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[0xCA2]

Yeah so I walked out of Calculus recitation (where we have to work in groups to solve problems that are 1000x more abstract / ambiguous than anything else we've done) because I was so confused and frustrated. Everyone is faster than me, and because of the time constraints I feel like I'd be holding everyone back by constantly asking questions, as well as a little embarrassed.* So I just sit there, looking at the problems and quietly trying to understand them, asking a question or contributing here or there, while everyone else In the group speeds me by.

I feel like in order to do the problems given I need to sit down, alone, and think about the questions for a long while and do some trial and error. I can't just blaze through them in an hour, and I hate the pressure of having to contribute to a group in that situation. I could just ask everyone for the answers just to get the points (the smart thing to do) but then I'd still feel confused and kind of awkward, plus recitation worksheets are only worth 50 points out of 750 total anyway.

I went to a tutor later on and got help, but I still don't know what to do about Calculus recitation. I fucking hate it. Anyone have any advice? NOTE: I'm fine with the material in the book and the homework, it's just these worksheets in recitation that get me.

*Because I have a disability, I grew reluctant to asking people for help, especially in school, because when you ask too often they began to assume that you're mentally handicapped (which, sure enough, the calc teacher seemingly thinks I am) . As an adult I've tried to care less because that's a petty reason to let yourself get a bad grade, but I still fall back into this pattern.

 

[Orbis Tabula]

It's hard to really give you advice because everyone learns differently and I could be telling you something that you've tried before. But I'll give you my recommendations. I TA'd and tutored college level calculus for a couple semesters and I think the most important thing that contributes to success in calc, more than any other math you do before, is practice. Being proficient at calculus is not the same as being good at say Algebra. For all the Algebra you usually do until this point, it's essentially an algorithm. You get a problem like "solve for x" or "find the y-intercept" and you know how to do it. Step 1, step 2, etc. It's all steps to follow with little to know variation. If you memorize the algorithm, you can solve any answer.

Calculus is taught differently. In calculus, you're rarely given an algorithm. Instead, you are given an enormous set of tools and with every problem, you have to figure out with tools work. And it's complicated because some tools are useless in a problem and you can't learn that until you try. And some problems require being tinkered with before any rule applies. The only way to become proficient at picking the right tools is to develop a sense for it, and that sense only develops with practice. Sometimes you might need product rule, sometimes chain rule, sometimes substitution. It's difficult. But as you keep doing problems you'll start to pick up patterns about what a product rule problem looks like, etc. Because of this change, calculus comes across a lot less "mathy" than most math students are used to. More intuitive and holistic than analytic. But the number one most important thing to learn in trying to develop this sense is this.

Do not be afraid to do the wrong thing.

Calculus is your first step into doing real mathematics. Breaking from the algorithm. And the only way you'll learn anything in mathematics is to fail frequently. In fact, when I do my math homework, I always have two sheets going. I spend one sheet trying a problem in every way I can, butting my head against it. And as soon as I find the way that works, I transpose that on to the other sheet, but only the successful method. So in the end, I'll turn in one sheet of success and actually have 6 pages of failed attempts. Failure is how you learn and develop this Calculus sense. If you're truly committed to getting this stuff, spend some leisure time going through problems from the book that the teacher didn't assign. The more practice you get, the better you will get at recognizing these patterns. Since you're not turning these problems in, you can fail to your hearts content.

If you have any examples of the kinds of problems you struggle with, we might be able to offer more specific help.

 

[Dramos]

Calling out Discrete Math gaffers,

I need someone to verify my direct proof for : ((p ∨ r) ∧ (q ∨ r)) ≡ ((p ∧ q) ∨ r)
My approach: LHS ⇒ RHS and RHS ⇒ LHS

LHS ⇒ RHS :
((p ∨ r) ∧ (q ∨ r)) ≡ r ∨ (p ∧ q) by Distributive Law
(p ∧ q) is true ⇒ p is true by Simplification
(p ∧ q) is true ⇒ q is true by Simplification
r ∨ (p ∧ q) is true, p is true and q is true ⇒ r is either true or false

RHS ⇒ LHS:
...

I need feedback, I'm not quite sure how to write an valid and appropriate proof.

 

[Eye for an Eye]

So this question was on a quiz, but because only one student got the answer correct, the teacher didn't mark it. He also didn't give us the answer, and told us to work on it as homework.

