<![CDATA[Math Is Fun Forum / Exercises]]> 2023-11-29T03:10:03Z FluxBB https://www.mathisfunforum.com/index.php <![CDATA[Compute the solution:]]>

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https://www.mathisfunforum.com/profile.php?id=682 2023-11-29T03:10:03Z https://www.mathisfunforum.com/viewtopic.php?id=21675&action=new
<![CDATA[A Motion Exercise]]> I will refer here to your original diagram (I guess you still have it).

I used to tell my students that, in general, finding how to solve a problem has two main directions, forwards and backwards.
To solve this exercise, walking backwards makes it a simple one.

Let us assume first that the full range (R) of the missile was used.
In this case, the warship had to be on the circle whose center is ‘Y’ and radius is (R).
If we draw this circle, it becomes clear that WX is the shortest distance from W to this circle.
Therefore, the mathematical solution of T=f(D,R,S1,S2,S3) could take advantage of Pythagoras theorem of the right triangle SWY.

One may think that it is not necessarily to use the full range of the missile, say (k*R) instead of (R), where 0<k<1.
The same reasoning above also applies in case (k*R). Only the radius of the circle becomes (k*R) instead of (R).
This means that T_new = f(D,k*R,S1,S2,S3). That is (R) is replaced with (k*R) in the final formula of (T).
And by analyzing the variation of the function (T_new) with the variable parameter (k), we will find out that T_new > T for all values of (k).

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https://www.mathisfunforum.com/profile.php?id=219354 2023-11-28T08:19:13Z https://www.mathisfunforum.com/viewtopic.php?id=30242&action=new
<![CDATA[An Unusual Tricky Exercise, Have Fun]]> It is clear, to me in the least, that you both, amnkb and you (Bob), are very good teachers.

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https://www.mathisfunforum.com/profile.php?id=219354 2023-11-26T08:50:35Z https://www.mathisfunforum.com/viewtopic.php?id=30240&action=new
<![CDATA[F(x) Having no Local Limits]]> I thought that such exercise is for students, not teachers Anyway, I guess you meant by your last line that the condition to be satisfied by the parameters A, B, C and D for f(x) doesn’t have local high and low limits is:
B^2 - 3*A*C ≤ 0
And the parameter (D) may have any value.

Thank you.

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https://www.mathisfunforum.com/profile.php?id=219354 2023-11-24T12:11:07Z https://www.mathisfunforum.com/viewtopic.php?id=30239&action=new
<![CDATA[Surface of a Triangle]]> Bob wrote:

hi KerimF

Apologies for not responding to this sooner.  I had an idea when I read your first post but it has knocked around in my brain for some days before immerging as a method.  It's got lots of steps and, if you wanted an actual formula, it could be done but it would be messy.  Instead I'll outline an algorithm that will work.

If you just knew the three angles you could draw a triangle with those.  But there are an infinite number of such triangles, all similar in shape, each scalable to scale up its size, but only one will have the right perimeter.

There's a formula, called Heron's formula after it's discoverer, that will calculate the area of a triangle if you know its perimeter. The semi perimeter = P/2 , sidea a, b, and c, and the formula is:

This Wiki page has an article about it https://en.wikipedia.org/wiki/Heron%27s_formula  but there P already stands for the semi perimeter.

So how to make use of this.

Choose a random value for a; I'll call it p.

You can then use the sine rule to work out the other two sides, q and r.

Work out P' the perimeter of this triangle by P' = p + q + r

If P' = P then you're done because you had a really lucky choice for a. Unlikely though. But what you have got is the sides of a similar triangle that is either a scaled down or scaled up version of the correct triangle.

So you can work out the scale factor P/P' for converting your triangle to the correct one.

So now you know a, b, and c and so can proceed with Heron's formula.

Bob

Very nice work, thank you.
As expected, there are more than one way to solve it.

Here is how I did it:
(back from step 1, to show below the complete solution)

The area's formula of a triangle ABC is:
S = base * height / 2

As a start, I chose the base AB. In this case:
height = CM (from C to AB, M on AB)
And,
S = AB * CM / 2

Now, we have two right triangles ACM and BCM (right angle at M)
From them, let us find:
AB = f(CM,A,B)

In the triangle ACM
AM = CM/tan(A)

And, in the triangle BCM
MB = CM/tan(B)

So AB=f(CM,A,B) is:
AB = AM+MB = CM*[1/tan(A) + 1/tan(B)]

===

Again, in the triangle ACM
AC=CM/sin(A)

And, in the triangle BCM
BC=CM/sin(B)

So,
P = AB+AC+BC = CM*[1/tan(A)+1/tan(B)+1/sin(A)+1/sin(B)]
P = CM*[cos(A)/sin(A)+cos(B)/sin(B)+1/sin(A)+1/sin(B)]
P = CM*{[1+cos(A)]/sin(A) + [1+cos(B)]/sin(B)}

===

Now, it is time to substitude CM with S from the area's formula:
S=AB*CM/2 ==> CM=2*S/AB

P = 2*S/AB*{[1+cos(A)]/sin(A) + [1+cos(B)]/sin(B)}
And we get:
AB = 2*S/P*{[1+cos(A)]/sin(A) + [1+cos(B)]/sin(B)}

===

Similarly, we can say:
AC = 2*S/P*{[1+cos(A)]/sin(A) + [1+cos(C)]/sin(C)}
CB = 2*S/P*{[1+cos(C)]/sin(C) + [1+cos(B)]/sin(B)}

