All Answers 358098 Answers

So......A = 30°

Sin 30°  =  1/2

.....The "Web Calc 2.0".......mail system......

( 11.000000, 0.000000, 0.000000, 0.000000, 0.000000,

0.000000, 11.000000, 1.000000, 1.000000, 1.000000,

0.000000, 1.000000, 11.000000, 1.000000, 1.000000,

0.000000, 1.000000, 1.000000, 11.000000, 1.000000,

0.000000, 1.000000, 1.000000, 1.000000, 11.000000 )

Matrix D: eigenvalue
( 10.000000, 0.000000, 0.000000, 0.000000, 0.000000,

0.000000, 10.000000, 0.000000, 0.000000, 0.000000,

0.000000, 0.000000, 10.000000, 0.000000, 0.000000,

0.000000, 0.000000, 0.000000, 11.000000, 0.000000,

0.000000, 0.000000, 0.000000, 0.000000, 14.000000 )

Matrix V: eigenvector
( 0.000000, 0.000000, 0.000000, 1.000000, 0.000000,

0.007685, -0.707107, 0.499941, 0.000000, 0.500000,

0.007685, 0.707107, 0.499941, 0.000000, 0.500000,

0.699338, 0.000000, -0.510809, 0.000000, 0.500000,

-0.714708, 0.000000, -0.489073, 0.000000, 0.500000 )

Vector D: eigenvalue - real
( 10.000000,

10.000000,

10.000000,

11.000000,

14.000000 )

Vector e: eigenvalue - imag
( 0.000000,

0.000000,

0.000000,

0.000000,

0.000000 )

Matix: eigenvalue - real,imag
( 10.000000, 0.000000,

10.000000, 0.000000,

10.000000, 0.000000,

11.000000, 0.000000,

14.000000, 0.000000 )

No she didn't "cheat" I doubt Kitty has ever heard of binomial expansions.

Shame on you Chris.  Never mind you are still cool with us.  Isn't he Kitty?  

Hi Reinout,

We all do.  From what I understand it is worse for the moderators.  

Do you get the new posts popping up on the bottom of your screen or do only the moderators get that?

Easiest is to put the matrix into a calculator, like my TI-89, and ask it for the eigenvalues, 14,10,10,10,11, but I suppose that's not what you're asking.

On paper they drop out easily by fiddling around with determinants, (rather than deriving and solving the characteristic equation).

I'm going to describe my calculation if that's OK. I don't have the patience to mess around with the LaTex.

Subtract L (read lambda) from each diagonal element and expand along the top row, that gets you (11-L) times a 4 by 4.

Add columns 2,3 and 4 to the first column, that get you 14-L down the first column which can then be removed as a common factor. You now have (11-L)(14-L) outside a 4 by 4 which has 1's down its first column.

Subtract the top row from rows 2,3 and 4, and you are left with a determinant which has 1's across the top row, 10-L down the remainder of the lead diagonal and zero's elsewhere. Expand that down the first column and then expand the resulting 3 by 3 and you finish up with (11-L)(14-L)(10-L)(10-L)(10-L), which you can now equate to zero.

BTW, I've just started to look at your earlier asymptotic stability problem. It's not something I've encountered in the past, but I'll see if I can get anywhere with it.

Actually, reinout......he lost me at the word "polytropic"...........

What's a "rectangular" carrot ???

Wait a minute...I think I've found it !!!......

What applies to this guy,

The post above is incorrect, it's actually rectangular carrot (90)

√90 = √9 *√10 = 3√10

carrot (90)

Thankyou alot, saving me a lot of saneness! Kudos my friend. 

You're welcome,

or

'graag gedaan' as they say in Dutch

We're looking for a line perpendicular to the line y = x/6

(edit: I accidently calculated a parallel line to x/6.)

Every line has the equation y = ax+b. where a is the slope and b is the y-intercept

Therefore the slope of the line y = x/6 is 1/6

Let's say line A and B are perpendicular,

then slope A = -1/(slope B)

Therefore the slope of a perpendicular line to y = x/6 is -1/1/6 = -6

All perpendicular lines to y = x/6 therefore have the equation

y = -6x+b

Let's see which line passes through (-2,4) by filling in x = -2 and y = 4

4 = -6*-2+b

4 = 12+b

b = -8

Hence the line you're looking for is

y = -6x-8

Reinout 

Would you rather have -$40 or $40 in your checking account ?? Can a negative be greater than a positive??

Again, thankyou for all your help. Late night study of which my teacher wants emailed by tomorrow morning. Kudos to you!

This helped so much actually thanks ! I got 65dB for the seccond answer

Thankyou for your time, I am currently struggling as it is a late night and you have legitimately managed to refresh my mind! Kudos. 

I think you care as much as I do 

All right, so we know the line intercepts the x-axis when x = 1/2. When a line intercepts the x-axis, y = 0 (because that's where the x-axis is). Therefore fill in x = 1/2 and y = 0 in your equation.

This gives

0 = b - 4*1/2

add 4*1/2 to both sides

4*1/2 = b

multiply 4*1/2 and flip the equation

b = 2

Fill this in in the equation

y = b-4x

and you'll find

y = 2-4x

I highly doubt this was a serious question Melody,

but my 'thumbs up' for you for answering it anyway.

Lateral SA_cylinder = 2*pi* r * h =  2 * pi * (6cm) * (7cm) = (84 * pi) cm² = 263.89cm²

SA base_cylinder = pi*(r)² = pi (6cm)² = (36*pi) cm² = 113.09cm²

So .....263.89cm²   +   113.09cm²   = 376.98cm² = 377cm²

YES!

Oh, yes....I see now....she actually "cheated," didn't she???.......

Shame on you, kitty<3 !!!!!!!!   (LOL!!!)