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When I first joined back in 2010 the forum had a lot of different categories of member. New members had limited options until they had made 10 posts when they became members. Then further 'promotions' came at post 1000 and I think 10000. This has led to the invention of several games that help to push up the post count. Moderators could also make a member 'real' which gave additional privileges.

But the forum had to move to a new platform and several things changed. Real members was the only special category that still exists. I'm not sure myself if anything changes when that happens. I'll conduct an experiment to try and find out.

edit: As far as I can tell the only change is you get to be called 'real'. Better than being imaginary I suppose.

Bob

You can forget about the sectors completely. If the cake is divided into 16 pieces then there's an inner circle of 8 sectors and an outer ring of 8 equal pieces. All you need is the radius of the inner circle. If the 16 pieces are welded back together to make an inner circle and an outer ring you just need to find the radius of the inner circle.

Divide 314.16 by 2 to get the area of the inner circle. Then:

You just need to solve that equation for r.

Bob

post the question and your answer here please.

Bob

hi jabah013.307

I agree; the questions are tricky but if you get one wrong you get clear notes on what you should have done. My advice is try all 10, then leave it a day and try them again. I think you'll find they are easily on day 2 and you'll get more (all?) right.

Bob

hi simonmagusflies

Firstly what a brilliant diagram! Did your Mum make that? It looks delicious. As you've had to slice it early do you need help eating the pieces before it goes dry? Can you send a slice using 3D uploading?

I'm happy with your answers to Q 1,2 and 3.

Q4. It's still the same cake. Unless you can allow for the loss due to slicing the answer is still 314.16

Q5. If you just divided the cake into an inner and outer with equal area then you can easily calculate the area of the inner circle. The outer would be a ring. So you can work backwards starting with that area. Divide by pi and the square root to get the radius.

But dividing like the diagram shows would give equal amounts of icing. To get it all you would need to cut horizontally and give your sister the bottom half.

Bob

Hi Milenne and jabah013.307

I too cannot figure out what you are after.

jabah: from your Ip it looks like you're in the UK. I've checked the government website and reflex angles are not in the year 4 programme of study. Simple identification questions like " Is this angle acute, obtuse or reflex?" come in year 5. Of course it's possible the US curriculum has this a year earlier but even so it's not a topic to get worried over.

Get a piece of paper and mark a point O as the centre of a circle and also put on a fixed point A. Now have a second point B that can freely roam around the circle.

If B is on top of A the angle AOB is zero.

If C is the point where COA = 90 then when B is between A and C the angle AOB will be acute (ie. less than 90)

If D is the point where AOD makes a straight line (the diameter of the circle) then any position for B between C and D make AOB on obtuse angle ie between 90 and 180.

If B moves even further around the circle then AOB will be reflex ie. over 180.

That is until B is once more on top of A. We could call the angle AOB 360 in that case but it's the same as zero. As B continues to rotate around the circle the angle goes over 360. eg. 370. But we don't have a special name for such angles as 370 looks the same as 10.

So the words acute, obtuse and reflex are just names given to different groups of angles.

On a protractor there are usually two scales, one running clockwise and the other anticlockwise. So one way it's 0 to 180 and the other way it's 180 to 0.

This gives the user the option to measure an angle clockwise or anticlockwise. You can get 360 protractors so you could measure a reflex angle with one of these. But you don't really need such a protractor as it's easy enough to add 180 to your measurement.

Milenne wrote:

how can you find the reflex angles of angles that are also reflex?

This question doesn't make sense. Reflex is just a describing word (like heavy or blue but for a type of angle) so you cannot find the reflex angle of anything any more than you can find the 'heavy' of something or the 'blue' of something.

So it looks like we have a translation issue here, and that's not what you mean. I googled reflex angles and there are loads of worksheets available. So if you are having trouble putting up a diagram, how about giving us the link to a worksheet that has a similar problem.

At the moment that's the most help I can offer.

Bob

hi Mark,

Well you surprise me. I'm sure you know Andrey already as you share the same computer! Is this just an elaborate way to promote a particular digital marketing course?