I believe I finally got the question correct, but want to make sure it's right. It took me about a good 30 minutes or so to work it out, there's no way I would've completed on the quiz in a limited time.

Sorry for the iphone picture, a little too lazy to get out the scanner. Ignore the numbers on the quiz sheet, my correct answers are in the second picture.

 

[Brakke]

WHAT IS 920 CHICKENS PLUS ONE GOLF CLUB EQUAL

Intruder uses golf club to kill 920 Foster Farms chickens.

 

[boviscopophobic]

So this question was on a quiz, but because only one student got the answer correct, the teacher didn't mark it. He also didn't give us the answer, and told us to work on it as homework.

Check again! It says that 120 members wash cars or bake cupcakes, not that the total number of all members is 120.

 

[0xCA2]

So I'm supposed to find the derivative of y = sqrt(x+y) (we're learning about implicit differentiation).

I understand that you're supposed to use the chain rule, the first part of it (f'(g(x)) is 1 / (2sqrt(x+y)). What I kind of don't get is why the derivative of x + y is 1 + dy/dx , and how exactly that leads to dx/dy being equal to 1 / (2sqrt(x+y)-1) (the solution).

Could someone explain this to me? thanks.

 

[Yrael]

So I'm supposed to find the derivative of y = sqrt(x+y) (we're learning about implicit differentiation).

I understand that you're supposed to use the chain rule, the first part of it (f'(g(x)) is 1 / (2sqrt(x+y)). What I kind of don't get is why the derivative of x + y is 1 + dy/dx , and how exactly that leads to dx/dy being equal to 1 / (2sqrt(x+y)-1) (the solution).

Could someone explain this to me? thanks.

y = sqrt(x+y)
-> y^2 = x + y

Take the derivative of both sides with respect to x:

LHS:
d/dx(y^2) = d/dy(y^2) * dy/dx (chain rule)
= 2y * dy/dx

RHS:
d/dx(x + y) = d/dx(x) + d/dx (taking derivative of each term separately)
= 1 + dy/dx (dx/dx = 1, d/dx = dy/dx)

Hence, we have 2y dy/dx = 1 + dy/dx - rearrange to get: dy/dx = 1/(2y - 1)
Since y = sqrt(x + y),

dy/dx = 1/(2sqrt(x+y) - 1)

 

[fireside]

So I'm supposed to find the derivative of y = sqrt(x+y) (we're learning about implicit differentiation).

I understand that you're supposed to use the chain rule, the first part of it (f'(g(x)) is 1 / (2sqrt(x+y)). What I kind of don't get is why the derivative of x + y is 1 + dy/dx , and how exactly that leads to dx/dy being equal to 1 / (2sqrt(x+y)-1) (the solution).

Could someone explain this to me? thanks.

since you don't know what the derivative of y with respect to x is, you are using dy/dx as a placeholder. it is simply notation. you are using the chain rule for the square root, so the the derivative of the inside is 1+dy/dx because the derivative of x with respect to x is just one, and the derivative of y with respect to x is the unknown. then you are just solving for dy/dx

here's a picture

 

[0xCA2]

Thanks guys.

 

[Eye for an Eye]

Check again! It says that 120 members wash cars or bake cupcakes, not that the total number of all members is 120.

But even then, the numbers overlap. So some that washed cars or baked cupcakes also sold candy. Even if a 120 members washed cars or baked cupcakes, 75 of those also sold candy. 15 is the starting point, because that's the group that did all three. So every number out from that deducts 15 to include those members that did all three.

 

[SenorArdilla]

So this question was on a quiz, but because only one student got the answer correct, the teacher didn't mark it. He also didn't give us the answer, and told us to work on it as homework.

I believe I finally got the question correct, but want to make sure it's right. It took me about a good 30 minutes or so to work it out, there's no way I would've completed on the quiz in a limited time.

Sorry for the iphone picture, a little too lazy to get out the scanner. Ignore the numbers on the quiz sheet, my correct answers are in the second picture.

Feel free to correct me if I got something wrong.

