Need Math Help?

[Jane]

thanks so much, druggedfox!

one more for ya ***!

Use the substitution method to find all solutions of the system of equations

x^2 + y^2 = 8
x + y = 0

i got the right answer... just by like plugging **** in and using common sense lol. but i need to know how to ACTUALLY do it

 

[GOD!]

Substitution is just the use of an equation to manipulate a problem.
If you have two variable and two equations, you can solve the equation for both variables.
Same with three variables and three equations
And four and four. Etc.

The trick is to get one variable as a function of other variables and constants.

In this problem, you can either solve for x in terms of other variables and constants
..or solve for y in terms of other variables and constants.

Taking the second equation and subtracting x from both sides, you obtain

x=-y

You also know that, if you multiply each side of the equation by -1, you will get

y=-x


These two equations have to be true; x always equals -y.

You have the variables isolated, so substitute:

(-y)^2 + y^2 = 8

2 y^2 = 8

y^2 = 4 --> y= +2, -2

Take this answer and substitute it into your next equation

x + y = 0

x + (-2) = 0
and
x + 2 = 0

So x = -2, 2

Hope this helps

 

[Druggedfox]

Edit: God! beat me to it. No reason for me to say anything =P

 

[Corpsecreate]

Well actually you wouldnt have to do the substitution after getting y since you already determined that x = -y; so if y = +-2 then x = -+ 2.

 

[GOD!]

but that is a substitution =D

 

[Jane]

i never thought id say this but,

gee, thanks god :3

 

[GOD!]

is that cause im a bad poster

or cause youre a postmodern individual :D

 

[Jane]

hahaha no it has nothing to do with you personally.

i just wanted to be a clever ***** and say that

 

[Kason Birdman]

ok so this question has a **** load of words but imma break it down into just what matters. grade 11 physics btw.

if pulling on a rope attached to a car makes the car travel at 3 m.s.. and then if you tie the rope to a tree and someone sitting in the car pulls it travels again at 3 m/s. If they both get out and pull on the rope how fast does it move?

... lolwut?

do.. do I just double the speed? or what. thats all the information it gives me..
I suppose it could be a thinking question which would explain all the extra details. Or it could just be obvious and straight forward and i'm not getting it..

 

[Kason Birdman]

ok actually they are both pulling on opposite sides. sooo uuuh.. I donno.
Would it be not moving at all in that case?

 

[Merkuri]

Hey is .5 and .500 the same thing? It's really important for this statistic project I'm doing so please let me know. I'll be really grateful.

 

[terr13]

I can't see why they wouldn't be the same thing, unless it's some kind of significant figure type of problem.

 

[Merkuri]

I can't see why they wouldn't be the same thing, unless it's some kind of significant figure type of problem.

Um it's related to statistical significance. You knowq since .05 is threshold is the threshold for whether or not a test is significant.

Does anyone here understand how to interpret Spatial Autocorelation from a Moran's I scatterplot and Randomize permutations? If so let me know ASAP. You'll be saving my project, help me out.

 

[Kason Birdman]

.5 and .500 are technically the same thing.. only .500 is more accurate and its saying that that number has been measured to the extra 00 accuracy, and .5 could have anything after it, its just not accurate enough to tell what they are. Also as terr mentioned .500 would mean you have more sig figures in your final answer

 

[Merkuri]

Great thanks. I've now figured out all the questions I asked before.

Hey when I do a software generated regression analysis. What statistic I am looking for to explain how strong the relationship between the two variables or how much of the deviation in one variable is explained by the other? My presentation is tomorrow so I really need to know this stuff.

 

[Corpsecreate]

Correlation Coefficient. Now for my question:

ln(999/x) / ln(y/100+1) + 1 = N

How can I minimise N and maximise x given these constraints:

1. 1 =< x =< 315
2. 1 =< y =< 110
3. if x >= 150 then y =< 50
4. if x >= 250 then y =< 20
5. if y >= 90 then x =< 60
6. x+y =< 340
7. N is a ceiling value, so 3.1 would be the same as 4.

By minimising N and maximising x i mean that lets say x = 200, y = 105 gives 3.24 for N while x = 210, y = 90 gives 3.43 for N. Since N is a ceiling value, both results are the same however the second set of answers is more desirable as x is 210 as opposed to 200. This particular set of values isnt valid since it ignores some of the restraints but it demonstrates what I mean by minimising N and maximising x.

I dont know how to solve this (is it even possible?), it seems like a linear programming type problem but I'm not sure.

 

[cjugs]

College algebra???????????????
(10-i)+(-2+8i)

 

[Corpsecreate]

add the real parts seperately to the imaginary parts.
(10 - 2) + (-1+8)i
= 8 + 7i

 

[cjugs]

ok thanks 4 your help!