As we don't allow advertising on this forum I have decided to remove your link. If you are serious about being a member with mathematical interests please don't do this again.

Thanks,

Bob

Yes.

This poster just copy/pasted the whole worksheet but without the diagrams. Q18 is not 'do-able' if you don't have a diagram.

Bob

hi Mark Dater

Welcome to the forum.

ajibola didn't say where the puzzle came from but I guessed it was on the main MathsIsFun website. Here's a ink:

https://www.mathsisfun.com/logic_puzzle2.html

Good luck solving it.

Bob

hi

The English language can produce some ambiguous sentences at times. The question asks for equations. What the questioner means is find one equation where a line is parallel and another where a line is perpendicular.

I would start with y + 4x = 7 and re-arrange it so that y is the subject ie. y = ......

So that would be y = -4x + 7

So the gradient for a parallel line has m = -4

So start with y = -4x + c. Substitute x = 1 and y = 5 to calculate c.

To get the perpendicular line make use of the fact that if gradients m and n are perpendicular then mn = -1

This rule is explained in post #3 in this thread: http://www.mathisfunforum.com/viewtopic.php?id=26175

So the second equation has the form y = (1/4) . x + c Once again you can use (1,5) to find c.

Bob

hi Hannibal lecter

(x,y) is the general point on the line and (a,c) is another known point.

To calculate a gradient you pick two points, find the difference in y coordinates and divide by the difference in x coordinates.

Then multiplying both sides by (x-a) gives the equation you have stated.

Bob

hi ajibola

Welcome to the forum.

Puzzles like this occur in lots of puzzle books and you are given a grid like this:

You could try making up a suitable set of grids but there are a lot of 'fields' required.

My own method involves using a spreadsheet.

I started with the position numbers 1,2,3,4, and 5 and then a column for each field. (eg "pets")

I copied the information into a text box which I pasted alongside my rows and columns.

I then put in information; systematically line by line; deleting each line as I added it to the sheet.

A fact like "Lucy is the first on the left" can go straight in as she must be in position 1.

A fact like "The person who eats Milky Bars owns a horse" canot be put in a position yet so it gets its own line further down my spreadsheet.

But "Hannah eats Dairy Milk" and later on "The person in the middle eats Dairy Milk" have enabled me to put those two bits of information together on my sheet.

Note how I entered the information "Jessica is on the left of Georgina".

Here's my progress so far:

Clearly, I've got some way to go but I find with these puzzles that everything gradually gets into place. As only four pets are mentioned, whoever is left out must own the crocodile*.

Bob

*This forum does NOT recommend keeping crocodiles as pets and can accept no responsibility for any injury resulting from you ignoring this advice!

hi again,

I now have a decent diagram for this.

I have assumed that ABCD is the base, and EFGH are above these points in that order.

J is the point where AC crosses BD.

As J lies on AC, the line HJ lies in the plane ACH.

Plane BFH includes point D.

As BD is in that plane, and J is on BD, HJ lies in the plane.

Thus HJ is the line of intersection of the planes.

In order to 'see' the true angle between the planes it is necessary to look along the line of intersection. In the diagram that follows I have shown H 'over' J so that we are looking along that line.

Notice that ACH and FBH appear as straight lines. That is as it should be as we are looking along the line of intersection.

The angle between the planes appears to be 90 degrees. Can we be certain of this, rather than 89 degrees say?

The plane AFHD is a plane of symmetry for the cube so an 'object' such as point A and it's 'image' C will be such that the angle betwen AC and the plane is 90. This follows from the laws of reflection. So the angle is 90.

Bob

Hi Rana,

Welcome to the forum. This question is quite difficult. I'll try to explain as I go.

Firstly, here's an experiment for you to try. I'm assuming you are in a room with walls at right angles. Get some card and cut out a scalene triangle ABC. That is a triangle with three different angles. Make one of them obtuse.

Now try to fit your triangle against a pair of walls so that A is in the corner, AB lying against one wall and AC lying against the other. Whatever shape your triangle, this is possible. Now repeat with B in the corner, then C.

