Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

hi Monox D. I-Fly

This is a type of calculus question. You are told dV/dt the rate of change of volume, and dr/dt, the rate of change of radius. So construct a chain rule type equation:

dV/dr = dV/dt times dt/dr

So you can calculate dV/dr. But there's a formula for V in terms of r. Differentiate it and set equal to the above and solve for r.

Note: dV/dr will change as r increases. If dV/dt is constant then dr/dt is not. So the given value (20) must be the rate of change at an instant and you will be calculating the radius at that time.

If the radius addition rate is 20 cm/s, the radius of the sphere after being blown is ....

So I would have worded this "When the radius is increasing at 20cm/s, what is the radius at this moment?"

Bob

hi WyoCowboy

H does not equal 2.5 divided by tan(67.5)?

I agree with your teacher. The triangle has an angle of 67.5 and it also has a right angle. The height is opposite to that angle and 2.5 is adjacent. opp = adj x TAN(angle) so you shouldn't be dividing. If you compare the answer you got by doing that with my diagram you'll easily see it's not the correct value.

Bob

Whenever you are doing a trig question work through these steps carefully:

step 1. Is the triangle right angled? The formulas only work if it is. There are more advanced formulas for cases of any triangle but these are not covered by your course.

step 2. Identify which angle you are using and which sides are Opp, Adj ad Hyp.

step 3. Choose the correct part of S=O/H C=A/H / T=O/A

step 4. Rearrange the formula to make the subject the side you don't know.

step 5. Do the calculation. Think about whether your answer 'looks right' for that triangle. If you can compare with an approximate drawing do so. It'll help you to see of you've done it correctly.

hi Kayla,

Apologies for the slow response. I've been helping to organise a dance show.

Okay so if s/2 is 2.5, then

h = 2.414 * s/2

h = 2.414 * 2.5

= 6.035

So, the height is 6.035 ft.

Yes, that's good for the height of one triangle (angles 67.5 - 67.5 - 90)

I took 6.035 * 5 (for the length of the bumpers)

h = 30.175 (this is the area of one polygon)30.175 (2) = 60.35 (The area of the whole octagon)

Whoops! Not quite! Area of a triangle is half base x height = 0.5 x 5 x 6.035

Then multiply by 8 as the octagon is made up of 8 of these triangles.

Would AK = 6 1/2 ? You said that AJ is twice of AK, so would that make AJ = 13 ? Then taking 10(13), then making the area of the whole rectangle = 130 ?

The length on my diagram does look like 6.5 but that's not the exact value if you use trig to calculate it.

In the right angled triangle with angle 67.5, AK is the hypotenuse and 2.5 is the adjacent. So you need to do AK = 2.5/COS(67.5) to get the exact value of AK.

Then work out the rectangle as you did and add on the octagon.

The solid for the volume is a prism. That means it has a constant cross section given by the surface shape. Volume of a prism is area of cross section times depth. You're told the depth is 6 ft.

Hope that helps you to finish this problem.

Bob

hi MellyBigD

Welcome to the forum.

A well-known business school in Cape Town conducted a survey in 2013 amongst its MBA applicants to determine whether students apply to only one business school. A sample of 2018 students was chosen and the following results were obtained: (rounded to 3 decimals)

Age Did you apply to more than one school?

Yes No Total

23 and under 207 201 408

24-26 299 379 678

27-30 185 268 453

31-35 66 193 259

36 and over 51 169 220

Total 808 1210 2018

1.1) What is the probability that a randomly selected applicant is younger than 27 [1]

Any probability is <number where the chosen event occurs divided by <total of all possible events>

So add up how many are younger than 27 and divide by the total number of applicants.

1.2) What is the probability that a randomly selected applicant is 24-26 years old given that he/she did not apply to more than one school? [3]

As we are told the applicant did not apply to more than one, you should confine the calculations to the numbers in the 'No' column. In all other respects this is the same as part 1.1

1.3) What is the probability that a randomly selected applicant is 24-26 years old or did not apply to more than one school? [3]

Now the event that satisfies the question includes two groups, so add together '24-26' group to the 'no' group.

1.4) Is the number of schools applied to independent of age? Let A = “Yes” and B = “23 and under”. [3]

To be 'independent' the probability taking age into account must be the same as the probability if we ignore age. If you're told to take A as 'yes' and B as '23 and under', I think the questioner intends that you just consider the B group and ask is the probability the same for this limited group. It looks like you've been given a formula for this involving A and B so you should be able to plug in the numbers.