W(total) = (# of people who washed cars) = 70
B(total) = (# of people who baked cupcakes) = ?
S(total) = (# of people who sold candy) = 80
N(total) = (# of people who did nothing) = 15

W(only) = (# of people who only washed cars) = 10
B(only) = (# of people who only baked cupcakes) = ?
S(only) = (# of people who only sold candy) = ?

WB = (# of people who washed cars and baked cupcakes) = 20
WS = (# of people who washed cars and sold candy) = ?
BS = (# of people who baked cupcakes and sold candy) = 35

WBS = (# of people who washed cars, baked cupcakes, and sold candy) = 15

It tells you that 120 people either washed cars or baked a cake but not both so we get:
120 = W(total) + B(total) - 2WB - 2WBS

Plug in the given variables and solve for the unknown:
120 = 70 + B(total) - 2(20) - 2(15)
B(total) = 120

From the Venn Diagram, we can conclude that the total number of members is the sum of all the parts. So, plug in the given variables and solve for the unknown:
B(total) = B(only) + WB + BS + WBS
120 = B(only) + 20 + 35 + 15
B(only) = 50

Similarly:
W(total) = W(only) + WB + WS + WBS
70 = 10 + 20 + WS + 15
WS = 25

And with WS, we can now solve for this equation:
S(total) = S(only) + WS + BS + WBS
80 = S(only) + 25 + 35 + 15
S(only) = 5

So you have:
W(total) = (# of people who washed cars) = 70
B(total) = (# of people who baked cupcakes) = 120
S(total) = (# of people who sold candy) = 80
N(total) = (# of people who did nothing) = 15

W(only) = (# of people who washed cars) = 10
B(only) = (# of people who only baked cupcakes) = 50
S(only) = (# of people who only sold candy) = 5

WB = (# of people who washed cars and baked cupcakes) = 20
WS = (# of people who washed cars and sold candy) = 25
BS = (# of people who baked cupcakes and sold candy) = 35

WBS = (# of people who washed cars, baked cupcakes, and sold candy) = 15

And finally, the total number of members is the sum of each component plus the non-participating members:
W(only) + B(only) + S(only) + WB + WS + BS + WBS + N(total)= ?
10 + 50 + 5 + 20 + 25 + 35 + 15 + 15 = 175

 

[simplayer]

Calling out Discrete Math gaffers,



I need feedback, I'm not quite sure how to write an valid and appropriate proof.

I'm not quite sure what you're looking for. Either which way you go (LHS->RHS or RHS->LHS) is straight up distributive law.

You can always use a truth table to show that they're equivalent too I guess, but that seems a bit like of a kludge.

 

[Dramos]

I'm not quite sure what you're looking for. Either which way you go (LHS->RHS or RHS->LHS) is straight up distributive law.

You can always use a truth table to show that they're equivalent too I guess, but that seems a bit like of a kludge.

Yeah truth table is certainly a way to prove the equivalency however I've just learned proofs recently in class and we need to use rules of inference to demonstrate.

 

[Jarmel]

Not Math Math but does anyone have a link or book that explains all the basic equations needed for a college level 101-102 Chem course?

Stuck doing this bullshit chem course for Engineering and the course is so damn fast that it's hard to keep everything straight.

 

[Granadier]

Brain fart time

Would someone explain to me why the input can be rewritten as the result here?

I understand that sqrt(x) == x^1/2 and where to go with the x^5/2 for the derivative,
but for some reason I my mind isn't working to figure out how to go from (x^3) / sqrt(x) to x^5/2

Thanks much!

edit: Is it because it can be rewritten as x^3/1 * x^2/1 -> x^5/2 ..... that looks wrong.... dumb stuff

 

[Eye for an Eye]

Not Math Math but does anyone have a link or book that explains all the basic equations needed for a college level 101-102 Chem course?

Stuck doing this bullshit chem course for Engineering and the course is so damn fast that it's hard to keep everything straight.

Does your school offer a "pre-chem" course? My school recommended to take that course if you didn't do well in high school or didn't take it high school. It was pretty beneficial for figuring out the math stuff of chem, plus some of the basic formulas/compounds.

Feel free to correct me if I got something wrong.

W(total) = (# of people who washed cars) = 70
B(total) = (# of people who baked cupcakes) = ?
S(total) = (# of people who sold candy) = 80
N(total) = (# of people who did nothing) = 15

W(only) = (# of people who only washed cars) = 10
B(only) = (# of people who only baked cupcakes) = ?
S(only) = (# of people who only sold candy) = ?

