I understood your first sentence. I can't even begin to grasp the second. Just trying to work out the dimensions for boxes with the most volume to send stuff home from Japan, and the post office keeps telling me I'm doing it wrong.
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The Math Help Thread
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I understood your first sentence. I can't even begin to grasp the second. Just trying to work out the dimensions for boxes with the most volume to send stuff home from Japan, and the post office keeps telling me I'm doing it wrong.
By largest you mean largest volume right?
Probably not difficult for any of you lot, but I'm trying to figure out the largest box I can send using the above rules.
Any ideas for vaguely "normal" dimensions?
You probably want to find an expression for the largest possible height and width in terms of the length to give you an expression for the volume in terms of length, probably a cubic. I think having the width and height equal will maximise the cross section which simplifies things a lot.
You can differentiate your expression for the volume to get the values of the length where it reaches a maximum and a minimum which should give you the result. I did it quickly and got length = 1m, width = height = 0.5m, volume = 0.25m.
Edit, oh so this isn't a school problem. Yeah try length 1m, width = height = 0.5m
I get it now.
So as long as the length is 1m and the width and height added together make 0.5m I'm golden; great, thanks!
Actually, he meant width is 0.5m, height is 0.5m. That is the optimal solution as well.
But doesn't 1m+(0.5+0.5) x2 = 4m?
Not if it follows BIDMAS. It'd be
1 + (0.5 + 0.5)*2
=1 + 1*2
=1 + 2
=3
Otherwise with the way they've put in brackets it wouldn't make sense.
Righto. I'll call up tomorrow and confirm just to make sure. Cheers!
A good poster
Banned
no i got it, thanks
youngwerther
Banned
I have 5 optimization problems for studying and I don't know how to do any of them.
The easiest one is probably this one
Find the point on the line -3x+6y+5=0 that is closest to the point (5,3).
Do you just use distance formula, solve for one variable in terms of another, differentiate...then what
The easiest one is probably this one
Find the point on the line -3x+6y+5=0 that is closest to the point (5,3).
Do you just use distance formula, solve for one variable in terms of another, differentiate...then what
Passive Assassin
Member
Anyone good with stats? I'm stuck on these two practice problems for a test that's going to happen tomorrow. These questions deal with confidence intervals for a large sample population.
19. An investigator computes a 95% confidence interval
for a population mean on the basis of a sample of
size 70. If she wishes to compute a 95% confidence
interval that is half as wide, how large a sample does
she need?
A 95% Za/2 value for a two sided CI is 1.96. The mean is unknown. The sample size is 70. The equation I'm supposed to use is (Za/2)(S)/sqrt
=E. Or N=((Za/2*S)/E)^2.
How would I find S? I'm going to assume E is 1 or .5 since the interval is half as wide.
21. Based on a large sample of capacitors of a certain
type, a 95% confidence interval for the mean capacitance,
in μF, was computed to be (0.213, 0.241).
Find a 90% confidence interval for the mean capacitance
of this type of capacitor.
95% CI is 1.96. 90% CI is 1.645. The n and sample size is unknown. But I can find the mean by (.213+.241)/2 = .227. It feels like I'm supposed to use .227 +- 1.645(S/sqrt) but can't find those values.
19. An investigator computes a 95% confidence interval
for a population mean on the basis of a sample of
size 70. If she wishes to compute a 95% confidence
interval that is half as wide, how large a sample does
she need?
A 95% Za/2 value for a two sided CI is 1.96. The mean is unknown. The sample size is 70. The equation I'm supposed to use is (Za/2)(S)/sqrt
=E. Or N=((Za/2*S)/E)^2.How would I find S? I'm going to assume E is 1 or .5 since the interval is half as wide.
21. Based on a large sample of capacitors of a certain
type, a 95% confidence interval for the mean capacitance,
in μF, was computed to be (0.213, 0.241).
Find a 90% confidence interval for the mean capacitance
of this type of capacitor.
95% CI is 1.96. 90% CI is 1.645. The n and sample size is unknown. But I can find the mean by (.213+.241)/2 = .227. It feels like I'm supposed to use .227 +- 1.645(S/sqrt
) but can't find those values.Never heard of BIDMAS. We teach PEMDAS here, actually it's Please Excuse My Dear Aunt Sally From Leaving The Room. FLTR is From Left To Right. What do the B and I stand for in BIDMAS? Nevermind. Googled it and it's Brackets and Indices. PEACE.
rearrange your equation to get y=(-5+3x)/6
Now a generic point on your line is (x,y) or (x, (3x-5)/6)
Use distance formula,
d^2=(x-5)^2 + (y-3)^2
Sub in y for what we found
d^2=(x-5)^2 + (((3x-5)/6) - 3)^2
Differentiate with respect to x
2dd'=2(x-5) + 2(((3x-5)/6) - 3)*(3/6)
Set d' to 0.