So B' is the number not '23 and under'

So is P(A given B) the same as P(A given not B)

Please have a try at these and post back your answers. Once you're ok with them we can move on to another question.

Bob

hi Kayla,

Ok, we have something we can work with. The triangle AFK (see an earlier post) is the key to this problem. It is isosceles so you'll have to split it in half to make a right angled triangle. The internal angles of a regular octagon are each 135 so angle FAK is 67.5. With that you can work out the height of the triangle, then its area and then the area of the whole octagon.

For the rectangle the length is AJ and the width is 10. AJ is twice AK which you can get with more trig in the half triangle already used.

Have a try and post back when you're ready.

Bob

hi Kayla,

That answer looks good to me.

Bob

We know **w** = |v|**u** + |u|**v**

So **u.w** = **u.**(|v|**u** + |u|**v**) = |v||u|^2 + |u|**u.v**

If this is |v| + |u|**u.v** then |u|^2 = 1. I don't see where it says u is a unit vector.

Am I missing something ?

Bob

Leesajohnson* is probably not a bot. This person put a advert for tutoring services in their signature, which I removed. The administrators and moderators have discussed this growing problem but decided not to ban automatically but rather to remove the advert and wait to see what the member does next.

Some years ago a person joined to tell us about some new mathematical research they had done and included a link to their book sale page. The work was genuine but the advert broke our rules. The person agreed to make a pdf of their work available to MIF members for free and that seemed to be a good compromise.

Some cases are hard to determine such as the recent bit coin maths puzzle post. Do I want to seem to encourage bit coin? But the puzzle was one that members might want to try. So I deleted the link but kept the puzzle.

Bob

ps. * Also I noticed 6 English grammar and spelling faults in that post. What sort of advert is it when the 'tutor' cannot write good English?

Make a habit of study

ingmathand thiswill make you proficient in this subject. It is good to studyfrombooks and get tuition from expert tutors,full stop / new sentence starting with 'This'is the indication ofreplace 'of' with 'for a'better future.

pps. On the other hand if this is from a bot, then it's entirely appropriate to post it in 'computational math'.

This site has many good examples:

https://courses.lumenlearning.com/finit … binations/

It also looks at the birthday problem, taking it slowly with easier cases first.

Bob

hi Kayla,

That just seems too big for a total area of 130. A square with those sides would be more than 144.

So I checked this by constructing an accurate diagram and it didn't come out correct. I'm so sorry ... it's my fault for not spotting you made an earlier error. It would be correct but that for that.

You started with 360 degrees and divided by 7 to get 51.428

That's right.

Then find half of it to get 25.714. That also is correct.

But it is not the angle OAH. This is AOH, the angle at the top of the green triangle. So subtract from 90 to get OAH, then as before.

Nearly there.

Bob

Well here's something to ponder. The DNA molecule is a double helix with billions of twists. I purchased a 50m coil of Ethernet cable and it was supplied tightly wound. When I unpacked it, it fell apart until it was a huge tangle of 'spaghetti' and it took me hours to untwist it.

So how does DNA uncoil, replicate, and recoil so easily?

Bob

hi Monox D. I-Fly

I'm not too sure about this, but here's my attempt.

I'll put vectors in bold and magnitudes not bold.

If **w** = v**u** + u**v** then **w** is a linear combination of **u** and **v**

so you can construct the following diagram:

Mark an origin at O. Draw representative vectors from O for **u** and **v**. Extend the u line to a point A and the v line to a point B.

Make a parallelogram OACB where **OC** = **w**

|OA| = |v|.|u| and |OB| = |u|.|v| so OA = OB.

So OACB is a rhombus.

The diagonals of a rhombus bisect the angles at the vertices so θ = φ

Is this correct?? Not sure.

Bob

is the number gender is the 6 possibilities or the 6 digit of the die?

These are the same thing. You consider how many equally likely events there are. The dice has six faces. There are six numbers. Either one gives 1/6.

If you are allowed to re-arrange the order of the throws there are six ways of getting 241. Altogether there are 216 possibilities. So the probability of throwing that combination is 6/216

Bob

hi Monox D. I-Fly

Try this:

Replace every x with y, and every y with x. Then change all the signs on both x and y.

