My math teacher is out to destroy me - help, pretty please!

[Svenskaflicka]

Linked out due to math symbols that can't be written at Lit.

 

[drksideofthemoon]

Linked out due to math symbols that can't be written at Lit.

...uhm...maybe the Egyptology department can help you with those arcane symbols...

 

[Vermilion]

...uhm...maybe the Egyptology department can help you with those arcane symbols...

*snerk*

If you still haven't got this solved by 10pm GMT I'll ask my smarty-pants fiance if he can do it
He's out till then.
x
V

 

[Misty_Morning]

Linked out due to math symbols that can't be written at Lit.

How bout hints?

on the first one...heres a hint...



the integral of sin u du = -cos u + C


think about it

 

[CharleyH]

Linked out due to math symbols that can't be written at Lit.

I thought math teachers were out to destroy everyone smarter than them?

 

[Svenskaflicka]

I think I know the answer to the red question - it's

h(x) = 5(x2 + 3x)(2x+3)4,

with the red numbers in superscript, right?

 

[Misty_Morning]

I think I know the answer to the red question - it's

h(x) = 5(x2 + 3x)(2x+3)4,

with the red numbers in superscript, right?

looks good to me....but it's been 24 years since I did this shit.....

 

[oggbashan]

Um

Svenskaflicka, I can't help with the problems but I think you would have to include the steps needed to get to the answers and UNDERSTAND them.

Especially if the Maths teacher is that picky.

I hope someone can explain the answers in detail.

Og

 

[Misty_Morning]

Svenskaflicka, I can't help with the problems but I think you would have to include the steps needed to get to the answers and UNDERSTAND them.

Especially if the Maths teacher is that picky.

I hope someone can explain the answers in detail.

Og

She got the last one.....it's pretty straightforward....

But I agree on the the ingrals of rational functions and indefinite integrals, that the prick will want to see everything.....


(I never found anything rational about rational functions...)

 

[rgraham666]

Rational functions are perfectly rational.

Most human beings aren't perfectly rational and so have a lot of trouble with rational functions.

 

[Misty_Morning]

why the fuck am I thinking about this shit......*sigh*



the first one:



the integral of sin 2x dx = 1/2 sin 2x(2dx) = -1/2cos2x + C

 

[Svenskaflicka]

I thought math teachers were out to destroy everyone smarter than them?

My theory is that he hates me because I pointed out a few faults he made during class...

SOME math problems I'm good at, these are greek.

 

[Svenskaflicka]

With the integrals, I did like this:

f(x) = sin 2x, then F(x) should be -cos x^2 == -cos (pi/2)^2 + cos (0) = -0,999 + 1 = 0,001;

and

f(x) = cos 2x, then F(x) should be sin x^2 == sin (pi/2)^2 - sin (pi/4)^2 = 7,51-1,87 ≈ 5,64.

But he said my integral equations were wrong.

 

[Misty_Morning]

With the integrals, I did like this:

f(x) = sin 2x, then F(x) should be -cos x^2 == -cos (pi/2)^2 + cos (0) = -0,999 + 1 = 0,001;

and

f(x) = cos 2x, then F(x) should be sin x^2 == sin (pi/2)^2 - sin (pi/4)^2 = 7,51-1,87 ≈ 5,64.

But he said my integral equations were wrong.

I don't think you are supposed to square, I think its supposed to be times 2 not to the 2nd power....but I'm flying by the seat of my pants here.....


fuck...your gonna make me get out my calculus books....fuck.....

 

[Svenskaflicka]

I don't think you are supposed to square, I think its supposed to be times 2 not to the 2nd power....but I'm flying by the seat of my pants here.....


fuck...your gonna make me get out my calculus books....fuck.....

I know that if you derive x^2, you get 2x...

 

[CharleyH]

My theory is that he hates me because I pointed out a few faults he made during class...

SOME math problems I'm good at, these are greek.

Indeed, math teachers are like that, I know - twice! I did not really take your initial thread seriously, apologies. Are you seriously having a math problem? Isn't Greek, well...the basis of? Well. Ask Lauren if it is THAT serious a math problem. (L)

(PS - She WILL hate me for that offer )

 

[Svenskaflicka]

I think I've got the grades figured out, with cosinus/sinus periods and all:

a) x = arc cos (1/2) = 60 + n* 360 grades. X can also be 300 + n* 360 grades.
b) 2x = arc cos (1/2) = 60, x = 60/2 = 30 + n* 360 grades. X can also be 300/2 = 150 + n* 360 grades.

c) sin x = (√3)/2 = x = arc sin (0.8660254038) = 60 + n* 360 grades. X can also be 300 + n* 360 grades.

 

[Misty_Morning]

why the fuck am I thinking about this shit......*sigh*



the first one:



the integral of sin 2x dx = 1/2 sin 2x(2dx) = -1/2cos2x + C

this is correct, it is given as an example in one of my calculus books.....and there's a note beside it in my handwriting the says....be able to explain this shit...


ETA:correction, I left something out:

the integral of sin 2x dx = (1/2) the integral of sin 2x(2dx) = -1/2 cos2x + C

 

[Svenskaflicka]

My teacher gave me these "tool tips" for calculating:

2) Primitive function of f(x)= cos(5x) is F(x)= (sin(5x)/5. Try this by deriving F(x).


3) cos(20+n*360) = cos(-20+n*360) if we use grades. What about sin(20+n*360) = sin ()?

 

[Pure]

the answer to the derivative question is

5(x**2 + 3x)**4 times (2x+3)(2)=
10(2x+3)(x**2+3x)**4

 

[Pure]

the answer for the first integral is 1.

integral (sin2x) dx = -1/2cos(2x)+ C

evaluate at pi/2 and at 0

-(1/2)(cos pi -cos 0)=-1/2(-2)=1

 

[Misty_Morning]

the answer to the derivative question is

5(x**2 + 3x)**4 times (2x+3)(2)=
10(2x+3)(x**2+3x)**4

oh, yeah....now it's coming back....5-1 = 4 (she had it in the wrong place)....I think I'm hauling ass to another thread before I turn back into a total fucking geek again......

 

[Pure]

the answer for the second integral is -1/2

integral cos (2x) dx is 1/2 sin 2x + C

evaluate at pi/2 and pi/4

1/2(0-1)=-1/2

 

[Pure]

cos x = 1/2,

main answers (principal values) x=pi/3 and x=2pi/3. then adding any multiples of 2pi (=360 degrees); i.e., k may have any integral value, positive or negative.
(i.e. 360 degrees)

general form of the answers
x=pi/3 + k(2pi)

{x=200/3 grads + k*400= 66.666666 repeating grads}

[[ for k=1 x=7/3pi ]]

x=2pi/3 + k(2pi)

{x=400/3 grads + k*400 = 133.33333 repeating grads, etc.}

[[for k=1 8/3 pi etc.]]

 

[Pure]

second trig problem

cos (2x) = 1/2

x= pi/6 or pi/3 and to these may be added as follows for all integral values of k (positive and negative)

pi/6+kpi
pi/3+kpi

NOTE: CONVERT TO GRADS, IF THAT's REQUIRED.

nobody ever uses grads, except engineers, i think.