Math Is Fun Forum

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

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 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.

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

Yes.

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 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 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 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:

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

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.

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

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.

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.

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.

The formula is

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

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 Esty bella

Welcome to the forum.

Bob

hi camicat

You had a fraction

and then it's gone!

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 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 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