Need Math Help?

[plasmawisp6633]

I could try and give it a shot...AltF4 can correct me if I'm wrong

perimeter = outer part of semi circle + 2 sides of rectangle + base of rectangle
area = area of semi-circle + area of rectangle

since the rectangle is topped by the semi-circle the base is as long as the diameter of the semi-circle. here I'll denote radius as r.

so the equations become:
perimeter = (1/2)(2πr) + 2x + 2r [where x is the length of the side of the rectangle]
area = (1/2)πr^2 + (2r)(x)

for the calculus part you need to use the first derivative test to see if the zero of the derivative is a local maximum of the original function and therefore the largest possible area. but since there are 2 variables this makes it a little difficult. we use the formula for perimeter to find x in terms of r

since perimeter = 288 in:
288 in = (1/2)(2πr) + 2x + 2r
288 - (1/2)(2πr) + 2r = 2x
[288 - (1/2)(2πr) + 2r]/2 = x

now we use this result and substitute into the formula for area:

area = (1/2)πr^2 + (2r)(x) => area = (1/2)πr^2 + (2r)([288 - (1/2)(2πr) + 2r]/2)

use the first derivative test to find the local maximum:

(area)' = πr + 288 - 4r - 2πr [simplified = 288 - 4r - πr]

set this equal to zero and find the value of r
skipping the algebra r = 288/(4 + π) -or- ~40.327

now to prove that this is a local maximum you must look at the values of the derivative function. to be a maximum the points to the left of the zero must be positive while the points to the right must be negative which is indeed the case

therefore the maximum area occurs when the radius of the semicircle is 288/(4 + π) -or- ~40.327


that's my reasoning...honestly I'd wait for AltF4 to respond since he's top math dog around here (meaning he'll know the answer better than I will)

Believe it or not, your reasoning seems very accurate. I forgot to incorporate the area into the problem, so I was stuck with only 1 equation. Thx for the help.

 

[KosukeKGA]

Please factor this out, Dan (I don't know how to get superscripts on here...But that's not the main problem.):

X^6 - 2x^5 + x^4 - x^2 + 2x - 1

 

[1048576]

There's no formula for a polynomial of degree 5 or greater, so you essentially have to guess solutions and then long divide. (I think)

for example x = 1 is a solution, so you divide (x-1)/|X^6 - 2x^5 + x^4 - x^2 + 2x - 1

I'm going to assume you know polynomial long division (or synthetic division, if it's easier)

you get x^5 -x^4 - x -1 Plug in 1 again long divide again. Now you have:

(x-1)(x-1)(x^4-1)

Now you can use an formula for polynomials of degree 4

You end up with:










(x-1)(x-1)(x-1)(x+1)(x^2+1)

I'm not a amth major or anything, so if there's an easier way to do it, I apologize.

 

[digitalmaster287]

Please factor this out, Dan (I don't know how to get superscripts on here...But that's not the main problem.):

X^6 - 2x^5 + x^4 - x^2 + 2x - 1

Well, the way I would do it is to split up the equation.

x^6 - 2x^5 + x^4 - (x^2 - 2x + 1)

x^6 - 2x^5 + x^4 - (x-1)^2

Now you find two things that multiply to get x^10 (product of 1st and last terms) and add to get -2x^5 (middle term).

x^6 - x^5 - x^5 + x^4 - (x-1)^2

(x^5)(x-1) - (x^4)(x-1) - (x-1)^2

(x-1)(x^5 - x^4) - (x-1)^2

x^4(x-1)(x-1) - (x-1)^2

(x-1)^2 (x^4 - 1)

(x-1)^2 (x^2 - 1)(x^2 + 1)

(x-1)^2 (x - 1)(x+1)(x^2 + 1)

(x-1)^3 (x^2 + 1) (x+1)

 

[JLynn943]

I don't understand Diff Eq

(aside from 1st order seperable)

 

[gongbomb]

Can someone help me with this?

Which of the following are analytic functions of the complex variable z in the entire z-plane (you may use Cauchy-Riemann conditions)?

abs(z)

Re(z)

e^(sin z)

Thanks

 

[Proud_Smash_N00b]

PHYSICS HELP PLZ!

1. A 0.140 kg baseball traveling 35.0 m/s strikes the catcher's mitt, which, in bringing the ball to rest, recoils backward 11.0 cm. What was the average force applied by the ball on the glove?

Answer is 780 N backwards but I want to know how you get that.