Clearly none of these angles is the true angle between the walls. To 'see' the true angle you need to be looking along the line of intersection. Above the walls looking down on the corner. Then the angle looks like 90 degrees.

So now to your question. I hope I've got the correct diagram. I'm assuming a square base ABCD, with E above A, F above B and so on.

The plane BFH is the plane BFHD. Join BD and also AC and call their intersection J. The line HJ lies in both planes so it must be the line of intersection. We now need to find a triangle that has one point on the line of intersection, let's choose J. And a line in the plane AJH (= ACH) that is at right angles to HJ. call it JK. And a line in the plane BHJ ( = BFH) that is at right angles to HJ call it JL. The angle KJL is the angle you want.

So draw the triangle AJH and use trig to calculate JK. And the triangle JHB and calculate JL.

Argh. Just realised I cannot calculate KL. Have to use vectors. Do you know the scalar product?

I'll have a longer think to see if I can simplify this. Back later.

Later edit:. I've got a sketch that makes it look like the answer is 90. But I haven't got access to my geometry software at the moment. When I have I'll post a diagram that maybe will demonstrate this.

Bob

hi zemole

Welcome to the forum.

Review or homework? what's the difference? This looks like a CompuHigh worksheet and you shouldn't be posting the whole set of questions like this.

I will try to help you to work out the answers yourself. I'm happy to check any answers you post.

Most of the sheet, I cannot help with anyway as the link you have provided is to your own computer's hard drive. Hence no pictures. You could try describing the diagram in words.

Here's the ones I can help with:

1. Two angles are complementary. The smaller angle is 35 degrees less than the larger angle. What is the measure of the larger angle?

Complementary means they add up to 90. If I call the larger one x, then the smaller is x - 35. So solve the equation:

x + x - 35 = 90

7. The apothem of a regular 9-sided figure is 6 feet and the area is 117.67 square feet. What is the perimeter of the regular 9-sided figure?

The angle at the centre of a regular nonagon is 360/9 = θ . If you join two adjacent vertices (A and B) to the centre ( O ) then AOB = θ. Split this isosceles triangle in half to make a right angled triangle with 6 feet as the height. From this and some trig. you can work out a side and hence the perimeter.

11. Which is the contrapositive of the following statement? If quadrilateral ABCD is a rectangle, then quadrilateral ABCD is a square.

If A => B, then the contrapositive statement is "not B => not A"

12. How does the sum of the exterior angles of a square compare to the sum of the exterior angles of a pentagon?

Imagine an ant crawling around the perimeter of a polygon. At each vertex it turns through an exterior angle before continuing its route. After completing the journey it has turned around 360, so the total for any polygon is the same.

15. What is the length of the line segment with endpoints (-8,6)and (-10, -1)?

To calculate this you need to use Pythagoras Theorem. Calculate the difference in x coordinates and square this number. Do the same with the difference in y coordinates. Add the two answers and square root to get the hypotenuse. If you plot the points and draw in a right angled triangle this might help.

16. A right circular cylinder has a base diameter of 12 meters and a height of 16 meters. What is the approximate total lateral surface area of the cylinder?

Imagine the cylinder is split down its length and unfolded to make a rectangle. The height is the same as the height of the cylinder and the width is the circumference of the circular end.

17. The volume of a cone is 218 cubic centimeters and the height of the cone is 13 centimeters. What is the radius of the cone to the nearest whole number?

The formula is

So use this to find the base area, then the radius squared and finally the radius.

19. A regular hexagon is inscribed in a circle with a radius of 10 meters (as shown). What is the length of each side of the hexagon?

If you divide a regular hexagon into 6 triangles made by joining each vertex to the centre, these are all equilateral triangles so their sides are equal. So this question is easy!

For help with any others I need the diagrams.

Bob

hi kerno9001

Welcome to the forum.

I'm not really understanding what you are asking. The number list doesn't seems to match the constraints.

In the top line 089 67111 41023 11018, 23 occurs at positions 12 and 13. Please provide more explanation including what it is for.