WB = (# of people who washed cars and baked cupcakes) = 20
WS = (# of people who washed cars and sold candy) = ?
BS = (# of people who baked cupcakes and sold candy) = 35

WBS = (# of people who washed cars, baked cupcakes, and sold candy) = 15

WB = 20, but you have to subtract the 15 from those that did all three. So starting from the center, you have to subtract as you move outwards.

So, WB = 20-15 = 5
BS = 35-15 = 20

Wash or Bake 120 minus wash 70 = 50 bake

I hope this shows my thought process.
The blue is the information provided. The equations on the right fill in the white sections.

 

[womfalcs3]

Brain fart time

Would someone explain to me why the input can be rewritten as the result here?

I understand that sqrt(x) == x^1/2 and where to go with the x^5/2 for the derivative,
but for some reason I my mind isn't working to figure out how to go from (x^3) / sqrt(x) to x^5/2

Thanks much!

x^3/x^0.5=x^(3-0.5)=x^2.5

 

[Orbis Tabula]

Brain fart time

Would someone explain to me why the input can be rewritten as the result here?

I understand that sqrt(x) == x^1/2 and where to go with the x^5/2 for the derivative,
but for some reason I my mind isn't working to figure out how to go from (x^3) / sqrt(x) to x^5/2

Thanks much!

So if sqrt(x) is the same as x^(1/2) than that means that 1/sqrt(x) is the same as x^(-1/2). Remember that when you have a term in the denominator you can bring it to the numerator by switching the sign. So now you go from (x^3)/sqrt(x) to (x^3)x^(-1/2). So then you add the powers and 3+(-1/2) is equal to 5/2. So you get x^(5/2).

Hopefully that's clear just written out in plan text. Fractional powers always look messy if you don't break out LaTeX.

 

[Granadier]

x^3/x^0.5=x^(3-0.5)=x^2.5

Damnit your right. The veil is lifted!

Thanks!

So if sqrt(x) is the same as x^(1/2) than that means that 1/sqrt(x) is the same as x^(-1/2). Remember that when you have a term in the denominator you can bring it to the numerator by switching the sign. So now you go from (x^3)/sqrt(x) to (x^3)x^(-1/2). So then you add the powers and 3+(-1/2) is equal to 5/2. So you get x^(5/2).

Hopefully that's clear just written out in plan text. Fractional powers always look messy if you don't break out LaTeX.

That's when math gets kinky.

Thanks for the detailed explanation. It makes sense to me now.

 

[SenorArdilla]

Does your school offer a "pre-chem" course? My school recommended to take that course if you didn't do well in high school or didn't take it high school. It was pretty beneficial for figuring out the math stuff of chem, plus some of the basic formulas/compounds.



WB = 20, but you have to subtract the 15 from those that did all three. So starting from the center, you have to subtract as you move outwards.

So, WB = 20-15 = 5
BS = 35-15 = 20

Wash or Bake 120 minus wash 70 = 50 bake

I hope this shows my thought process.
The blue is the information provided. The equations on the right fill in the white sections.

Alright,

W(total) = (# of people who washed cars) = 70
B(total) = (# of people who baked cupcakes) = ?
S(total) = (# of people who sold candy) = 80
N(total) = (# of people who did nothing) = 15

W = (# of people who only washed cars) = 10
B = (# of people who only baked cupcakes) = ?
S = (# of people who only sold candy) = ?

WB = (# of people who only washed cars and baked cupcakes) = 20 - WBS = 5
WS = (# of people who only washed cars and sold candy) = ?
BS = (# of people who only baked cupcakes and sold candy) = 35 - WBS = 20

WBS = (# of people who washed cars, baked cupcakes, and sold candy) = 15

120 = W(total) + B(total) - 2WB - 2WBS
120 = (70) + B(total) - 2(5) - 2(15)
B(total) = 90

It says that 120 people either washed cars or baked but not both. So I assume that means that you would just subtract the number of people who did both (20) from both amounts.
70 (who washed) - 20 (who washed and baked) = 50 (who washed but didn't bake)
B(total) (who baked) - 20 (who washed and baked)
120 (who washed or baked) = 50 (who washed but didn't bake) + B(total) (who baked) - 20 (who washed and baked)
B(total) = 90 (who baked but didn't wash)

B(total) = B + WB + BS + WBS
(90) = B + (5) + (20) + (15)
B = 50

W(total) = W + WB + WS + WBS
(70) = (10) + (5) + WS + (15)
WS = 40

S(total) = S + WS + BS + WBS
(80) = S(only) + 40 + 20 + 15
S(only) = 5

 

[Orbis Tabula]

Alright, someone see if they can help me figure this out because I know I'm just not setting it up right, but I've been butting my head against this problem.

