Asimov Pupil

what are we anchoring it to?

oo i never though of that anyone got an answer

[math]W=Fd[/math]

[math]P=\frac{W}{t}[/math]

[math]1hp=743w[/math]

there is no distance for the second ship

the first ships hp was measured from engine work

none

testing [math]\frac{dy}{dx}=y^{2x-1}[/math]

cool!

[math]v=\frac{meters}{seconds}[/math]

thanks dave!

uhh i think he didn't take anything out of highschool. his comments are redundant and through the use of pure logic that, yes even us high schoolers posess, are proven wrong. I think he just reads things in national geographic and tries to prove them wrong. quite pityable though

wasn't it archimetese who solved pi by puting something like 20 different multisided polygons inside and outside a circle?

but he said without using sine or cosine or heron laws

if you can come up with the equation of the circle at any point. then the tangent is just the derivative. so just pick a point.

i thought it was 9604 for number 3

is it a right triangle? i think i solved it using the designs of our good buddy pythagorus

with those dimensions would it be 6.25 for the radius?

circumfrence of 12.5pi?

i'm afraid that only works if it's a right triangle though

my algebra comes out to 1/2 let me keep trying

thank you I'll remember that

Very good and if your teacher's are stingy and don't want you to leave it in double angles or if you solve it in three pages of work like i did then it is 2sin(a)cos(a)

by the way where did you learn to do it like that

it took me three pages doing addittion/subtraction formulas

that's a good idea but i cannot prove that right and that is not the answer i am looking for

what does

[math]\frac{\tan(\frac{\pi}{4} + a) - \tan(\frac{\pi}{4} - a)}{\tan(\frac{\pi}{4} + a) + \tan(\frac{\pi}{4} - a)}[/math]

equal to?

thank you

plus you say you are bout 9th grade.

ooooo just wait! you'll be far from staight lines soon!

I think it should be broken down into teasers and homework help that are just for math.

can someone give a list of some inert metals?

much obliged

BAH! it's 4cos4a

(take out a negative four, then use pythagorean therom!)

thanks for the help by the way

thank you

8((sin2a)^2)-4

8(sin2(a))(sin2(a))-4

8(2sin(a)cos(a))(2sin(a)cos(a))-4

32(sin^2(a))(cos^2(a))-4

32sin^2(a)(1-sin^2(a))-4

32sin^2(a)-32sin^4(a)-4

4(8sin^2(a)-8sin^4(a)-1)

now what pls

ps. just a sin(a) or cos(a) funtion not to any power

2Sin(a)Cos(a) = Sin^2(a)

 

So the formula at the end becomes:

 

16 - 16(Sin^2(a))

 

How's that?

 

If you want to put it in terms of cos then:

16(1-Sin^2(a)) = 16 Cos^2(a)

hey i thought it was 2sin(a)cos(a)= Sin2(a)

hmm i get that and thank you, but how do you put it as just a sin(a) function

THank you