---

2. A particular race car can cover a quarter mile track (402m) in 6.40 seconds starting from a standstill. Assuming the acceleration is constant, how many "g's" does the driver experience? If the combined mass of the driver and the race car is 485 kg, what horizontal force must the road exert on the tires?

Answer: 2.00 g's; 9.51 x 10^3 N

---

3. The two forces F1 and F2 shown in Figure 4-43a and b act on a 27.0 kg object on a frictionless tabletop. If F1=10.2 N and F2=16.0 N, find the net force on the object and its acceleration for 4-43a and 4-43b.
(Since I do not have a pic of it, I will give details.

4-43a
F1 pulls exactly from the west (270 degrees when starting from top going clockwise)
F2 pulls exactly from the south (180 degrees when starting from top going clockwise)

4-43b
F1 pulls exactly 120 degrees when starting from top going clockwise
F2 pulls exactly from the north (0 degrees when starting from top going clockwise)

no answer given in book

---

4. A 15.0 kg box is released on a 32 degree incline and accelerates down the incline at 0.30 m/s^2. Find the friction force impeding its motion. What is the coefficient of kinetic friction?

Answer: 73 N, mu(coefficient of friction)=0.59

---

5. An 18.0 kg box is released on a 37 degree incline and accelereates down the incline at 0.270 m/s^2. Find the friction force impeding its motion. How large is the coefficient of kinetic friction?

Answer: 101 N, mu=0.719

---

6. A 28.0 kg block is connected to an empty 1.35 kg bucket by a cord running over a frictionless pulley. The coefficient of static friction between the table and the block is 0.450 and the coefficient of kinetic friction between the table and the block is 0.320. Sand is gradually added to the bucket until the system just begins to move. (a) Calculate the mass of sand added to the bucket. (b) Calculate the acceleration of the system

no answer given

---

Thank you for your time

 

[TPoint1BUA]

PHYSICS HELP PLZ!

1. A 0.140 kg baseball traveling 35.0 m/s strikes the catcher's mitt, which, in bringing the ball to rest, recoils backward 11.0 cm. What was the average force applied by the ball on the glove?

Answer is 780 N backwards but I want to know how you get that.

Use Work=Force*Distance

2. A particular race car can cover a quarter mile track (402m) in 6.40 seconds starting from a standstill. Assuming the acceleration is constant, how many "g's" does the driver experience? If the combined mass of the driver and the race car is 485 kg, what horizontal force must the road exert on the tires?

Answer: 2.00 g's; 9.51 x 10^3 N

Use d=1/2*a*t^2
a "g" is 9.8 m/s^2, so 2 g's is 19.6 m/s^2
You can then just use F=ma

3. The two forces F1 and F2 shown in Figure 4-43a and b act on a 27.0 kg object on a frictionless tabletop. If F1=10.2 N and F2=16.0 N, find the net force on the object and its acceleration for 4-43a and 4-43b.
(Since I do not have a pic of it, I will give details.

4-43a
F1 pulls exactly from the west (270 degrees when starting from top going clockwise)
F2 pulls exactly from the south (180 degrees when starting from top going clockwise)

4-43b
F1 pulls exactly 120 degrees when starting from top going clockwise
F2 pulls exactly from the north (0 degrees when starting from top going clockwise)

no answer given in book

This is just adding vectors.
For 43a, vector F1 is (-10.2, 0)
vector F2 is (0,-16)
You add them and get (-10.2,-16)
Then you can use the Pythagorean Theorem and F=ma to solve the rest of the problem.
44a works the exact same.

4. A 15.0 kg box is released on a 32 degree incline and accelerates down the incline at 0.30 m/s^2. Find the friction force impeding its motion. What is the coefficient of kinetic friction?

Answer: 73 N, mu(coefficient of friction)=0.59

You can separate the force of gravity into a component along the slope and perpendicular to the slope.
Fparallel = mg*sin(32 degrees)
Fperpendicular = mg*cos(32 degrees) = Fn
Ffr=mu*Fn
The net force is mg*sin(32 degrees)-mu*mg*cos(32 degres) = ma
You can cancel out m and you know a = .3m/s^2
so
g*sin(32 degrees)-mu*g*cos(32 degrees) = .3 m/s^2
Then you can solve for mu.
Then use Ffr=mu*mg*cos(32 degrees) to find the force of friction.

5. An 18.0 kg box is released on a 37 degree incline and accelereates down the incline at 0.270 m/s^2. Find the friction force impeding its motion. How large is the coefficient of kinetic friction?

Answer: 101 N, mu=0.719

Exact same as the last one.