The City of Rulertown consists of the unit interval [0,1]. Rulertown needs to determine where to build the city's only fire station. It knows that for small delta_x, the probability that a fire occurs at a location between x and delta_x is 2*x*delta_x. Rulertown wants to minimize the average distance between the fire station and a fire. Where should the fire station be located?

The question is from my Stochastic Models class, the section on One-Period Models. The other examples involve two companies bidding one a job, with a rival company's likely bid represented as a random variable and the goal is to find out the best bid to maximize expected profit. So I guess in my example, the random variable is the location of the fire. I just don't know how to write up the integrals I need.

For reference in case anyone has the book, it's from Operations Research: Applications and Algorithms by Winston, fourth edition. On page 889.

 

[Rich Uncle Skeleton]

It says that 120 people either washed cars or baked but not both. So I assume that means that you would just subtract the number of people who did both (20) from both amounts.

It doesn't say "but not both." I'm pretty sure "or" here is an inclusive or.

 

[boviscopophobic]

Alright, someone see if they can help me figure this out because I know I'm just not setting it up right, but I've been butting my head against this problem.

The City of Rulertown consists of the unit interval [0,1]. Rulertown needs to determine where to build the city's only fire station. It knows that for small delta_x, the probability that a fire occurs at a location between x and delta_x is 2*x*delta_x. Rulertown wants to minimize the average distance between the fire station and a fire. Where should the fire station be located?

The question is from my Stochastic Models class, the section on One-Period Models. The other examples involve two companies bidding one a job, with a rival company's likely bid represented as a random variable and the goal is to find out the best bid to maximize expected profit. So I guess in my example, the random variable is the location of the fire. I just don't know how to write up the integrals I need.

For reference in case anyone has the book, it's from Operations Research: Applications and Algorithms by Winston, fourth edition. On page 889.

I presume that when you wrote "between x and delta_x" you meant "between x and x + delta_x". Also, in the paragraph below, insert "expected distance" for "distance" whenever that is appropriate; it's too onerous to go back and correct it.

(see correction below)

 

[Orbis Tabula]

I presume that when you wrote "between x and delta_x" you meant "between x and x + delta_x". Also, in the paragraph below, insert "expected distance" for "distance" whenever that is appropriate; it's too onerous to go back and correct it.

Okay, this is pretty close to what I've got right now, but I'm still tweaking it. The problem is that the back of the book has sqrt(2)/2 as the answer and I can't get that.

Edit: Also, you have your f_X(x) equal to 2*x where the problem says 2*x*delta_x. Is the change part of the problem?

Edit2: Alright, now I'm getting sqrt(2)/3 which makes sense. I hate assuming the book is wrong but I know it happens.

 

[paparazzo]

As someone who struggled with college algebra but passed with a C+ (in spring 2011), would Precalc and Calc 1 be hugely difficult? I also had a tutor help me pass college algebra, as I'd failed it the previous semester.

I'd probably be taking the classes on their own or while taking 1-2 other classes max and am willing to hire a tutor.

I ask because I'm graduating this semester in Communication, but I'm already thinking about going back next year to major in CIS.

 

[SenorArdilla]

It doesn't say "but not both." I'm pretty sure "or" here is an inclusive or.

Alright. So it's:

120 = 70 + B(total)
B(total) = 50

B(total) = B + WB + BS + WBS
(50) = B + (5) + (20) + (15)
B = 10

W(total) = W + WB + WS + WBS
(70) = (10) + (5) + WS + (15)
WS = 40

S(total) = S + WS + BS + WBS
(80) = S(only) + 40 + 20 + 15
S(only) = 5

 

[boviscopophobic]

Okay, this is pretty close to what I've got right now, but I'm still tweaking it. The problem is that the back of the book has sqrt(2)/2 as the answer and I can't get that.

