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#51 Re: Help Me ! » Math, Reviews about finding areas of Cylinders and Polygons » 2021-04-25 02:01:04
Hello Clara,
I have no idea if you realize this, but we are unable to access the images in question(it appears there is a login system). Also, there is a dedicated tag for images, img, so use that in the future.
Thanks,
Mathegocart.
#52 Re: Introductions » Hello there » 2021-04-23 07:09:17
Hello and welcome, Not Dav,
This forum has been around for a good 14 years or so, and while we have had our ups and downs(currently the activity is not necessarily so promising,)it's still around! Which is something that (unfortunately) can't be said about many other forums started around MIFF's time.
I recommend you help out in the Help Me! section and try to interact with others in the Members Only and the Puzzles and Games section. Those sections have remained relatively quiet for the past couple of months.
#53 Re: Help Me ! » How to complete the square with x² having negative coefficient » 2021-04-23 07:03:12
This is relatively easy, all things considered. Divide the whole equation by -1 so the x^2 coefficient is positive 1, then add 3 to both sides, and boom, just plug and chug.
#54 Re: Help Me ! » Product Rule » 2021-04-23 02:12:45
Given g(s) = 2s/(s + 1), find g'(s).
Let me see.
g'(s) = [(s + 1)d/ds (2s) - (2s)d/ds (s + 1)]/(s + 1)^2
g'(s) = [(s + 1)(2) - (2s)(1)]/(s + 1)^2
g'(s) = (2s + 2 - 2s)/(s + 1)^2
g'(s) = 2/(s + 1)^2
Correct?
What I found is the slope of the given curve at any point.
Correct?
Your answer and your working is all OK. Wonderful work.
The derivative is the slope of the tangent line. So yes, that is the slope of the given curve at any point.
#55 Re: Help Me ! » Sketch A Scatter Plot » 2021-04-22 14:28:26
Bob wrote:What does the rest of the question ask?
If you're just asked for the scatter plot then it sounds like you've done it.
Bob
I posted the entire question. I think it is simply an exercise in terms of plotting points.
That's exactly what it is.
#56 Re: Help Me ! » Determining Quadrant(s) for a Point » 2021-04-22 14:24:51
Bob wrote:9-12 all ok.
13. Try plotting this as a straight line. It goes across two quadrants
14. So does this as xy can be positive when both x and y are negative.
Bob
Ok. For 13, I say quadrants 1 and 2.
For 14, I say quadrants 1 and 3.
Yes?
For 14, yes. Try looking at 13 a little closer.
#57 Re: Maths Is Fun - Suggestions and Comments » Math Video Answers » 2021-04-21 14:23:02
Mathegocart wrote:Who is this aimed at? Us forum users or MIF himself?
I just want to know if this site ever plans to include a video clip session?
The forum doesn't have video clips; users may be free to add additional video clips if they feel it serves in facilitating the explanation. The MIF website does have a couple mathematics videos(though they are relatively basic topics.)
#58 Re: Help Me ! » Rate of Change » 2021-04-21 14:16:12
Mathegocart wrote:mathland wrote:Are you saying that dV/dr = -2kR for part (c)?
Yes.
Easier than I thought. Are you a math teacher?
I did use to tutor some kids in mathematics. I do some study of higher level mathematics and calculus in my free time - I'm currently programming and my work could use a massive dose of more convoluted mathematics.
#59 Re: Dark Discussions at Cafe Infinity » Chauvin Trial » 2021-04-21 14:15:01
Whenever an user starts a question on a rather important world event, I'd like the user themself to answer the question. So what about you, mathland?
#60 Re: Maths Is Fun - Suggestions and Comments » Math Video Answers » 2021-04-21 14:13:51
Who is this aimed at? Us forum users or MIF himself?
#61 Re: Help Me ! » Population of Animals » 2021-04-20 03:20:13
Mathegocart wrote:mathland wrote:Going back to the picture, can you provide the steps?
Sure. To upload a personal image of yours onto this forum,
1. Upload the image to an image hoster such as postimage. Others have suggested imgur in the past, but it has now been infested with commercialism and other such idiocies of the Internet. Hence why I recommend postimage.2. On the page after you upload your image, you will be presented with a myriad of links. Copy and paste the Direct Link into an tag, i.e [img]DIRECT LINK GOES HERE[/img].
I have to join postimage first, right?
You may if you wish, but it's not necessary for uploading images.
#62 Re: Help Me ! » Rate of Change » 2021-04-20 01:26:01
Mathegocart wrote:mathland wrote:You are right. This is the correct answer. Can you set up part (c)?
