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The math help thread - Horusmaster - 2009-09-28

Please post here if you need help with your math homework, and I and other smart people will help you.

The math help thread - Plague - 2009-09-28

I'm in Geometry, and I'm doing proofs. The tatement is that angle 1 and angle 2 are a linear pair. What is the reason?

The math help thread - Noah - 2009-09-29

Plague Wrote:I'm in Geometry, and I'm doing proofs. The tatement is that angle 1 and angle 2 are a linear pair. What is the reason?

When the sum of the angles are 180 degrees/ π radians, they are a linear pair. That's the definition.

Noah

The math help thread - Anzamyu - 2009-09-29

Speaking of math, who here is in college cause i need help preparing for placement tests for a community college and i havent done math in like 4 yrs. I need help brushing up on math Sad

The math help thread - KajitiSouls - 2009-09-29

You not telling us what specifics you need help on doesn't help you.

*is studying Statics this quarter*

The math help thread - Kalovale - 2009-09-29

Anzamyu Wrote:Speaking of math, who here is in college cause i need help preparing for placement tests for a community college and i havent done math in like 4 yrs. I need help brushing up on math Sad

I believe more than a few here are both capable of and interested in helping.

The math help thread - tzk221 - 2009-09-29

What's wrong with this proof?

integral tan(x) dx = integral sin(x) / cos (x) dx = integral sin(x) sec(x) = -cos(x) sec(x) - integral -cos(x) sec(x) tan(x) dx = -1 + integral tan(x) dx

integral tan(x) dx = -1 + integral tan(x) dx

0 = -1

The math help thread - apathy - 2009-09-29

tzk221 Wrote:What's wrong with this proof?

integral tan(x) dx = integral sin(x) / cos (x) dx = integral sin(x) sec(x) = -cos(x) sec(x) - integral -cos(x) sec(x) tan(x) dx = -1 + integral tan(x) dx

integral tan(x) dx = -1 + integral tan(x) dx

0 = -1

Need to consider constant of integration

The math help thread - sky54264 - 2009-09-29

tzk221 Wrote:What's wrong with this proof?

integral tan(x) dx
= integral sin(x) / cos (x) dx
= integral sin(x) sec(x)
= -cos(x) sec(x) - integral -cos(x) sec(x) tan(x) dx
= -1 + integral tan(x) dx

integral tan(x) dx = -1 + integral tan(x) dx

0 = -1

Wow, break up lines and use more parentheses. So you used integration by parts and it looks like you let

u = sec(x)
dv = sin(x)
du = sec(x)tan(x)dx
v = -cos(x)

∫udv = vu - ∫vdu

Keep track of your du, dx and what you're substituting.

From here (the bolded part):
-cos(x) sec(x) - ∫-cos(x)du

= -cos(x)sec(x) + [cos(x)] ∫du
= -cos(x)sec(x) + [cos(x)] ∫sec(x)tan(x)dx
= -cos(x)sec(x) + [cos(x)] (sec(x) + c)

The math help thread - apathy - 2009-09-29

sky54264 Wrote:Wow, break up lines and use more parentheses. So you used integration by parts and it looks like you let

u = sec(x)
dv = sin(x)
du = sec(x)tan(x)dx
v = -cos(x)

∫udv = vu - ∫vdu

Keep track of your du, dx and what you're substituting.

From here (the bolded part):
-cos(x) sec(x) - ∫-cos(x)du

= -cos(x)sec(x) + [cos(x)] ∫du
= -cos(x)sec(x) + [cos(x)] ∫sec(x)tan(x)dx
= -cos(x)sec(x) + [cos(x)] (sec(x) + c)

You cannot pull cos(x) out of the integration sign (Note that cos(x) is actually equal to 1/u).

Nothing's wrong with his integration/differentiation steps, just missing integration constant. Also, you can solve this problem by using substitution (I know it's not the question asked, but just wanna make it clear)

The math help thread - tzk221 - 2009-09-30

Okay, thanks. Russt explained it to me before but I forgot. =P

Part B of the problem: Basically the same thing but with limits of integration from pi/4 to pi/6.

Yes, I know I can use substitution, but this "proof" was on the exam and we had to figure out what was wrong.

The math help thread - lolitsme - 2009-10-01

A friend of mine got this question on his assignment, basically x can be any digit. Somebody in our grade (9) got the answer but I don't know how. Could somebody explain this?

......___.xxx
xxx/xxx.
.....xxx.x
....-xxx.x
........0.xxx
.........-.xxx
.........0.000

(dots on the left are place-holders)

The math help thread - KajitiSouls - 2009-10-01

lolitsme Wrote:A friend of mine got this question on his assignment, basically x can be any digit. Somebody in our grade (9) got the answer but I don't know how. Could somebody explain this?

