8}\right)\times \dfrac{4}{5}\times \dfrac{5}{3}\\ =5 + \dfrac{3}{8}\times\dfrac{4}{5}\times \dfrac{5}{3}\\ =5 + \dfrac{1}{2}\\ =5\dfrac{1}{2}\text{ OR }\dfrac{11}{2}\text{ OR }5.5\)

5*5+5-5/5

I like to do some spacing :D

5*5 + 5 - 5/5

= 25 + 5 - 1

= 30 - 1

= 29.

\(\dfrac{3a}{5}-6 = 3\\ \dfrac{3a}{5}=9\\ \dfrac{a}{5}=3\\ a=15\)

Find:

(1) \(\text{LCD of }\dfrac{7}{y}\text{ and }\dfrac{21}{x^2-y}\)

and

(2) \(\dfrac{7}{y}-\dfrac{21}{x^2-y}\)

First question:

\(\text{LCD}(\dfrac{7}{y},\dfrac{21}{x^2-y})\\ =\text{LCM}(y,x^2-y)\\ =x^2y-y^2\)

Second question:

\(\dfrac{7}{y}-\dfrac{21}{x^2-y}\\ =\dfrac{7(x^2-y)}{x^2y-y^2}-\dfrac{21y}{x^2y-y^2}\\ =\dfrac{7x^2-7y-21y}{x^2y-y^2}\\ =\dfrac{7x^2-28y}{x^2y-y^2}\)

That way Maths cannot exist!

"No more Maths homework because Maths does not exist!!"

"Hurray!!"

~The smartest cookie in the world.

Is a^2 - 4 prime?

If a is an integer > 3, it will be prime. Because if the least value to make it not prime = y, then

y - 2 > 1, because we can't let y-2 = 1.

y > 3.

If a is an integer < 3 it will be either prime or negative.

You are Form 2? I am Form 3! :D

(t-13) + (t-6) + 99 = 180

(t - 13) + (t - 6) = 180 - 99 = 81

t - 13 + t - 6 = 81

2t - 19 = 81

2t = 81 + 19 = 110

t = 110/2 = 55 <- final answer.

\(3a^2x-2a^3 = x^3\\ -x^3 + 3a^2x - 2a^3 = 0\\ x^3 - 3a^2x + 2a^3 = 0\\ (x^3-a^3)+(3a^3-3a^2x)=0\\ (x-a)(x^2+ax+a^2)+3a^2(a-x)=0\\ (x-a)(x^2+ax+a^2)-3a^2(x-a)=0\\ (x-a)(x^2+ax-2a^2)=0\\ (x-a)(x^2-a^2+ax-a^2)=0\\ (x-a)((x-a)(x+a)+a(x-a))=0\\ (x-a)(x-a)(x+a+a)=0\\ (x-a)^2(x+2a)=0\\ x-a=0\quad \text{OR}\quad x+2a = 0\\ a = x\quad\text{OR}\quad a=-\dfrac{x}{2}\)

I have more detailed solution on the factorization :)

1/4 x 1/2 = 1/(4x2) = 1/8

\(\boxed{i^{4n-1} = -i}\)

\(1/i\\ =i^{4\cdot 0 - 1}\\ =-i\)