6. A 28.0 kg block is connected to an empty 1.35 kg bucket by a cord running over a frictionless pulley. The coefficient of static friction between the table and the block is 0.450 and the coefficient of kinetic friction between the table and the block is 0.320. Sand is gradually added to the bucket until the system just begins to move. (a) Calculate the mass of sand added to the bucket. (b) Calculate the acceleration of the system

no answer given

The force of friction when it is not moving has a maximum of mu*mg. (mu is static and m is for the brick)
So, you need the bucket to weigh at least mu*mg for the block to move.
So, the bucket+sand needs to have a mass of mu*m. And mu*m-1.35 is the mass of sand you need to add.
For part b, the net force on the system is mu(static)mg-mu(kinetic)*mg. (m is the mass of the block)
You can then use F=ma (where m=mass of block+mass of bucket and sand) to find the acceleration.

 

[1048576]

y'' + ty' -y = 0

y(0) = 0

y'(0) = 3

I've been trying it with Laplace transforms, but it's hard to find L{ty'}(s) and even if I have that right, solving for Y(s) using Linear equations is giving me an enormous integral that I can't decipher.

 

[Grandeza]

ok some math help for a dumb 8th grader

simplify;
(X+5/16X)/ (3X-2/4X^2)

(2-X/7X)* (14/X-2)

(7X/X^2+4X)/ (35X^2/X+4)

oh and how do i factor
2m^2+5m-3

 

[Gosu_Engineer]

ok some math help for a dumb 8th grader

simplify;
(X+5/16X)/ (3X-2/4X^2)

(2-X/7X)* (14/X-2)

(7X/X^2+4X)/ (35X^2/X+4)

oh and how do i factor
2m^2+5m-3

well 2m^2+5m-3 = (2m -1)(m + 3) I'm not really sure what conventions or algorithms that you're supposed to know

as for your fractions I'm not clear with them. the first one for example in the numerator did you mean (X+5/16X) or [(X+5)/(16X)]

just making sure...sometimes people forget order of operations

 

[1048576]

Please help. HW due tomorrow!

 

[Gosu_Engineer]

I don't think I can help you with this one in time...I don't remember Laplace transform that well nor Fourier transforms...

 

[DTKPch]

Would Fourier transforms happen to have anything to do with Fourier series?

Because I know Fourier series. So if they're related, then I'll do some Fourier transform crap. And then I'll take it one step further, and do some Laplace transform crap.

Yeah, I actually have no idea how to help. Is that class Differential Equations?

 

[Gosu_Engineer]

Would Fourier transforms happen to have anything to do with Fourier series?

Because I know Fourier series. So if they're related, then I'll do some Fourier transform crap. And then I'll take it one step further, and do some Laplace transform crap.

Yeah, I actually have no idea how to help. Is that class Differential Equations?

they're very different. A fourier series is usually a decomposition of a periodic function into other oscillating functions where a fourier transform transforms a function of one variable into another.

They're both named after the French mathematician Joseph Fourier

 

[JrdnS]

i know this is chem. but it has some math i think. help please.

1.what happens to carbon and oxygenin the compound Na2CO3 when you mix it with H2SO4? justify the answer.

2. write the balanced chemical equation for the reaction between sulfuric acid and sodium carbonate.

thank you./

 

[Grandeza]

how do you convert .99999... into a decimal.

every time i do
10x=9.999...
- x=.999...
____________
9x=9
so .999...= 9/9
now 9/9 is clearly 1 so what did i dp wrong?

and when i do .0999 with only the 9's repeating it come out as 1/10. now thats clearly .1 so what am i doing wrong?

 

[1048576]

nothing .999... is 1.

Think of it this way: 1/3 is .333, 2/3 is .666 3/3 is .999 = 1

 

[TPoint1BUA]

You, sir, just got trolled.

And to keep my post math related:

Anyone else not like Negative Schwartzian Derivatives?

 

[blueKarat]

how do you convert .99999... into a decimal.

every time i do
10x=9.999...
- x=.999...
____________
9x=9
so .999...= 9/9
now 9/9 is clearly 1 so what did i dp wrong?

and when i do .0999 with only the 9's repeating it come out as 1/10. now thats clearly .1 so what am i doing wrong?

if you mean 0.999 into a fraction you can use geometric series

*does math* *ends up with 9/10 / 9/10*

omg it comes out to 0.999 = 1

 

[Proud_Smash_N00b]

Moar Physics Help Plz!!