Thanks,

Bob

hi Esty bella

Welcome to the forum.

Bob

hi simonmagusflies

If x = 30, then the third angle is 180 - 90 - 30 = 60

So I think you do the same but with x rather than 30.

Get an algebraic expression and simplify it.

Check that when you replace x by 30, you get the answer above.

Bob

hi camicat

You had a fraction

.....(1)and then it's gone!

......(2)is correct so substitute (2) into (1) and you will still have a fraction.

Trig functions are periodic so that, for example sin(370) = sin (360 + 10) = sin(10)

So when working with inverse trig, such as arctan, there are many angles that would all give rise to tan(angle) = -1.44

So mathematicians have introduced the idea of principal values to make the inverse into proper functions ie. a single input has just one output.

The domain of the principal values is given here:

https://www.mathsisfun.com/algebra/trig … s-tan.html

You'll have to scroll almost to the bottom of the page but you'll find graphs that show the inverse trig functions (If a 'mapping' has multiple values it isn't called a function).

When you use a calculator, as it has to give a single value, it will have been set up to give the principal value. For arctangent the domain is -90 to + 90.

Bob

hi camicat

In Q11 you have made use of the trig. identity:

There are several versions of this. They come directly from Pythagoras Theorem. If you scroll down this page:

https://www.mathsisfun.com/algebra/trig … ities.html

you'll find them just after the place where the theorem is first mentioned.

So from 1/cot = sin/cos you can express cos in terms of sin and you're done.

Bob

hi nycmathguy

Just discovered this post.

Surely the cost goes up for each step up in w.

I've got a 13 line function definition covering (0,1], (1,2], (2,3] ...(12,13] where an open bracket means don't include the endpoint and a square bracket means do include it. And then over 13 a statement that the function is undefined.

But it would be good practice for you to try and and express lines two up to thirteen in a single expression using a suitable step definition. The choices are **floor**, in which the charge is truncated down to the first lower integer, and **ceiling** in which the charge is rounded up.

Decide which of these two is the correct one and also find a suitable algebraic expression for the charge applicable. You need to include the fixed charge for the first oz, and then the m value to reflect how the charge increases for each subsequent oz added.

In another reply I gave you a link to mathword where these two functions are illustrated. The MIF function plotter will handle floor and ceiling so you can test your function by making a graph.

Bob

hi aley

Welcome to the forum.

Let's say you have a plan to buy a new flat screen TV. You are going to mount it on a wall and have measured the space available. The critical measurement is the width of the wall space.

But TVs are advertised as a certain size where the given measurement is the diagonal of the screen.

The screens are usually 'landscape' with proportions 16:9

That's enough information to calculate the optimum size for your new TV (assuming you've got the money of course )

Bob

hi Huzaifa Abedeen

Welcome to the forum. Everyone here tries to be friends!

But it is not wise to give away your email address like that. Some unpleasant people may take advantage of you. Please reply now so we can make you safe. Oh dear, not quick enough and you have logged off. Please use the forum to talk to other members; not your private email. It is in your records at the forum but I will delete it from your public post so it does not fall into the wrong hands.

Bob

hi phro,

Trouble is I don't think it's the phone's fault. Imgur don't yet have a BCCode image stored or addressable for phone app users. It looks like the app designer started afresh with the user storage space; dis-regarding what already exists for web browser users. Left hands and right hands. The world of computing is full of them!

The word gallery in the address suggests they have set up the user-space differently for app users. I may find out later when I join via the phone app.

Ho hum. Annoying. Funny story coming your way via another route.

later edit : update.

I have downloaded the app. I can sign in using my existing username and pw. I can view all my previous uploads. I can only get an https:// link not a BCCode link. I've uploaded a new picture using my phone and I can view it using my laptop. From the laptop I can get the right code so here is my new pic:

If you are familiar with this puzzle, please note that the two parts are beginning to separate!!! If you haven'y yet solved your own I took this pic to show the exact number of windings needed to get the threads in the right place for separation. Before that I spent hours gradually tweaking the threads and 'feeling' how loose things were becoming. Happy puzzling!

Bob