That gives:

Hope that helps,

Bob

hi Hannibal lecter

If repetitions are allowed:

If the five numbers were all different it would be 5! = 5x4x3x2x1.

But we have repeated 1s. You couldn't tell the difference between 16122 and 16122 where I have swapped the 1s around. So very possibility gets counted twice because of the repeated 1s and similarly the 2s. S we need to calculate 5! ÷ 2 ÷ 2

If repetitions are not allowed the set is effectively reduced to {1,2,6} so it's just 3! = 3x2x1.

126, 162, 261, 216, 612, 621.

Bob

hi Hannibal lecter

I'll have a go with dice. Let's say you have three dice, numbered 1-6 as usual. And let's say the faces have the actual digits rather than dots. If you throw the first, then the second, placing it to the right of the first and finally the third placing it to the right of the second, you will be looking at a number with hundreds, tens and units.

What's the probability the number is 241 ?

The first number has to be a 2 (P= 1/6) the second a 4 (P=1/6) and the third a 1 (P=1/6) so the probability of getting this number is P = 1/6 x 1/6 x 1/6 = 1/216

Now what about if we just throw all three at once, and we're allowed to re-arrange them to try and make that number. That's easier to get because any of the three can be the 2, the 4 and the 1. So 1, 2, 4 would be ok.

There are still 216 possible outcomes. But how many will give us those numbers in any order:

We could throw 124, 142, 214, 241, 412, or 421 and re-arrange to get our target number. 6 possibilities is the number of **combinations** of three digits ie. 3x2x1.

In my example it was easy to write out all six possibilities but if we had lots more digits to consider (supposing it were ten dice) so it's useful to know there's a formula to get us there quickly.

Bob

hi Kayla,

130 / 7 = 18.571. (This is the area of one triangle)

51.428 / 2 = 25.714 = angle OAH

18.571 = 0.5 x s x s/2 x TAN (OAH)

18.571 = 0.5 x s x s/2 x TAN (25.714)

18.571 = 0.5 x s x s/s (0.481)s^2= 18.571 x 0.5 x 0.481

s^2= 4.466 (I the took the square root)

s= 2.113

Correct up to this line "18.571 = 0.5 x s x s/**s** (0.481)"

except that bold 's' should be a 2.

But then you haven't re-arranged it properly.

times by 4 to get

4x18.571 = s^2 x 0.481

So s = square root of (4x18.571/0.481)

Bob

hi WyoCowboy

This problem is also causing difficulties for another MIF member. Here's a copy of what I have just sent to Kayla. I'll answer your latest post below the quote.

hi Kayla,

I'm not surprised you are having difficulty. I think the person who set the question made a mistake and until we get that person to change the question you won't be able to do it. I'll give my reasoning and send a copy of this post to WyoCowboy who is also struggling to complete this.

Here's the picture you sent me:

and here's the one WyoCowboy sent. I've added some to the picture to show the pool split into two shapes.

from this question you wrote:This 12 sided pool is not a regular polygon because the inner angles are not all equal. You will need to break the base of the pool down into 2 regular polygons.

As you can see the two ends do not make a hexagon. They make an octagon and it certainly looks regular so no problem so far.

But what's left is a rectangle, not a square! So it's not a regular shape. Here's my accurate construction of one end. Each square of my grid represents one foot. You can clearly see that the diagonal of the octagon, which would be one of the middle shape measurements is not 10 feet. It's closer to 13.

Please raise this with your teacher and ask them for clarification. You may copy from this post if it helps but I would recommend that you do not tell them it's from MIF.

Bob

you wrote:

Area of octagon (Be sure to state what the shape is.)

360/8=45

2*side+45=180

2*side=135

Side=67.5

Tan(67.5)=H/(5/2)

Tan(67.5)=H/2.5

You have not correctly solved this equation for H.

You need to calculate the area of triangle AFK and then multiply by 8 for the whole octagon.

base = 5

base angle = 67.5

So height, H = 0.5 x 5 x tan(67.5) (because in the right angled triangle 0.5 x 5 is the adjacent and H is the opposite)

You have written "Tan(67.5)=H/2.5" which is correct!

Hope this helps.

Bob

hi Kayla,

I'm not surprised you are having difficulty. I think the person who set the question made a mistake and until we get that person to change the question you won't be able to do it. I'll give my reasoning and send a copy of this post to WyoCowboy who is also struggling to complete this.