If I give an answer, I would like you to tell me how you get that answer.
I am gonna need the info asap. It is due tomorrow.
Thanks in advance!!


1) A 1300 N crate rests on the floor. How much work is required to move it at a constant speed 4.0 m along the floor against a friction force of 230 N.

Answer: 9.2x10^2

---

2) A sled is initially given a shove up a frictionless 28.0 degree incline. It reaches a maximum vertical height of 1.35 m than where it started. What is its initial speed?

Answer: 5.14 m/s

---

3) In the high jump, Fran's kinetic energy is transformed into gravitational potential energy without the aid of a pole. With what minimum speed must Fran leave the ground in order to leave her center of mass 2.10 m and across the bar with a speed of 0.7 m/s?

No answer given

---

4) A projectile is fired at an upward angle of 45.0 degrees from the top of a 265 m cliff with a speed of 185 m/s. What will be its speed when it strikes the ground below?

No answer given

---

5) Two railroad cars, each of mass 7650 kg and traveling 95 km/h in opposite directions, collide head-on and come to rest. How much thermal energy is produced in the collision?

Answer: 5.3x10^6 N

---

6) In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.6 cm. A second projectile causes the same pendulum to swing twice as high, h2=5.2 cm. The second projectile was how many times faster than the first?

Answer: sqrt 2

---

7) A 920 kg sportscar collides into the rear end of a 2300 kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.8 m before stopping. The police officer, knowing that the coefficient of kinetic friction is 0.8, calculates the speed of the sports car at impact. What was that speed?

Answer: 23 m/s

---

Please help and thanks
=]

 

[Rici]

Lol, this is turning out to be a:"Do my homework for me, NOW!' thread.

 

[Killfest]

That book with the letter addition is Sideways Arithmetic from Wayside School. At least, that's where I learned it.

 

[A_man13]

how do you convert .99999... into a decimal.

every time i do
10x=9.999...
- x=.999...
____________
9x=9
so .999...= 9/9
now 9/9 is clearly 1 so what did i dp wrong?

and when i do .0999 with only the 9's repeating it come out as 1/10. now thats clearly .1 so what am i doing wrong?

A) I believe .99999 is a decimal already. if all else fails, use 99/100

B)you need to divide by 10, therefore canceling the coeficiants, leaving you with .99999

 

[Wardub]

A) I believe .99999 is a decimal already. if all else fails, use 99/100

B)you need to divide by 10, therefore canceling the coeficiants, leaving you with .99999

I believe he is talking about .9999~ repeating, which as someone already pointed out is equal to 1.

.99999 in fraction would be 99999/100000, not 99/100.

 

[AltF4]

Yea, this isn't the "Do My Homework" thread.

If it's due tomorrow, don't expect an answer. You have to have a real question. Or at least appear to have tried, and need help on something specific. You can't come in here and post your entire homework sheet. (I'm looking at you, Smash Noob)

 

[Proud_Smash_N00b]

Yeah, sorry

I just wanted to know how to get the answers..
oh well

 

[1048576]

y'' + ty' -y = 0

y(0) = 0

y'(0) = 3

I've been trying it with Laplace transforms, but it's hard to find L{ty'}(s) and even if I have that right, solving for Y(s) using Linear equations is giving me an enormous integral that I can't decipher.

Sorry for repost, but Alt seems to be posting again, and he's a genius.

 

[Death]

I've been having a bit of trouble with inverse functions and whatnot. I have a question:

How would you graph: y = 1/9-x^2

Ok, so you have to factor the bottom which gives (3-x)(3+x) but where does the negative go? And what would the resulting graph look like?

 

[1048576]

Why not plot points to see what your graph does. You know from factoring the denominator that you have two vertical asymptotes at x = 3 and x = -3. Between 3 and -3 you have a minimum at x = 0, which is 1/9. As x approaches 3 from the left and -3 from the right, the limit of your function goes to infinity, so you have a form resembling an upwards parabola. As x approaches 3 from the right and -3 from the left, the graph goes to minus infinity, and when x = positive or negative infinity, the function = 0.

 

[Death]

Does the negative affect the graph? It should shouldn't it?

So the graph would then look something like this?


___x|____ |y
____ |____|______

So those are the two asymptotes and one curve would go from "x" down and to the left and the other would go from "y" down and to the right??

Oh, btw the equation is y = 1(in the numerator) divided by 9-x^2 (in the denominator).

