Melody

Hi Max :)

\cos^4x-(\sin 2x)(\sin x)(\cos x)-2\cos^2x+\sin^4x+2\sin^2x+1=0

\(\cos^4x-(\sin 2x)(\sin x)(\cos x)-2\cos^2x+\sin^4x+2\sin^2x+1=0\\ \cos^4x+\sin^4x-(\sin 2x)(\frac{sin2x}{2})-2(\cos^2x-\sin^2x)+1=0\\ \cos^4x+\sin^4x+2cox^2xsin^2x-2cox^2xsin^2x-(\sin 2x)(\frac{sin2x}{2})-2(cos2x)+1=0\\ (\cos^2x+\sin^2x)^2-\frac{4cox^2xsin^2x}{2}-(\sin 2x)(\frac{sin2x}{2})-2(cos2x)+1=0\\ 1-\frac{(2cosxsinx)^2}{2}-(\sin 2x)(\frac{sin2x}{2})-2(cos2x)+1=0\\ -\frac{(sin2x)^2}{2}-\frac{(sin2x)^2}{2}-2(cos2x)+2=0\\ -(sin2x)^2-2(cos2x)+2=0\\ -[ 1-cos^2(2x) ]-2(cos2x)+2=0\\ -1+(cos2x)^2 -2(cos2x)+2=0\\ (cos2x)^2 -2(cos2x)+1=0\\ let \;\;y=cos2x\\ y^2-2y+1=0\\ (y-1)^2=0\\ y=1\\ cos2x=1\\ 2x=2\pi n \qquad n\in Z\\ x=\pi n \qquad n\in Z\\ \)

check

\(\cos^4(2\pi n)-(\sin 2*(2\pi n))(\sin (2\pi n))(\cos(2\pi n))-2\cos^2(2\pi n)+\sin^4(2\pi n)+2\sin^2(2\pi n)+1=0\\ \quad 1 \qquad - \qquad 0 \qquad - \qquad 2 \qquad + \qquad 0 \qquad + \qquad 0 \qquad + \qquad 1 \qquad = \qquad 0 \qquad GOOD \)

Thanks Max, I am just going to do it a different way.

What's the derivative of \(\sqrt{\sin x \cos x}\;?\)

\(sin2x=2\sin x \cos x\\so\\ 0.5sin2x=\sin x \cos x\\ \frac{d}{dx}\;[0.5sin2x]^{0.5}\\ =\frac{d}{dx}\;\sqrt{0.5} \;\;[sin2x]^{0.5}\\ =\;\sqrt{0.5}*0.5 \;\;[sin2x]^{-0.5}*2cos(2x)\\ =\;\sqrt{0.5} \;\;\frac{cos(2x)}{ \sqrt{sin(2x)} }\\ = \;\;\frac{cos(2x)}{ \sqrt{2sin(2x)} }\\ = \;\;\frac{cos(2x)}{\sqrt{2*2sinxcosc} }\\ = \;\;\frac{cos(2x)}{2\sqrt{sinxcosc} }\\ \)

There you go - our answers are the same ...   hopefully that means we are both correct  !!!

Find the equation of the hyperbola with vertices (4,0) (-4,0) and focus (7,0)

Here is a video that will help you    

This is a very basic question Ballhopper.

You are posting a million question on here and I see no evidence that you understand anything.

I think you need to stop posting so many question and start putting more effort into understanding - starting from the basics!

Just because WE answer does not mean that you understand or that you will even be able to parrot the answer later.  You need to put in some serious effort!

Give the coordinates of the circles center and its radius

x^2 + y^2 - 16 = 0

Give the coordinates of the circles center and its radius

\(x^2 + y^2 = 4^2\)

centyre (0,0)  radius =4

Determine (if it exists) by evaluating the corresponding one-sided limits.

lim                          | x-3 | 

x --> 3                 ---------------

                              x^2 -9

As x tends to 3 from above x-3 is positive so  |x-3|=x-3

so

\(\displaystyle\lim_{x\rightarrow3^+}\;\frac{|x-3|}{x^2-9}\\ =\displaystyle\lim_{x\rightarrow3^+}\;\frac{x-3}{x^2-9}\\ =\displaystyle\lim_{x\rightarrow3^+}\;\frac{1}{2x}\\ =\frac{1}{6}\)

BUT  as x tends to 3 from below x-3 is negative so  |x-3|=3-x

\(\displaystyle\lim_{x\rightarrow3^-}\;\frac{|x-3|}{x^2-9}\\ =\displaystyle\lim_{x\rightarrow3^-}\;\frac{3-x}{x^2-9}\\ =\displaystyle\lim_{x\rightarrow3^-}\;\frac{-1}{2x}\\ =\frac{-1}{6}\)

This means that the graph is not continuous at x=3 so the limit does not exist.

Here is a picture of the graph.

Find by direct substitution

lim                              x-8

x --> y                  ------------------

                           x^2 + 13x + 42

a. limit does not exist

b. -1/240

c. 1/240

d. 0

e. infinity

Are you sure you want x tends to y ??

If you do then just replace the x's with y's 

Find 

lim                  'squareroot' 14+y - 'squareroot' 14

y --> 0            -----------------------------------------------

                                                 y

I am using L'Hopital's rule to sove this:

If you do not know the rule you should watch this clip.

I did some almost identical yesterday - I expect they were your questions as well

It means to round to 1 decimal place :)

Already answered

solve (С*(((2.7^-(t/(R*C)))-G)^2))/2=A for t

\(\begin{align}(С*([(2.7^{-(t/(R*C))})-G]^2))/2&=A\\~\\ \frac{С\;[(2.7^{-(t/(R*C))})-G]^2}{2}&=A\\~\\ \;[(2.7^{-(t/(R*C))})-G]^2&=\frac{2A}{C}\\~\\ (2.7^{-(t/(R*C))})-G&=\pm\sqrt{\frac{2A}{C}}\\~\\ 2.7^{-(t/(R*C))}&=G\pm\sqrt{\frac{2A}{C}}\\~\\ log[2.7^{-(t/(R*C))}]&=log\left[G\pm\sqrt{\frac{2A}{C}}\;\right]\\~\\ -(t/(R*C))log[2.7]&=log\left[G\pm\sqrt{\frac{2A}{C}}\;\right]\\~\\ t/(RC)&=\frac{-log\left[G\pm\sqrt{\frac{2A}{C}}\;\right]}{log(2.7)}\\~\\ t&=\frac{-RClog\left[G\pm\sqrt{\frac{2A}{C}}\;\right]}{log(2.7)}\\~\\ \end{align}\)