Melody

((log2(n^4)+5*n)/(n*log2(n))) 

\(n\ne0\)

\(\frac{log_2(n^4)+5n}{nlog_2(n)}\\ =\frac{4log_2(n)+5n}{nlog_2(n)}\\ =\frac{4}{n}+\frac{5}{log_2(n)}\\\)

Are you sure the brackets were in the right spots?

You need to include brackets so that the meaning is clear.

hi i was wondering how you graph a linear function

i.e.

d(x)=-1/2x-3

d is rarely used in a spot like this - I suppose it can be but f(x) or g(x) or just plane y would be more normal.

y=-1/2x-3

The linear equation is of the  form     y=mx+b

so

m=-1/2       and    b=-3

b is the y intercept, it is a point so you can plot it straight away.  It is where the line cuts the y axis.

m is the gradient which is the slope.

\(m=\frac{rise}{run}=\frac{-1}{2}\)

m is negative so the line must head to the TOP LEFT corner of the graph.

start at the y intecept   go up 1 and  left 2 places,  this means that the second point will be (-2,-2)

use   asin   it is the same as inverse sine :)

Hi,

There are lots of sites depending on the type and level of question that you are trying to solve.

Here is one site that might be helpful

There is no question here. 

I am sorry but I do not know either.    

I also hope that he is alright.   

Tut mir leid, aber ich weiß nicht, entweder.

Ich hoffe auch, dass er in Ordnung ist.

\(y=\frac{(x^2+4)(x-3)}{2x}\qquad x\ne0\)

y=u/v                                u=(x^2+4)(x-3)                                         v= 2x              

                                         u'= 2x(x-3)+1(x^2+4)                               v'=2

                                        u' = 3x^2 -6x +4

\(y=\frac{u}{v}\\ y'=\frac{vu'-uv'}{v^2}\\ y'=\frac{(2x)(3x^2-6x+4)-(x^2+4)(x-3)*2}{(2x)^2}\\ y'=\frac{\not{2}(3x^3-6x^2+4x)-(x^2+4)(x-3)*\not{2}}{\not{4}^{\;2}x^2}\\ y'=\frac{(3x^3-6x^2+4x)-(x^2+4)(x-3)}{2x^2}\\ y'=\frac{(3x^3-6x^2+4x)-(x^3-3x^2+4x-12)}{2x^2}\\ y'=\frac{3x^3-6x^2+4x-x^3+3x^2-4x+12}{2x^2}\\ \frac{dy}{dx}=\frac{2x^3-3x^2+12}{2x^2}\\\)

\(y=\frac{(x^2+4)(x-3)}{2x}\\ when\;\; x=-1\\ y=\frac{(1+4)(-1-3)}{-2}\\ y=\frac{(1+4)(-1-3)}{-2}\\ y=\frac{(5)(-4)}{-2}\\ y=10\\ (-1,10)\\~\\ \)

\(y'(-1) = \frac{-2-3+12}{2}=\frac{7}{2}\quad \text{This is the gradient of the tangent at x=-1}\)

To find the equation of the tangent you need to simplify this

\(\frac{y-10}{x--1}=\frac{7}{2}\)

Why would school be cancelled because of an election ?   This makes no sense to me. ://

cos(x)= - 0.7. If x and y are complementary, what is sin(y)?

Think about this

You have sine(angle) and you have cosine(angle) 

Did you ever stop to think about what the co stands for?

Co stands for complementary!

so     cos(x)=sin(90-x)                   and      sin(x)=cos(90-x)

in your question     x+y=90   so    y=90-x

you should be able to finish it now :)