Edit: Also, you have your f_X(x) equal to 2*x where the problem says 2*x*delta_x. Is the change part of the problem?

Edit2: Alright, now I'm getting sqrt(2)/3 which makes sense. I hate assuming the book is wrong but I know it happens.

When they say the probability of a fire between x and x + delta_x is 2x * delta_x, they're saying something like dF_x = 2x dx. Thus I wrote f_X(x) = dF_X/dx = 2x, so that difference is not significant.

I found the issue, which is that minimizing the expected squared distance is not the same as minimizing the expected distance. (I thought something was fishy when I got such a strong conclusion...) What does hold true is that for arbitrary distributions which have a second moment, putting the firehouse at the expected location of a fire minimizes the expected squared distance.

So if you compute \int_0^1 2x*abs(x_0-x) dx, where abs(...) denotes absolute value, you will get (2 - 3x_0 + 2x_0^3)/3. Differentiating with respect to x_0 and setting equal to zero yields the minimizer x_0 = 1/sqrt(2).

 

[Orbis Tabula]

When they say the probability of a fire between x and x + delta_x is 2x * delta_x, they're saying something like dF_x = 2x dx. Thus I wrote f_X(x) = dF_X/dx = 2x, so that difference is not significant.

I found the issue, which is that minimizing the expected squared distance is not the same as minimizing the expected distance. (I thought something was fishy when I got such a strong conclusion...) What does hold true is that for arbitrary distributions which have a second moment, putting the firehouse at the expected location of a fire minimizes the expected squared distance.

So if you compute \int_0^1 2x*abs(x_0-x) dx, where abs(...) denotes absolute value, you will get (2 - 3x_0 + 2x_0^3)/3. Differentiating with respect to x_0 and setting equal to zero yields the minimizer x_0 = 1/sqrt(2).

Thanks a ton for the help. This is one of those problems that I don't think I should have struggled with nearly as much as I did, but for some reason I couldn't get my head around it the right way. Thanks again.

 

[0xCA2]

I fucking suck at word problems. Could some please explain this? Thamks.

Suppose there's a spherical bubble that's growing at 2 cubic cm/second. Find the rate at which the radius grows when the radius is 5cm.

 

[Kazerei]

V = 4/3 * pi * r^3
dV/dt = 2
Find dr/dt when r = 5

Basically differentiate with respect to time, plug some numbers in and solve

 

[DeBurgo]

so, dumb/vague question maybe from early calculus, but: the limit of sin(x)/x is 1 as x approaches zero. I'm pretty sure I get that, and I roughly understand the proof for that. However, why is its reciprocal, the limit of x/sin(x) as x approaches zero also one? Also, is there any way to prove this given the knowledge that the limit of sin(x)/x is 1?

Keep in mind this is early calc so we can't use L'Hopital's rule, which is just what wolfram alpha gives me when I ask it to solve it.

I thought you could simply prove it just using limit laws (since it seems intuitive by the result, since the reciprocal of 1 is itself and x/sin(x) is the reciprocal of sin(x)/x) but I guess that's not correct.

 

[jnWake]

so, dumb/vague question maybe from early calculus, but: the limit of sin(x)/x is 1 as x approaches zero. I'm pretty sure I get that, and I roughly understand the proof for that. However, why is its reciprocal, the limit of x/sin(x) as x approaches zero also one? Also, is there any way to prove this given the knowledge that the limit of sin(x)/x is 1?

Keep in mind this is early calc so we can't use L'Hopital's rule, which is just what wolfram alpha gives me when I ask it to solve it.

I thought you could simply prove it just using limit laws (since it seems intuitive by the result, since the reciprocal of 1 is itself and x/sin(x) is the reciprocal of sin(x)/x) but I guess that's not correct.

Actually you can, given that lim f(x) and lim g(x) exist, it's known that

lim f(x)/g(x) = lim f(x) / lim g(x)

Now, consider f(x) = 1 and g(x) = sin(x)/x and you're done!

 

[DeBurgo]

Actually you can, given that lim f(x) and lim g(x) exist, it's known that

lim f(x)/g(x) = lim f(x) / lim g(x)

Now, consider f(x) = 1 and g(x) = sin(x)/x and you're done!

okay! thanks! I was doing it with the power law of limits, though i thought it was wrong since my textbook says that law only works if the exponent is a positive integer (though looking around on the internet I seem to get different answers on that).

