Melody

Hi Titanium

Correct, a diagonal cut will produce 2 right angled triangles.

After that your air waves got a bit scrambled.   LOL.  :)

I'm trying to find the amplitude, period, phase shift, and midline of the following problem: y=2sinθ-2

It depends on where you want the brackets put :)

 y=(2sinθ)-2

amplitude=2, period=2pi, phase shift=0, and midline y=-2

 y=2sin(θ-2)

amplitude=2, period=2pi, phase shift=+2, and midline y=0

Thanks Alan  :)

$$\\12^2+2ab^2+4(a^6)^{-3}+5ab^2\\\\
144+2ab^2+4a^{-18}+5ab^2\\\\
144+4a^{-18}+5ab^2+2ab^2\\\\
144+4a^{-18}+7ab^2\\\\
or\\\\
144+7ab^2+\frac{4}{a^{18}}\\\\$$

$${\sqrt{{\mathtt{4}}}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}} = {\mathtt{8}}$$

or

$${{\sqrt{{\mathtt{4}}}}}^{\,{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}} = {\mathtt{10}}$$

$$\\(5a^2 b - 3bc^2)^4\\\\
=^4C_0(5a^2b)^4+^4C_1(5a^2b)^3(-3bc^2)^1 +^4C_2 (5a^2b)^2 (-3bc^2)^2 +^4C_3(5a^2b)^1 (-3bc^2)^3 +^4C_4(-3bc^2)^4\\\\
=(5a^2b)^4+4(5a^2b)^3(-3bc^2)^1 +6 (5a^2b)^2 (-3bc^2)^2 +4(5a^2b)^1 (-3bc^2)^3 +(-3bc^2)^4\\\\
=(625a^8b^4)+4(125a^6b^3)(-3bc^2)+6 (25a^4b^2)(9b^2c^4)+4(5a^2b)(-27b^3c^6) +(81b^4c^8)\\\\
=(625a^8b^4)-(1500a^6b^4c^2)+(1350a^4b^4c^4)-540(a^2b^4c^6)+(81b^4c^8)\\\\
=625a^8b^4\;-\;1500a^6b^4c^2\;+\;1350a^4b^4c^4\;-\;540a^2b^4c^6\;+\;81b^4c^8\\\\$$

Good one Titanium :)

@@ End of Day Wrap    Tues 14/7/15     Sydney, Australia Time   11:20pm   ♪ ♫

Hi all,

Over the past 2 days our forum users have been enlightened by Radix, Geno3141, TitaniumRome, Alan, Heureka, Sir-Emo-Chappington, asinus and fiora.  Our question askers are very fortunate to have such fine minds available to them.  Thank you.

Thanks Alan, I wasn't thinking of anything.  I know very little physics :)

So strain does not have anything to do with breaking point then?

I mean elastic can double in length without breaking but most things break much more quickly :/