Here's the picture you sent me:

and here's the one WyoCowboy sent. I've added some to the picture to show the pool split into two shapes.

from this question you wrote:

This 12 sided pool is not a regular polygon because the inner angles are not all equal. You will need to break the base of the pool down into 2 regular polygons.

As you can see the two ends do not make a hexagon. They make an octagon and it certainly looks regular so no problem so far.

But what's left is a rectangle, not a square! So it's not a regular shape. Here's my accurate construction of one end. Each square of my grid represents one foot. You can clearly see that the diagonal of the octagon, which would be one of the middle shape measurements is not 10 feet. It's closer to 13.

Please raise this with your teacher and ask them for clarification. You may copy from this post if it helps but I would recommend that you do not tell them it's from MIF.

Bob

hi Hannibal lecter

Some probability problems will require one or other of these.

For example: How many ways can you make a five letter 'word' (it doesn't have to be in the dictionary) from the letters of MATHEMATICS. What is the probability that the word has two Ms in it.

There are 11 letters and we are choosing 5, so that's the number of permutations of 5 from 11.

Now consider the words made. Some will have no Ms, some will have 1 M and some will have 2 Ms. So consider choosing 3 non M words first, then add two Ms but bearing in mind the Ms may be either way round (M1 before M2 or M2 before M1). That's a combination question because order of Ms doesn't matter.

Bob

hi Kayla,

So the yellow area is half base x height = 0.5 x s x s/2 x TAN(OAH)

So start with 130 for the whole polygon, divide by 7 to get one yellow triangle and set it equal to the above. Then you've got to re-arrange that equation to get s squared and then s.

Bob

hi WyoCowboy

The middle section isn't a square even though the question says 'two regular polygons'. It'll have to be a rectangle as it's length and width are not the same.

Bob

hi Kayla,

Thanks for the picture. That's a big help. I've provided a link in one of your other posts but now I've seen the picture I don't think what I've said there is correct. I had assumed that the ends are each half a regular octagon but the diagram there has the distance across the pool as about 13, whereas, if the middle part is a square it should be 10 (=2 x 5). Does the original wording of the question make it clear which is correct. I'll make a new diagram based on a square middle section but I think it won't then give regular ends so the question may not be do-able.

I've read WyoCowboy's post again. It doesn't say the middle is a square ... that poster has just assumed it is. But both their diagram and your picture show a rectangular middle, width AJ in my diagram and length 10 as it's made form two 5 sections. SO let's assume the correct version is octagonal ends and rectangle in the middle. So my help to WyoCowboy will still work. I'll add a post there too.

Bob

hi Kayla,

That's the answer I get too.

Bob

ps. another member has posted the pool question and I've given some help here:

http://www.mathisfunforum.com/viewtopic … 81#p402481

I forgot to say that you'll need to calculate the side of the square, AJ. It's twice AH which you can get from triangle AHF split in two to make a right angle.

hi WyoCowboy

That's great. Thanks for the link. I've re-made part of the diagram so I can see what's going on. I've left my construction lines as it will help us work out the sizes.

First I constructed AB and made a circle, centred on A with radius AC = 5

And I made AD perpendicular to AB

Then I bisected angle DAB to fix E with EAB = 45, and bisected DAE to fix F with FAE = 22.5

Then I made a second circle, centred on F with radius 5 again to fix G.

Finally I fixed H and J by reflection in the line from G, perpendicular to AB. I should have labelled the other end of this line. Let's say GK where K is halfway between A and J.

So that gives us half an octagon AFGHJ. The other end will have a similar shape.

The question doesn't say that the ends are half of a **regular** octagon but without this assumption there is not enough information to do the problem.

I'm assuming you have already done simpler problems involving regular polygons. If so you'll know how to calculate the area of triangle KFA. Multiply by 8 and you'll have the area of the octagonal part. Add on the area of the square and you've got the total surface area.

As this pool is a prism the volume is then just surface area x depth of pool.

Hoe that helps,

Bob

I've also assumed the surface area means the area of the top (water) surface. If the question means the total surface area of the prism you'll have to double the answer you have for the 'top' and add on 10 rectangular areas for the side walls. You might point out the ambiguities in the question wording so it can be improved for future students. It should state that the ends are each half a regular shape and that surface area means all surfaces including the bottom and sides or just the top surface of the pool.