 

[1048576]

Correlation Coefficient. Now for my question:

ln(999/x) / ln(y/100+1) + 1 = N

How can I minimise N and maximise x given these constraints:

1. 1 =< x =< 315
2. 1 =< y =< 110
3. if x >= 150 then y =< 50
4. if x >= 250 then y =< 20
5. if y >= 90 then x =< 60
6. x+y =< 340
7. N is a ceiling value, so 3.1 would be the same as 4.

By minimising N and maximising x i mean that lets say x = 200, y = 105 gives 3.24 for N while x = 210, y = 90 gives 3.43 for N. Since N is a ceiling value, both results are the same however the second set of answers is more desirable as x is 210 as opposed to 200. This particular set of values isnt valid since it ignores some of the restraints but it demonstrates what I mean by minimising N and maximising x.

I dont know how to solve this (is it even possible?), it seems like a linear programming type problem but I'm not sure.

Can't you just brute force it? There's only 315*110 combinations to test, and that's applying constraints 3-6 after the fact. Also, it might be easier to consider constraint 5 as if x >=60, y <= 90, which is equivalent but follows the same format as the others.

 

[kidneybeans]

Correlation Coefficient. Now for my question:

ln(999/x) / ln(y/100+1) + 1 = N

How can I minimise N and maximise x given these constraints:

1. 1 =< x =< 315
2. 1 =< y =< 110
3. if x >= 150 then y =< 50
4. if x >= 250 then y =< 20
5. if y >= 90 then x =< 60
6. x+y =< 340
7. N is a ceiling value, so 3.1 would be the same as 4.

By minimising N and maximising x i mean that lets say x = 200, y = 105 gives 3.24 for N while x = 210, y = 90 gives 3.43 for N. Since N is a ceiling value, both results are the same however the second set of answers is more desirable as x is 210 as opposed to 200. This particular set of values isnt valid since it ignores some of the restraints but it demonstrates what I mean by minimising N and maximising x.

I dont know how to solve this (is it even possible?), it seems like a linear programming type problem but I'm not sure.

Yeah, I'd think you should use linear programming. The if/then statements make it questionable though.

 

[1048576]

Also you can ignore constraint 6, since it's taken by the other ones.

Edit: Didn't spend too much time on it, so I could be wrong, but I get 149 with this:


import java.lang.Math;
public class Mathproblem
{
public static void main(String[] args)
{
int N = 10;
int xmax = 0;
for (int x = 1; x <= 59; x++)
{
for (int y = 1; y <= 110; y++)
{
double dummy1 = ((double)(999)/x);
double dummy2 = ((double) y)/100 + 1;
N = (int) Math.min(N,Math.ceil(Math.log(dummy1)/Math.log(dummy2) + 1));
if (Math.ceil(Math.log(dummy1)/Math.log(dummy2) + 1) == N )
{
xmax = Math.max(x, xmax);
}
}
}

for (int x = 60; x <= 149; x++)
{
for (int y = 1; y <= 90; y++)
{
double dummy1 = ((double)(999)/x);
double dummy2 = ((double) y)/100 + 1;
N = (int) Math.min(N,Math.ceil(Math.log(dummy1)/Math.log(dummy2) + 1));
if (Math.ceil(Math.log(dummy1)/Math.log(dummy2) + 1) == N )
{
xmax = Math.max(x, xmax);
}
}
}
for (int x = 150; x <= 249; x++)
{
for (int y = 1; y <= 50; y++)
{
double dummy1 = ((double)(999)/x);
double dummy2 = ((double) y)/100 + 1;
N = (int) Math.min(N,Math.ceil(Math.log(dummy1)/Math.log(dummy2) + 1));
if (Math.ceil(Math.log(dummy1)/Math.log(dummy2) + 1) == N )
{
xmax = Math.max(x, xmax);
}
}
}
for (int x = 250; x <= 315; x++)
{
for (int y = 1; y <= 20; y++)
{
double dummy1 = ((double)(999)/x);
double dummy2 = ((double) y)/100 + 1;
N = (int) Math.min(N,Math.ceil(Math.log(dummy1)/Math.log(dummy2) + 1));
if (Math.ceil(Math.log(dummy1)/Math.log(dummy2) + 1) == N )
{
xmax = Math.max(x, xmax);
}
}
}
System.out.print(xmax);
}

}

N = 4

 

[Merkuri]

Can someone please explain to me the difference between correlation plot slope's value and the R-squared of a regression line. What makes these two different?

It'd be really cool if we could try to respond to each others' post more quickly. By the time some has answered the asker usually has little need for it anymore.

 

[Derkis]

Can someone please explain to me the difference between correlation plot slope's value and the R-squared of a regression line. What makes these two different?

It'd be really cool if we could try to respond to each others' post more quickly. By the time some has answered the asker usually has little need for it anymore.

As far as I know...

The slope of the plot's linear regression tells you nothing more than it's direction. R-squared, on the other hand, is a measurement of how reliably said regression fits the data, ranging from 0 to 1 (with 1 being a perfect fit), and calculated: R^2 = 1 - (error^2 / dataTotal^2). My stats is kind of rusty, so this might not be totally correct.