Now you can solve for x, find x.
Then plug in x into y=(-5+3x)/6 to get y.
This gives the coordinate you need.
im having some difficulty figuring out the inverse of functions .. so i have to find the inverse of
f(x) = x + 8.
the book is telling me the correct answer is
f^-1(x) = x - 8
so .. just looking at that simple one, it's the opposite! yay! but what steps do i actually have to take to get there?
f(x) = x + 8.
the book is telling me the correct answer is
f^-1(x) = x - 8
so .. just looking at that simple one, it's the opposite! yay! but what steps do i actually have to take to get there?
Think of f(x) as y and rewrite so it's x=gwhere g is some function of y. Then switch y to x and x to f^-1.
so starting with f(x) = x + 8 .. switch it f(x) to y so i have
y = x + 8 .. then solve for x? so
y - 8 = x
then just .. switch the variables?
y = x - 8
then that's called f^-1(x) = x - 8?
Hey, trig question here. Does anyone know how to graph equations like
y=3sec(2x-pi/4)?
I know to graph cosx, then 3secx, then use the numbers I get from that, and I know to put it in shifted form (y = 3sec(2(x-pi/8))) but I have no idea what to do after that. Do I shift my Xs and then multiply them 2? Do I shift and then divide by 2? How do I use shifted form here? I'm confused.
EDIT: Okay I figured this out. I should've done the horizontal compression near the beginning and not at the end.
.
y=3sec(2x-pi/4)?
I know to graph cosx, then 3secx, then use the numbers I get from that, and I know to put it in shifted form (y = 3sec(2(x-pi/8))) but I have no idea what to do after that. Do I shift my Xs and then multiply them 2? Do I shift and then divide by 2? How do I use shifted form here? I'm confused.
EDIT: Okay I figured this out. I should've done the horizontal compression near the beginning and not at the end.
.
kidtamagotchi
Member
Not sure if this is a trick question:
A carpenter is installing a baseboard around the a room that has a length of 20ft and width of 15 ft. The room has four doorways and each doorway is 3 feet wide. If no baseboard is to be put across doorways and the cost of the baseboard is $1.40 per foot, what is the cost, to the nearest dollar, of installing the baseboard around the room?
The area of the room comes out to 300 square feet, so would I have to square the $1.40, or leave it as it is before I multiply it with the area of the room?
A carpenter is installing a baseboard around the a room that has a length of 20ft and width of 15 ft. The room has four doorways and each doorway is 3 feet wide. If no baseboard is to be put across doorways and the cost of the baseboard is $1.40 per foot, what is the cost, to the nearest dollar, of installing the baseboard around the room?
The area of the room comes out to 300 square feet, so would I have to square the $1.40, or leave it as it is before I multiply it with the area of the room?
hemtae
Member
I'm not sure what you're doing with the area, but a baseboard is the thing that wraps around the bottom of your walls. So you would use the perimeter minus the doorways.
kidtamagotchi
Member
That is right! I am confusing baseboard with floor board! Thanks for the heads up!
Hey gaf I need help with this:
Let D be the set of all finite subsets of positive integers, and define T: Z+ --> D by the rule: for all integers n, T = the set of all the positive divisors of n.
Is T one-to-one? prove/counterexample
Is T onto? prove counter example
It gives the hint that T is 1-1 and not onto but I'm still having trouble.
Let D be the set of all finite subsets of positive integers, and define T: Z+ --> D by the rule: for all integers n, T
= the set of all the positive divisors of n.Is T one-to-one? prove/counterexample
Is T onto? prove counter example
It gives the hint that T is 1-1 and not onto but I'm still having trouble.
Hey gaf I need help with this:
Let D be the set of all finite subsets of positive integers, and define T: Z+ --> D by the rule: for all integers n, T
Is T one-to-one? prove/counterexample
Is T onto? prove counter example
It gives the hint that T is 1-1 and not onto but I'm still having trouble.
It's one to one because if a and b are distinct positive integers with a < b, then b is a divisor of b but not of a, so b is in T(b) but not in T(a), so T(a) doesn't equal T(b).
It's not onto because 1 is a divisor of every positive integer, so any subset of Z+ which doesn't include 1 cannot be in the image of T.
hey math dudez. i could use some help.
The amount of money in an investment is modeled by the function
A(t) = 800(1.065)^t
it asks me a few questions, which i've answered correctly , but this one is throwing me off. it sez: the annual rate of change in the balance is r = _______ or r = _______ %.
i need to fill in both of these blanks. what is it asking? isn't the rate of change 1.065? what goes in the box with the % sign? what is happening!?
