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Who wants to know??

I think the real solution for this. Is giving what Melody said, 'half-answers'. We could do some of the work for them, but not all.

You are correct guest - I made a slight error, it should be 

+tan(pi/4)

and the final answer is +2

y= (sin(7pi)/6+tan(5pi/4))/cos(5pi/3)

Hi Max,

I think one of your brackets was in the wrong place ://

Otherwise yours was a great answer.  

\(y=\frac{ \frac{sin(7\pi)}{6}+tan(\frac{5\pi}{4})}{cos(\frac{5\pi}{3})}\\ y=\left( \frac{sin(7\pi)}{6}+tan(\frac{5\pi}{4})\right)\div cos(\frac{5\pi}{3})\\ y=\left( \frac{sin(\pi)}{6}-tan(\frac{\pi}{4})\right)\div cos(\frac{\pi}{3})\\ y=\left( 0-1\right)\div cos(\frac{\pi}{3})\\ y=-1\div \frac{1}{2}\\ =-2 \)

(1/6 sin(7 pi) + tan(5×pi/4)) / (cos(5×pi/3))

=     0            +        1          /       1/2

=1 / (1/2)

=2

Find the equation of the tangents to the circle x2 +y2=9 with slope 1

I thought I would also try to do it with the help of calculus.

\(x^2 +y^2=9\\ 2x+2y\frac{dy}{dx}=0\\ 2y\frac{dy}{dx}=-2x\\ y\frac{dy}{dx}=-x\\ \frac{dy}{dx}=-\frac{x}{y}=1\\ y=-x\\ etc\\ \mbox{The rest is the same as my other answer} \)

\(\frac{(sin\frac{7\pi}{6}+tan\frac{5\pi}{4})}{cos\frac{5\pi}{3}}\)

=\(\frac{(-sin\frac{\pi}{6}+tan\frac{\pi}{4})}{cos\frac{\pi}{3}}\)

= \(\frac{-\frac{1}{2}+1}{\frac{1}{2}}\)

=\(\frac{\frac{1}{2}}{\frac{1}{2}}\)

= 1

Find the equation of the tangents to the circle x2 +y2=9 with slope 1

This is a circle centred at (0,0) with a radius of 3

It will have a tangent with a gradient of 1 where  the circle intersects with y=-x

\(x^2 +y^2=9\\ x^2 +(-x)^2=9\\ 2x^2=9\\ x^2=4.5\\ x=\pm\sqrt{4.5}\\ y=\pm\sqrt{4.5}\\ The\; points\; of \; intersection \;are \;(-\sqrt{4.5},\sqrt{4.5})\;\;and\;\;(\sqrt{4.5},-\sqrt{4.5})\\ x+\sqrt{4.5}=y-\sqrt{4.5}\\~\\ y=x+2\sqrt{4.5}\\ \mbox{and the other one will be}\\ y=x-2\sqrt{4.5} \)

Perhaps this would be more nicely presented as

\(y=x+3\sqrt2\;\;\;and\;\;\;y=x-3\sqrt2\)

\(250\sqrt{250}\)

=\(1250\sqrt{10}\)

4 , 8 and 16.

2, 4, 8, 16, 32.

i is an imaginary number

\(i=\sqrt{-1}\)

It is the basic unit of the complex number system.

---------------------------------

es un número imaginario
 

Es la unidad básica del sistema de número complejo.

Thanks Chris, I get confused with these. :)

what is -y squared +y ?

\(-y^2+y\;\;\mbox{cannot be simplified}\\ (-y)^2+y=y^2+y \)

250 = 2 x 5^3 =5sqrt(10)

There is no question here.

Hi Jasmine,    I like this question - I had to think a bit.  :)

Starting from rest at the ground floor, an elevator takes 19.0 seconds for a vertical trip of 70.0 meters. It begins its journey with 4.50 seconds of constant acceleration, then moves for 9.00 seconds at constant velocity, and finally, moves for 5.50 seconds of constant negative acceleration that brings the elevator to a stop. (a) What is the maximum velocity of the elevator in m/s? 

t=19

s=70

4.5sec  then  9 sec  then 5.5sec

Here are the satandard motion in a straight line equations

I only used equation 3  And I let v be the maximum velocity

Acceleration section

\(s_1=0.5(0+v)*4.5\\ s_1=0.5*4.5v\\ s_1=2.25v\)

middle section

\(s_2=0.5(v+v)*9\\ s_2=9v\\\)

deceleration section

\(s_3=0.5(v+0)*5.5\\ s_3=0.5*5.5v\\ s_3=2.75v\\\)

Now the total distance traveled is 70m

\(s_1+s_2+s_3=70\\ 2.25v+9v+2.75v=70\\ 14v=70\\ v=5m/s\)

Maximum velocity = 5m/s

5 radical negative 400 squared

\(\sqrt[5]{(-400)^2}\\ =\sqrt[5]{160000}\\ =\sqrt[5]{1.6*10^5}\\ =10\sqrt[5]{1.6}\)

100+20(5 1/4)

100+20*(5+1/4) = 205

|25|=25

how much is 10% from 867€

10/100 * 867 = 86.7 euros

use the  \(\sqrt{x}\;\;\;button\)

250x^3-128y^3

\(250x^3-128y^3\\ =2(125x^3-64y^3)\\ =2[(5x)^3-(4y)^3]\\~\\ \qquad \boxed{a^3 - b^3 = ( a-b )(a^2 + ab + b^2)}\\~\\ =2(5x-4y)(25x^2+20xy+16y^2) \)

More info needed

I have no idea what you are talking about.