Same deal as part (b). The distance from the center of the tube to the wall is R. Plug that in for r(i.e, r=R).
Are you saying that dV/dr = -2kR for part (c)?
Yes.
#63 Re: Help Me ! » Rate of Change » 2021-04-19 14:08:56
Bob wrote:The graph of constants is a horizontal straight line so their gradient function is zero. You have treated R as a variable. It isn't. It is the fixed radius of the tube. Little r is the only variable.
So dv/dr = -2kr
Bob
You are right. This is the correct answer. Can you set up part (c)?
Same deal as part (b). The distance from the center of the tube to the wall is R. Plug that in for r(i.e, r=R).
#64 Re: Help Me ! » Swimming Pool » 2021-04-19 14:02:29
Bob wrote:Find dW/dt and then put in the given t value.
B
Hello Bob.
Let me see.
dW/dt = d/dt [35,000 − 20t^2]
dW/dt = -40t
What do I do with dW/dt?
Plug t=2 hours into dW/dt as the problem tells you to.
#65 Re: Help Me ! » Tangent Line » 2021-04-19 13:58:33
Bob wrote:work out dy/dx for the gradient of the tangent line and hence find the equation using y = mx + c.
This line will cross the curve again so work with a pair of simultaneous equations (tangent equation) and (curve equation).
The resulting equation in x will look complicated but you have one extra clue ... you know (1/2 , 1/8) lies on both so the expression must factorise with (x-1/2) or (2x-1) as a factor. That leaves a quadratic implies two more solutions. Wait a mo though. If the line makes a tangent at that point then (2x-1) will be a double factor leaving only a linear equation left for the solution you want.
B.
This is one is a bit unclear.
I hope this explanation of mine may help in clarifying Bob's explanation to you.
To find that specific point Q, we need to find the equation of the tangent line T(y = mx + b) at the point(1/2, 1/8).
So, let's find the derivative of the function y = x^3 so we can find the slope of said line.
y' = 3x^2(per the power rule.)
Plug in x = 1/2 and we get y' = 3(1/2)^2 = 3/4. So the slope, or "m," in the tangent line is 3/4.
Now we have y = (3/4)x + b. Plug in the point that the tangent line contains so you can find b, and with that equation, of the tangent line you want to find where it intersects with y = x^3 so you can determine point Q.
Set the ys equal and you will receive
The Tangent Line Equation = x^3.
As Bob mentions, you will receive a cubic equation that can be reduced down to a quadratic equation via dividing the cubic by (2x-1)(as you already know that x = 1/2 is a solution to the problem above.)
You will find the other solution to the quadratic, and then determining the slope of the tangent line at Q to y=x^3 at x = other solution is relatively easy.
#66 Re: Help Me ! » Poiseuille’s Equation » 2021-04-19 13:40:48
Bob wrote:Here again dV/dr is your starting point.
B.
This will take some time to do. Bob, the questions posted are the questions for which the author does not provide a sample for.
(a). Take the derivative of V(R) with respect to R. I.e use the power rule or the limit definition of the derivative.
(b). Plug in R = 0.03 and R = 0.04 into the derivative of V(R).
(c). Determine V'(0.04) - V'(0.03) to find the increase (or decrease) of the rate at which blood is flowing.
#67 Re: Help Me ! » Population of Animals » 2021-04-19 13:34:40
Mathegocart wrote:A great, detailed reply. Please, look for my questions from now on. I know that Bob is always on the look out for my threads. Maybe you guys can share in terms of reply. If I don't show my work, it simply means that I haven't the slightest idea how to begin to post an answer. Otherwise, I show work or effort. Is there a way to upload pictures on this site? I am thinking geometry,, trigonometry and calculus questions involving a geometric interpretation. You say?
I always attempt to do a general review of the forum twice a day(when I can, of course.).
Try the [img]IMAGEURL[/img] tag. For example, this is an url of a bunch of blue balloons: https://i.pinimg.com/736x/78/e1/49/78e149af9501adae51dae96faa2307dd.jpg .Place this url where the IMAGEURL is.
As for a geometric approach to tackling down problems, why not? Mathematics is beautiful in its multifactorial ways to represent the same problem, concept, or idea.
https://i.pinimg.com/736x/78/e1/49/78e149af9501adae51dae96faa2307dd.jpg
Going back to the picture, can you provide the steps?
Sure. To upload a personal image of yours onto this forum,
1. Upload the image to an image hoster such as postimage. Others have suggested imgur in the past, but it has now been infested with commercialism and other such idiocies of the Internet. Hence why I recommend postimage.