 

[Therion]

okay! thanks! I was doing it with the power law of limits, though i thought it was wrong since my textbook says that law only works if the exponent is a positive integer (though looking around on the internet I seem to get different answers on that).

The version with positive integers is easier to present early on, and also easily prevents results like taking the square root of a negative or dividing by zero. If you've learned something about how limits work with the composition of a continuous function, you can take x^r to be your continuous function (on the right domain anyway) to get a more general version of the power law. If you haven't heard about that yet, I'd expect it to come up soon.

 

[Rich Uncle Skeleton]

Alright. So it's:

120 = 70 + B(total)

No, 120 = 70 + (B(total) - WB(total)). Otherwise, you're double counting WB because it's already included in W(total) = 70.

The problem's assumptions can be written:
(1) 120 = W or B = W(only) + B(only) + WB(only) + WS(only) + BS(only) + WBS
(2) 15 = WBS
(3) 20 = WB(only)
(4) 10 = W(only)
(5) 35 = B and S = BS(only) + WBS
(6) 70 = W = W(only) + WB(only) + WS(only) + WBS
(7) 80 = S = S(only) + WS(only) + BS(only) + WBS
(8) 15 = lazy

(a) Combining (2) and (5) gives BS(only) = 20.
(b) Combining this with (7) gives S(only) + WS(only) = 45.
(c) Combining (3), (4), and (6) gives WS(only) = 70 - 10 - 20 - 15 = 25.
(d) Combining (b) and (c) gives S(only) = 20.
(e) Finally, comining (1) with (2), (3), (4), (a), (c), and (d) gives B(only) = 120 - 10 - 20 - 25 - 20 - 15 = 30.

So we have
W(only) = 10
B(only) = 30
S(only) = 20
WB(only) = 20
BS(only) = 20
WS(only) = 25
BWS = 15
lazy = 15

 

[Skittles]

So does anyone know of some good online videos for discrete mathematics? The book I have for the class is legit awful for learning.

Speaking of online videos, this dude: https://www.youtube.com/channel/UCoHhuummRZaIVX7bD4t2czg is legit the best calculus teacher I've learned from. Watching his videos is making calculus 2 a hell of a lot simpler

 

[Eye for an Eye]

This is what I finally came up with. I'll find out next week if I got it right. Starting from the middle I worked my way to the outside, filling in each circle with what information was provided.

The Blue sections are the ones with the information provided. The purple sections are filled in with what can be deduced from the information provided.

I started with Wash because that only had one section missing, moved over to Candy Sales, then completed the Bake circle.

 

[harriet the spy]

So does anyone know of some good online videos for discrete mathematics? The book I have for the class is legit awful for learning.

Speaking of online videos, this dude: https://www.youtube.com/channel/UCoHhuummRZaIVX7bD4t2czg is legit the best calculus teacher I've learned from. Watching his videos is making calculus 2 a hell of a lot simpler

What level of discrete math are you looking at?

 

[Rich Uncle Skeleton]

This is what I finally came up with. I'll find out next week if I got it right. Starting from the middle I worked my way to the outside, filling in each circle with what information was provided.


The Blue sections are the ones with the information provided. The purple sections are filled in with what can be deduced from the information provided.

I started with Wash because that only had one section missing, moved over to Candy Sales, then completed the Bake circle.

The problem states "20 members only wash cars and bake cupcakes." Your diagram shows only 5 people who wash cars and bake cupcakes but don't sell candy.

 

[Skittles]

What level of discrete math are you looking at?

Introductory, will probably go over the first 3-4 chapters in the book over the semester(currently on chapter 2 at mid terms).

 

[Eye for an Eye]

The problem states "20 members only wash cars and bake cupcakes." Your diagram shows only 5 people who wash cars and bake cupcakes but don't sell candy.