The amount of money in an investment is modeled by the function
A(t) = 800(1.065)^t
it asks me a few questions, which i've answered correctly , but this one is throwing me off. it sez: the annual rate of change in the balance is r = _______ or r = _______ %.
i need to fill in both of these blanks. what is it asking? isn't the rate of change 1.065? what goes in the box with the % sign? what is happening!?
Got a question
10. A rock is thrown up from the surface of the moon at time t = 0 and its height above the moons surface is given by feet after t seconds. What is the highest elevation of the rock attained above the moon?
A) 16 feet
B) 22 feet
C) 24 feet
D) 27 feet
E) 32 feet
Thanks in advance
10. A rock is thrown up from the surface of the moon at time t = 0 and its height above the moons surface is given by feet after t seconds. What is the highest elevation of the rock attained above the moon?
A) 16 feet
B) 22 feet
C) 24 feet
D) 27 feet
E) 32 feet
Thanks in advance
There isn't enough information to answer this question. What is the rock's initial upward velocity?
thequickandthedead
Member
Can anyone help me with this?
The quadratic equation x^2+kx+2k = 0, where k is a non zero constant, has roots α and β. Find a quadratic equation with roots α/β and β/α
The quadratic equation x^2+kx+2k = 0, where k is a non zero constant, has roots α and β. Find a quadratic equation with roots α/β and β/α
start with a quadratic equation with roots α ,β:
(α - c)(x - β
= x^2 - (α + βx + cβ. Now, identify coefficients. Since α and β are your roots, what must k be?
For a quadratic polynomial whose leading coefficient is one, the coefficient of 'x' is always the negative of the sum of the roots, and the constant term is the product of the roots. So α + β = -k and α * β = 2k. Now use this to write a quadratic with leading coefficient = 1 which has roots α/β and β/α. The constant term has to be 1 because α/β * β/α = 1, and the coefficient of 'x' is -(α/β + β/α
. You just need to rewrite α/β + β/α so that you can write it in terms of k. To do this, give them a common denominator of αβ and then rewrite the numerator as a function of (α + β and αβ, so that you can rewrite everything in terms of k.Yeah I think it might have to plug in freefall gravity which is -32 something but I forgot.
Anyways, I have one more.
12. Car A and Car B started at the same intersection, X. Car A drove North at a speed of 40 mph and Car B drove East at a speed of 32 mph. How fast is the distance between them changing ½ hour after they left the intersection?
Capt.Haddock
Member
The distance between them at 1/2 an hour is 25.612m.increasing at the rate of 0.853m/min
hemtae
Member
I get 51.22 mph by differentiating the Pythagorean theorem with respect to time.
thequickandthedead
Member
Thanks for your help guys. Does anyone have any experience with proof by induction?
Capt.Haddock
Member
which proof.
thequickandthedead
Member
An example would be the first question on this paper:
http://www.ocr.org.uk/Images/61394-question-paper-unit-4725-01-further-pure-mathematics-1.pdf
How did you guys get the answer though
Im just having trouble putting down the variables for a squared + b squared = c squared. jeje sorry
Capt.Haddock
Member
That is quite easy :v
let n=1
1(1+1)=2=[1*(1+1)*(1+2)]/3 true for n=1
let n=2
1(1+1)+2(2+1)=8=[2*(2+1)*(2+2)]/3 true for n=2
let n=3
1(1+1)+2(2+1)+3(3+1)=20=[3*(3+1)*(3+2)]/3 true for n=3
going by this let it is ture for n=m;
∑r(r+1)=[n(n+1)(n+2)]/3 r=1(1)m m>=1
now we have to show that it is true for n=m+1
let b= ∑r(r+1) ;r=1(1)m
so for n=m+1
∑r(r+1) r=1(1)m+1
=b+[m+1(m+2)]
=[m*(m+1)*(m+2)]/3+[m+1(m+2)]
=(m+1)(m+2)[(m/3)+1]
=(m+1)(m+2)[(m+1+2)/3]
=[(m+1)(m+1+1)(m+1+2)]/3
so it is proved for n=m+1,m>=1
so it is proved for all n by the method of induction.
a=distance covered by car A at 1/2 hr=40/2=20
b=second car=32/2=16
No, you shouldn't need any linear algebra for calc 1. Later on you may do vector calculus but that's later.
I'm not sure if this is a valid approach or what they're looking for but one way (and easiest...given if it's correct) I think is just to consider two right angled vectors of magnitude 40 and 32 (so velocity vectors) and then just find the hypotenuse vector magnitude from pythagoras' theorem. I'm not sure if this is what they're looking for though since I do remember doing similar questions in high school and I think we used a totally different approach.
