Melody

$$\\\left(\frac{3}{4}\right)^{-2}=\left(\frac{4}{3}\right)^{+2}=\frac{16}{9}=1\frac{7}{9}$$

I suggest that you use this cacl for graphing.

Maybe you will find this old post useful :)

$$\\y=asin(bx)\\
amplitude=a, \\
wavelength=2\pi/b$$

$$\\3log B = 12 \\
log B = 4 \\
B=10^4=10,000$$

$$(3.2)(10^{18})*1000*1000=3.2*10^{24}$$

if log3=a and log4=b, express log72 in terms of a and b

$$log72\\
=log(9*8)\\
=log9+log8\\
=log3^2+log4^{(3/2)}\\
=2log3+1.5log4\\
=2a+1.5b$$

Find the H.C.F of(x²+y²)(x²+2xy+y²),(x²-y²)³ and (x⁶-y⁶) 

$$\\(x^2+y^2)(x^2+2xy+y^2)\\
=(x^2+y^2)(x+y)^2\qquad (1)\\\\\\
(x^2-y^2)^3\\
=(x-y)^3(x+y)^3\qquad (2)\\\\\\
(x^6-y^6)=(x^3-y^3)(x^3+y^3)\\
=(x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)\qquad(3)\\\\\\
(x^2+y^2)(x+y)^2\qquad (1)\\
(x-y)^3(x+y)^3\qquad (2)\\
(x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)\qquad(3)\\\\\\
HCF= x+y$$

I think that is correct. 

YES, I want to know where Freddy went.   

$$2^{1/3}=\sqrt[3]{2}\approx 1.2599$$