Because when you're filling in a Venn Diagram, you have to take away the 15 people that Baked, Sold Candy, and Washed Cars. 20-15=5
Same goes for the 35 members only Baked and Sold Candy. 35-15=20

 

[Rich Uncle Skeleton]

Because when you're filling in a Venn Diagram, you have to take away the 15 people that Baked, Sold Candy, and Washed Cars. 20-15=5
Same goes for the 35 members only Baked and Sold Candy. 35-15=20

Again, the problem states "20 members only wash cars and bake cupcakes," meaning there are 20 people who wash cars and bake cupcakes but don't sell candy. The 15 people who did all three are not included in the 20 so you should not subtract them away. (Maybe you're reading this as "only 20 members wash cars and bake cupcakes," which means something different.)

In the other case, he writes "35 members sell candy and bake cupcakes" (note the lack of the word "only") so that includes people who did all three. In that case you should subtract the 15.

 

[Eye for an Eye]

How about we agree teachers shouldn't be making up nonsense questions. The wording is horrible, and the information is all over the place. The more I read it, the less it makes sense.

 

[Rich Uncle Skeleton]

How about we agree teachers shouldn't be making up nonsense questions. The wording is horrible, and the information is all over the place. The more I ready, the less it makes sense.

Well, I wouldn't call it a nonsense question, but yeah I personally wouldn't put this on a quiz, and I'm not surprised your class bombed it. In your instructor's defense, it happens sometimes. When you're writing the problem everything seems much clearer to you than to someone else reading it.

 

[MisterLuffy]

Specify all the properties: reflexive, symmetric, antisymmetric, transitive

Z is the set of integers. Relation R: Z x Z is defined as:
x,y ∈ Z; (x,y) ∈ R, x is a multiple of y;

Can anyone help me with this problem?

 

[Skittles]

To clarify, my book's chapters look something like this:
Chapter 1: Foundations of logic and proofs
Chapter 2: Basic Structures:Sets, Functions, Sequences,Sums, and Matrices.
Can provide a more detailed breakdown if needed

 

[jnWake]

Specify all the properties: reflexive, symmetric, antisymmetric, transitive

Z is the set of integers. Relation R: Z x Z is defined as:
x,y ∈ Z; (x,y) ∈ R, x is a multiple of y;

Can anyone help me with this problem?

It may help you to note that:

x is a multiple of y (that is, (x,y) ∈ R) if and only if x = k*y, where k is an integer.

For example, is R reflexive? That is, is there an integer such that x = k*x?

Well, x = 1*x so x is a multiple of itself and then (x,x) ∈ R.

You can check the other properties this way!

 

[concordia]

Calling out Discrete Math gaffers,



I need feedback, I'm not quite sure how to write an valid and appropriate proof.

1. "((p ∨ r) ∧ (q ∨ r)) " is true
2.Then "(p v r)" is true (why?)
3. And so is "(q v r)".
4. Now, "r" is either false or true.
4.a) If "r" is true, so is....
4.b) If "r" is false, then by modus tollendo tollens and 2. and 3. ....


can you complete the left- to- right proof? The proof from right to left is similar.

Specify all the properties: reflexive, symmetric, antisymmetric, transitive

Z is the set of integers. Relation R: Z x Z is defined as:
x,y ∈ Z; (x,y) ∈ R, x is a multiple of y;

Can anyone help me with this problem?

Ask yourself:
Is every integer a multiple of itself?
Does it hold for every pair x,y of integers that, if x is a multiple of y, then is y also a multiple of x?
If x is a multiple of y AND y a multiple of x, does it follow that x=y?
If x is a multiple of y AND y a multiple of z(!,) does it follow that x is a multiple of z?

 

[MisterLuffy]

It may help you to note that:

x is a multiple of y (that is, (x,y) ∈ R) if and only if x = k*y, where k is an integer.

For example, is R reflexive? That is, is there an integer such that x = k*x?

Well, x = 1*x so x is a multiple of itself and then (x,x) ∈ R.

You can check the other properties this way!

I never thought of doing it this way, especially using k. So we substitute k for 1 because it's the only integer that would multiply itself with any other integers we substitute for x that would equal itself. Thanks. I would've never figured this out but this makes sense.

Ask yourself:
Is every integer a multiple of itself?
Does it hold for every pair x,y of integers that, if x is a multiple of y, then is y also a multiple of x?
If x is a multiple of y AND y a multiple of x, does it follow that x=y?
If x is a multiple of y AND y a multiple of z(!,) does it follow that x is a multiple of z?

I'll try to answer these questions, even though jnWake answered the first one.