Thanks, I'll wait then.
No, I solved it and it was a pain in the ass because we did not go over that particular way of solving it.
12. Car A and Car B started at the same intersection, X. Car A drove North at a speed of 40 mph and Car B drove East at a speed of 32 mph. How fast is the distance between them changing ½ hour after they left the intersection?
so we know (dsA/dt)=40mph and (dsB/dt)=32mph
Distance=rate x time
Car A = 40 x 1/2 = 20
Car B = 32 x 1/2 = 16
Dsquared = Asquared + B squared
Dsquared = (20)^(2) + (16)^(2)
D=squareroot((20)^(2) + (B)^(2))
We want to find (d/dt)
Dsquared = (20)^(2) + (16)^(2)
f'(x) =2D(d/dt)=2(20)(dsA/dt)+2(16)(dsB/dt)
f'(x)=2(squareroot((20)^(2) + (16)^(2)))(d/dt)=2(20)(40)+2(16)(32)
(d/dt)=2(20)(40)+2(16)(32)/2(squareroot((20)^(2) + (16)^(2)))
2s cancel
=1312/squareroot(656)
=51.225m/h
Thanks to everyone that helped me out this semester. I really appreciate it.
Edit:
this one was tricky too.
You get the function take derivative set equal to zero. Get value and plug it back into original equation.
12. Car A and Car B started at the same intersection, X. Car A drove North at a speed of 40 mph and Car B drove East at a speed of 32 mph. How fast is the distance between them changing ½ hour after they left the intersection?
so we know (dsA/dt)=40mph and (dsB/dt)=32mph
Distance=rate x time
Car A = 40 x 1/2 = 20
Car B = 32 x 1/2 = 16
Dsquared = Asquared + B squared
Dsquared = (20)^(2) + (16)^(2)
D=squareroot((20)^(2) + (B)^(2))
We want to find (d/dt)
Dsquared = (20)^(2) + (16)^(2)
f'(x) =2D(d/dt)=2(20)(dsA/dt)+2(16)(dsB/dt)
f'(x)=2(squareroot((20)^(2) + (16)^(2)))(d/dt)=2(20)(40)+2(16)(32)
(d/dt)=2(20)(40)+2(16)(32)/2(squareroot((20)^(2) + (16)^(2)))
2s cancel
=1312/squareroot(656)
=51.225m/h
Thanks to everyone that helped me out this semester. I really appreciate it.
Edit:
this one was tricky too.
You get the function take derivative set equal to zero. Get value and plug it back into original equation.
thequickandthedead
Member
Out of curiosity, has anyone here done the infamous STEP exams?
For Cambridge admissions?
I did STEP I to get into Warwick.
Main piece of advice would be to use these booklets written by the guy who used to run it, I think his name was Stephen something. Can't remember his surname exactly, I think it was Greek. Incredibly useful.
thequickandthedead
Member
Yeah that's the one. I pretty much just used that and the past papers.
hey gaf
i have this problem
i know to factor out 2 and divide both sides by it to get -sqrt(2)/2, but after that I'm stuck. I've logged in and out of the program to try to spawn different questions to get a better idea of what it's actually asking me (i.e double angle identity or angle multiplied by a constant), at first it was 2 (cos2x, sin2x etc) instead of four, so I jerked around a bit with the double angle identity and I got nowhere. please help. Also, in other spawns it was a mixup of cos and sin, so 2cos2xsinx-2sin2xcos = -1, etc.
i have this problem
i know to factor out 2 and divide both sides by it to get -sqrt(2)/2, but after that I'm stuck. I've logged in and out of the program to try to spawn different questions to get a better idea of what it's actually asking me (i.e double angle identity or angle multiplied by a constant), at first it was 2 (cos2x, sin2x etc) instead of four, so I jerked around a bit with the double angle identity and I got nowhere. please help. Also, in other spawns it was a mixup of cos and sin, so 2cos2xsinx-2sin2xcos = -1, etc.
Use the identity cos(A - B) = cosAcosB + sinAsinB
oh wow, thanks. i'd almost completely forgot about that.
kidtamagotchi
Member
It's me again! I need help with this statement:
If I am tired or hungry, then I cannot concentrate.
I can concentrate.
_________________________________________
Therefore I am neither tired nor hungry.
It would be: [(p or q) -> ~r ^ r] -> ~(p ?q)
I don't know what symbol to use for nor.
If I am tired or hungry, then I cannot concentrate.
I can concentrate.
_________________________________________
Therefore I am neither tired nor hungry.
It would be: [(p or q) -> ~r ^ r] -> ~(p ?q)
I don't know what symbol to use for nor.
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