 

[1048576]

I'm not going to try to draw a picture on this. the left resembles the right half of an upside-down parabola. The middle resembles a normal parabola. The right resembles the left half of an upside-down parabola. Remember that none of the curves ever cross the asymptotes, however.

When in doubt, plot a bunch of points.

Obv. if you have access to a graphing calculator...

 

[Neon Ness]

...Wow.

Looking at this thread makes me glad I'm done with high school. Luckily, the Fine Arts major requires zero math. :bigthumbu Math hurts my head...

 

[1048576]

Math is way harder for me than any other subject. don't know why I continue to stick with it. I guess it's because it can provide a secure job, since most people suck at math.

 

[Neon Ness]

Math is way harder for me than any other subject. don't know why I continue to stick with it. I guess it's because it can provide a secure job, since most people suck at math.

True that. You definitely won't have a problem getting a job; me on the other hand... I was really good at math way back when, I just really hated it. Not because I thought it was pointless like I heard my classmates complaining, but just because there was so much to remember and it was so complex.

 

[NintendoMan07]

I'm not exactly sure where this goes, seeing as although the course it pertains to is Discrete MATH, I'm a Computer SCIENCE major, so the two help topics for this kinda thing confuses me, but I decided this is more math oriented.

Anyway, is anyone familiar with Huffman coding, which assigns bit strings to symbols?

Well, I'm constructing trees using Huffman coding for a homework problem.

For example:
Construct the binary tree with prefix codes representing this coding scheme:
A: 1
E: 01
T: 001
S: 0001
N: 00001

Now, my question about this is: Is the resulting tree supposed to have an empty leaf (one with no label)? Considering the few examples of this I have to go by show instances where every leaf has a label, I really have no idea what's correct in this instance.

Also, would happen to be the only college sophomore using this topic? I feel kinda dumb in asking this, but I'm a stickler for accuracy.

 

[AltF4]

Hey! Cool, more Computer Scientists!

Well, if that's all you're given, then I guess you have to have empty leaves. That or it's not strictly a binary tree.

Plus, you should be getting 5 empty leaf nodes, right? One at 00, one at 010, 0110, etc...

It HAS been a while since I've done that, though. So I don't feel 100% on all of that anymore.

y'' + ty' -y = 0

y(0) = 0

y'(0) = 3

I've been trying it with Laplace transforms, but it's hard to find L{ty'}(s) and even if I have that right, solving for Y(s) using Linear equations is giving me an enormous integral that I can't decipher.

Sorry for repost, but Alt seems to be posting again, and he's a genius.

Oh, god. Laplace Transformations. That was, like, one of those things you do in upper calc that you never see ever again.

What exactly are you trying to solve for anyway?

Sorry, I've been so slow to respond. I may not be of the best help for this one, though! You'd get a faster answer from your TA or prof.

 

[NintendoMan07]

Hey! Cool, more Computer Scientists!

Well, if that's all you're given, then I guess you have to have empty leaves. That or it's not strictly a binary tree.

Plus, you should be getting 5 empty leaf nodes, right? One at 00, one at 010, 0110, etc...

It HAS been a while since I've done that, though. So I don't feel 100% on all of that anymore.

Oh ok, well, I was just checking my work here. It just seemed kinda odd when I looked at the examples I had to work with. Thanks for the prompt response!

It's kinda weird seeing as I'm actually further in my math classes (I've been through all my calculus courses for the moment, I hope) than I am in CS. I really didn't have ANY computer science background when I chose this as my major, so it's been interesting learning stuff. But I kinda feel at this point that I'm just BARELY grasping the material. I've been doing very well, but I hope there won't be a point where everything goes over my head.

Speaking of BARELY grasping, I may consider bringing some linear algebra stuff here in about a week or so... subspaces and spanning and a professor that doesn't really speak clearly is making the class a pain in the neck to sit through.

 

[AltF4]

Lol, yea. We all say around here (ASU that is, not the smashboards!) that getting a Computer Science degree also gives you a minor in Broken English!

I wouldn't be too worried, though. I was fairly similar when I started my undergraduate degree. I had a strong math background, but didn't want to make that my career. I was more interested in computers. I wound up getting doing a math minor with compSci major.

I thought my computer classes were much harder than the math ones, honestly, too. It took a lot of late nights and lost sleep, for sure! Just keep working at it, and if you want it enough, you'll be fine.

 

[Matt]

I read a few posts in this topic and my head exploded. I've spent all afternoon trying to put the pieces back together again. There should be a disclaimer in the title: STEER CLEAR ALL YE ENGLISH MAJORS

Owwwwwwwwwwww