===

By adding the above three formulas of AB, AC and CB
P = AB+AC+CB = 2*S/P*{2*[1+cos(A)]/sin(A) + 2*[1+cos(B)]/sin(B) + 2*[1+cos(C)]/sin(C)}
P = 4*S/P*{[1+cos(A)]/sin(A) + [1+cos(B)]/sin(B) + [1+cos(C)]/sin(C)}

And the required formula/function is:
S = f(P,A,B,C) = P^2 / {[1+cos(A)]/sin(A) + [1+cos(B)]/sin(B) + [1+cos(C)]/sin(C)} / 4

Kerim

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https://www.mathisfunforum.com/profile.php?id=219354 2023-11-22T19:53:48Z https://www.mathisfunforum.com/viewtopic.php?id=30215&action=new

Bob

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https://www.mathisfunforum.com/profile.php?id=67694 2023-11-20T12:07:30Z https://www.mathisfunforum.com/viewtopic.php?id=30210&action=new
<![CDATA[Task Pack 1(equations and system of equations)]]> Shiver wrote:

1. Solve this equation: 3x-2+-6=4(x²+2x)2.

what is 2. at end? a square?

Shiver wrote:

2. Solve this equation: ln(4x)=(x-cos(x²))(3x-2)3.

what is 3. at end? (i dont think u can solve this w/ algebra; have to use numeric approx)

Shiver wrote:

3. create an linear system of equations where x=2,y=0 and z=2.5  4.

pick numbers for x,y,z. use what they gave you to get values.

e.g. 3x+2y+4z=3*2+2*0+4*2.5 = 6+0+10 = 16 so eqn is 3x+2y+4z=16

now do two more to get system

Shiver wrote:

Let a³x²+b²x+c=0. What values are a,b,c if a < b and c = 4a

put 4a in for c: a^3x^2+b^2x+4a=0

do quad fiormula for x in terms of a&b

do cases. e.g. if x=0 then what? (a=0 and b>0 so infinite sol'n set)

try other cases

(https://www.wolframalpha.com/input?i=a%5E3+x%5E2+%2B+b%5E2+x+%2B+4a+%3D+0%2C+a+%3C+b)

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https://www.mathisfunforum.com/profile.php?id=245511 2023-09-20T06:32:25Z https://www.mathisfunforum.com/viewtopic.php?id=29834&action=new
<![CDATA[nine squares]]> very good phrontister

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https://www.mathisfunforum.com/profile.php?id=8775 2023-06-01T19:02:16Z https://www.mathisfunforum.com/viewtopic.php?id=29486&action=new
<![CDATA[convex hexagon]]> Hi Bob...the guitarist is Gerhard Gschossmann.

And here's the link to the recording I used: Gerhard Gschossmann - 'Summertime' (George Gershwin) guitar solo fingerstyle

That version is played at the song's correct tempo, with good feeling and expression...which makes my Geogebra soundtrack sound rather chipmunkeyish.

Here's a vocal clip of the song: GEORGE & IRA GERSHWIN 'Summertime' HAROLYN BLACKWELL ('Porgy & Bess'). Very moving!!

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https://www.mathisfunforum.com/profile.php?id=40741 2023-05-16T22:58:08Z https://www.mathisfunforum.com/viewtopic.php?id=29210&action=new
<![CDATA[isosceles right-angled triangle]]> hi Tony123

It's been a while  ... my excuse is I've been away bird watching.  Had this method just before I went ... only just got back.

Will check it later for typos.

Bob

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https://www.mathisfunforum.com/profile.php?id=67694 2023-05-02T13:44:37Z https://www.mathisfunforum.com/viewtopic.php?id=29334&action=new
<![CDATA[Find the perimeter of the square]]> Well if one side is (x+7) and its a square, then the perimeter is that side times 4.

4(x+7)

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https://www.mathisfunforum.com/profile.php?id=244237 2023-03-24T22:30:04Z https://www.mathisfunforum.com/viewtopic.php?id=27932&action=new
<![CDATA[the largest possible]]> Thanks Bob great job

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https://www.mathisfunforum.com/profile.php?id=8775 2023-03-11T04:12:40Z https://www.mathisfunforum.com/viewtopic.php?id=29118&action=new
<![CDATA[Solve the equation]]> How can I get started?

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https://www.mathisfunforum.com/profile.php?id=243925 2023-02-20T06:53:14Z https://www.mathisfunforum.com/viewtopic.php?id=29007&action=new
<![CDATA[Group Theory]]> hi liker777

Welcome to the forum.

Well done, looks like you have found the solution ok.

Were you setting these as exercises or because you wanted help with doing them?

Bob

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https://www.mathisfunforum.com/profile.php?id=67694 2023-02-09T13:05:05Z https://www.mathisfunforum.com/viewtopic.php?id=4056&action=new
<![CDATA[interval]]> hi tony123

It's worth trialling different k values using the grapher here:
https://www.mathsisfun.com/data/function-grapher.php

You'll find that x = 1/4 is always the vertical line of symmetry, so one limit will be found by finding the equation when the graph goes through (1,0)

And as k becomes more positive the curve shifts up and eventually fails to cross the x axis at all. That should give you the other limit.

Bob

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https://www.mathisfunforum.com/profile.php?id=67694 2022-12-02T08:48:19Z https://www.mathisfunforum.com/viewtopic.php?id=28448&action=new