 

[ballin4life]

Yeah slope doesn't matter. Ten points in a line with slope .00001 and ten points in a line with slope 100 both would have correlation coefficient 1. The correlation coefficient is about how far on average the points are from the best fit line.

Also post your questions earlier if you want to make sure they get answered by the time you need it. No one has an obligation to answer anyone's questions.

 

[MintyFlesh]

What's the difference between y =tan(x) and y=2tanx?

also, is the recirpicol of y = sin(x) the same thing as y=arcsin(x)?

 

[terr13]

2tan(x) is twice tan(x). so y = 2tan(x) => y/2 = tan(x)

arcsin is the inverse of sin, so y =sin(x) => arcsin(y) = x.
reciprocal of sin(x) is 1/sin(x).

 

[Airgemini]

So, I'm stuck on my course selections for a math class senior year. I don't HAVE to take a math class, only three years are required, but I want to look impressive for college applications. Now, while I want to look impressive I don't want to fail the class due to difficulty lol. Math isn't my strongest point, specifically geometry, but algebra is pretty easy.

Anyways, I have a choice of either Pre-Calculus or Trigonometry/Discrete Math (combined course). I know in some places trig and precal are combined or whatever but for some reason they're not here. Anyways, which is easier is mainly what I'm asking. In your experience.

 

[terr13]

Maybe you're not that bad at math if you think algebra is easy. Geometry (the one with proofs) is a bull**** class that should never be taught. If you want a more solid background fir college then go for precalc, otherwise trig/discrete is probably more interesting

 

[Ocho(*8*)]

I think trig/ discrete will probably be the easiest of those.

and if you want to go further in math youll need multiple of those so you might as well just take the trig one now.

 

[Kason Birdman]

heyro.. phsics..

A portable radio is designed to operate at 8.0 V and a current of 400 mA. However, the onlu source available is 12V. What resistance must be added in series to make the radio operate properly?

 

[Rici]

U= I . R

4 = 0.4 . R

R= 4/0.4=10 ohm.

 

[shiva39]

This is supposed to be code for forward euler. I'm really overlooking something simple here though. Anyone know what I'm missing?

function yout = feuler(F, y0, t0, tfinal, N)

%step size
dt = (tfinal-t0)/N;
yold = y0;
t = t0;

%iterate for N steps.
for j=1:N
ynew = yold + dt*F(t,yold);
t = t+dt;
yold = ynew;
end

yout = y;
end




Also, I'm really having a lapse in how to solve second order differentials, e.g.
y'' = -y

Is x = (x2, -x1) the correct solution?

 

[Ocho(*8*)]

what language is that? I only know some c++ but ill tell you what i notice and it will probably be wrong but just maybe it will be helpful:

the 2nd to last line - yout = y;
it seems to me like you never defined y/ told the computer what y was.

also, in c++ at the end of a for loop you have to tell the computer to increase the counting variable (in this case you used j). So at the end of the loop i'd do something like "j = j+1;"
but maybe thats not necessary at all, idk

 

[shiva39]

Uh, well it's whatever language Matlab uses, I believe that's objective C? Anyway, I don't think it has anything to do with the syntax etc. That line (yout = y) should have been (yout = ynew), it was just a typo. Something's wrong with the algorithm.

Thanks though.

Edit

Welp, good news on one count. Turns out the euler algorithm is fine exactly as is.

 

[Merkuri]

Anyone know what the ! symbol means when it's used in a math equation? Please let me know and thank you.

 

[Druggedfox]

Its factorial.

3! (read: Three factorial) = 3*2*1 = 6

5! = 5*4*3*2*1 = 120

I gave an example rather than actually formally define it cuz im lazy

 

[forward]

......forward euler.......

That'll be my math name

 

[shiva39]

Ha lol, you namesearcher you

 

[ballin4life]

Also, I'm really having a lapse in how to solve second order differentials, e.g.
y'' = -y

Is x = (x2, -x1) the correct solution?

Two equivalent ways to solve this. The first is to write z=y' and then solve the matrix differential equation

y' = (0 1) (y)
z' = (-1 0) (z)

Which can be solved using matrix exponentials. This is the more general way of doing it.

But there's also a slightly faster way for these simple linear ones.

We know y'' + y = 0 (this way works primarily when you can get the right side to be 0).

This gives a characteristic polynomial of t^2 + 1 = 0. This polynomial has roots of i and -i.

Then the solution is always of the form y = ae^(ix) + be^(-ix).

Which gives a(cosx + isinx) + b(cosx - isinx) = (a+b)cosx + (a-b)isinx

You can simplify this to

y = d(cosx)+c(sinx)

By choosing a and b so that a+b=d and a-b=c/i i.e. a = .5(d-ic) and b = .5(d+ic)


So your final solution is y = dcosx+csinx where d and c are arbitrary constants.

This makes sense too since we know the second derivative of cos is -cos and the same for sin.