......___.xxx
xxx/xxx.
.....xxx.x
....-xxx.x
........0.xxx
.........-.xxx
.........0.000

(dots on the left are place-holders)

Work backwards =P It's a logic puzzle basically

Code:
___.xxx
xxx/xxx.
    xxx.x
   -[U]xxx.x[/U]
      0.xxx
      -[U].xxx[/U]
       .000

abc is a multiple of ijk. efg.h is an even bigger multiple of ijk.
Code:
___.xxx
ijk/xxx.
    xxx.0
   -[U]efg.h[/U]
      0.abc
      -[U].abc[/U]
       .000
mno is smaller than ijk, otherwise there's no need to tack on an extra decimal digit BEFORE subtracting once.
Code:
___.x0x
ijk/mno.
    mno.0
   -[U]efg.h[/U]
      0.abc
      -[U].abc[/U]
       .000
Notice how bc is also tacked on at the end in abc. Those can only be 0's. Also, mn = ef since mn-ef = 0.
Code:
___.x0x
ijk/mno.
    mno.0
   -[U]mng.h[/U]
      0.a00
      -[U].a00[/U]
       .000
g is 1 less than o. Something like 1230 - 1227 = 3 or something (in this case, g = 2, o = 3).

Now, we know that h =/= 0, since 10 - h = a, and that a =/= 0 or else the division would have ended already. We know that ijk multiplied by some number = a00. We also know that ijk multiplied by a bigger number = mng.h. Now we ask, what number t below 10 can we use to multiply jk so that it equals 100 or 200, etc? k cannot be 0 due to h being non-zero. Also, we must make sure that ijk multiplied by t doesn't go over 1000. Here are the possibilities:

Code:
k = 8, 6, 5, 4, 2
t = 5, 5, 2, 5, 5
          4
          6
          8

jk = 25
t  = 4


At this point, we can probably guess the numbers.

Code:
___.904
125/113.
    113.0
   -[U]112.5[/U]
      0.500
      -[U].500[/U]
       .000

Sorry if it was confusing >.<

The math help thread - Noah - 2009-10-02

Finding a good function for this would certainly be loved:

 Spoiler

Looks like this:

[Image: weirdo.png]

A good approximation is [Image: yeuervb.png]

Problem is the jumps at certain locations. Happens roughly for every 50 x. I'd assume it's a function looking like this:
[Image: ybk4ppa.png]

But I need to find the expressions. Any way without trying/failing?

The math help thread - apathy - 2009-10-02

tzk221 Wrote:Okay, thanks. Russt explained it to me before but I forgot. =P

Part B of the problem: Basically the same thing but with limits of integration from pi/4 to pi/6.

Yes, I know I can use substitution, but this "proof" was on the exam and we had to figure out what was wrong.

The limits would make everything easier, since -1 evaluated at pi/4 is equal to -1 evaluated at pi/6.

∫tan(x) dx = -1 - (-1) + ∫tan(x) dx = ∫tan(x) dx

(all integral signs here have upper and lower limits)

The math help thread - Tempus - 2009-10-02

No idea what information this is giving me. It's Mechanics, more specifically moments.

[imgspoiler]http://solarisftw.com/temp/pics/Ex%205B,%20Q3.png[/imgspoiler]

The math help thread - KajitiSouls - 2009-10-02

A moment is calculated by M = (r x F), or for the lazy way, d*F, where d is the distance perpendicular to the line of action F is acting on.

Since the problem is being ebil by not giving us the length of the jib arm, we'll have to utilize an additional variable which I'll call d.

M = [cos(25)*d]*5000*9.81 N*m

I believe the problem is saying that the maximum safe load based on moment is always the value above. θ is the independent variable, and F is the dependent variable.

The math help thread - Tempus - 2009-10-02

Still not understanding I'm afraid. The maximum load is always

[Image: eq.latex?5000\cos%2025\sec%20\theta]

but even working backwards, I can't seem to see where that's been got from...

The math help thread - KajitiSouls - 2009-10-02

What's the reason behind sec(θWink? I'm not sure where you got that from.

Also, keep in mind that 5000kg is a MASS. It is not a FORCE, which is in Newtons.

The math help thread - Tempus - 2009-10-03

I'm well aware of the difference, and that is the answer to the question. The maximum load (in kg) is always 5000cos(25)sec(theta), where theta is the angle of the jib to the horizontal. Likewise, I've no idea how they've come up with that answer, hence my problem. :\