2. On the page after you upload your image, you will be presented with a myriad of links. Copy and paste the Direct Link into an tag, i.e [img]DIRECT LINK GOES HERE[/img].
#68 Re: Help Me ! » Population of Animals » 2021-04-19 07:26:46
A great, detailed reply. Please, look for my questions from now on. I know that Bob is always on the look out for my threads. Maybe you guys can share in terms of reply. If I don't show my work, it simply means that I haven't the slightest idea how to begin to post an answer. Otherwise, I show work or effort. Is there a way to upload pictures on this site? I am thinking geometry,, trigonometry and calculus questions involving a geometric interpretation. You say?
I always attempt to do a general review of the forum twice a day(when I can, of course.).
Try the [img]IMAGEURL[/img] tag. For example, this is an url of a bunch of blue balloons: https://i.pinimg.com/736x/78/e1/49/78e149af9501adae51dae96faa2307dd.jpg .Place this url where the IMAGEURL is.
As for a geometric approach to tackling down problems, why not? Mathematics is beautiful in its multifactorial ways to represent the same problem, concept, or idea.
#69 Re: Help Me ! » Aaron’s Total Charge » 2021-04-19 07:20:03
Mathegocart wrote:You're rather close here, but note that the question states that a tax of 8% is applied(i.e, added onto) the room rate. Your equation as currently written would apply a 92% tax on the hotel rate.
So it should be T(x) = 1.08(99.95x) + 5.00.Where did 1.08 come from?
1.08 simply means that the rate was increased by a tax of 8%. 1.00 increased by 8% is 1.08.
#70 Re: Help Me ! » Two Drinks » 2021-04-19 07:14:24
Mathegocart wrote:Hello mathland,
yup, that's all correct - meet or exceed is indeed >=.How about that? It was a guess, really. It's important to understand the WHY in math.
I agree. There wasn't much more - you just stated the problem in mathematical form. I.e, 299mg * cups of milk + 261mg * cups of juice (meets or exceeds, i.e greater than or equal to) 1000 mg.
Great stuff.
#71 Re: Help Me ! » Population of Animals » 2021-04-17 11:24:16
A general formula that I think applies here is y = ab^x.
Let a = 50
Let b = double or 2
We get P = 50(2)^12n. Here 12 represents the months of a year.
A friend told me that the correct answer is P = 50(2)^(n/12).
Is my friend right? If so, why is my set up wrong?
Try plugging in a couple of values into your equation after developing it - it'll help a ton, especially when it comes to these equations modelling real life.
Since the population of animals at Central High School doubles every 12 years, the period is 12. So when n = 12, we should expect the population to double. If we plug n=12 years into your equation, we see the population skyrockets to an eyewatering 1.11*10^45. Clearly not a doubling in a year.
As we want the population of animals to double in this exponential model, we want P=100 when n=12, so let's solve a simple mathematical equation to determine the exponent's power.
Let k be a real-valued number.
population = base*(how much the population increases every period number of years)^(n*k)
100 = 50*2^(12*k) (divide by 50)
2 = 2^(12*k)
Obviously, to equalize both sides, k must be 1/12. Hence, your friend is right - the exponential function modelling the population of animals is indeed 50*(2)^(n/12).
#72 Re: Help Me ! » The Company » 2021-04-17 11:14:27
Hi mathland,
you're all good here. 12 bucks times n items is indeed 12n as written, and 7n + 350 is indeed the cost of producing n items.
#73 Re: Help Me ! » Aaron’s Total Charge » 2021-04-17 11:13:10
You're rather close here, but note that the question states that a tax of 8% is applied(i.e, added onto) the room rate. Your equation as currently written would apply a 92% reduction of the hotel rate.
So it should be T(x) = 1.08(99.95x) + 5.00.
EDIT: READ ABOVE AGAIN.
#74 Re: Help Me ! » Two Drinks » 2021-04-17 11:11:29
Hello mathland,
yup, that's all correct - meet or exceed is indeed >=.
#75 Re: Help Me ! » Limit of Rational Function...3 » 2021-04-03 17:46:21
Find the limit of 5/(x^2 - 4) as x tends to 2 from the right side.
Approaching 2 from the right means that the values of x must be slightly larger than 2.
I created a table for x and f(x).
x...............2.1.....2.01................2.001
f(x)...........12......124.68............1249.68I can see that f(x) is getting larger and larger and possibly without bound.
I say the limit is positive infinity.
Yes?
Approaching 2 from the right means that the values of x must be slightly larger than 2.
Indeed.
Looks all good to me here.