Ronan

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14 years, 155 days
East Grinstead, United Kingdom

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These are answers submitted by Ronan

On my pc Maple docs are here. Or are you asking about windows File History?

If memory serves me correctly, I think I specified Home User. As I have a Personal Edition license.

Hope this is correct
 

restart

L:=1

1

(1)

E:=[0,0]

[0, 0]

(2)

A:=[-1/2*L,L*sqrt(3)/2]

 

[-1/2, (1/2)*3^(1/2)]

(3)

D1:=[L,0]

[1, 0]

(4)

C:=[L,L]

[1, 1]

(5)

AC:=C-A

[3/2, -(1/2)*3^(1/2)+1]

(6)

AB:=L

1

(7)

BC:=L

1

(8)

expn1:=2*L*cos(theta)=sqrt(AC[1]^2+AC[2]^2)

2*cos(theta) = (1/2)*(9+4*(-(1/2)*3^(1/2)+1)^2)^(1/2)

(9)

theta:=solve(expn1,theta)

arccos((1/4)*(9+4*(-(1/2)*3^(1/2)+1)^2)^(1/2))

(10)

b:=2*Pi-(Pi-2*theta)

Pi+2*arccos((1/4)*(9+4*(-(1/2)*3^(1/2)+1)^2)^(1/2))

(11)

simplify( (11) );

2*Pi-2*arcsin((1/2)*(4-3^(1/2))^(1/2))

(12)

expn2:=L-L*cos(c)=L*sin(2*Pi/3)-L*sin(a+Pi/3)

1-cos(c) = (1/2)*3^(1/2)-sin(a+(1/3)*Pi)

(13)

expn3:=L-L*sin(c)=L*cos(2*Pi/3)+L(a+Pi/3)

1-sin(c) = 1/2

(14)

c:=solve(expn3,c)

(1/6)*Pi

(15)

a:=solve(expn2,a)

-(1/3)*Pi+arcsin(3^(1/2)-1)

(16)

simplify( (16) );

-(1/3)*Pi+arcsin(3^(1/2)-1)

(17)

B[1]:=L-L*sin(c)

1/2

(18)

B[2]:=L-L*cos(c)

-(1/2)*3^(1/2)+1

(19)

B:=[B[1],B[2]]

[1/2, -(1/2)*3^(1/2)+1]

(20)


plots:-display(plottools:-polygon([E,D1,C,B,A]),colour=yellow)

1

 

 

 


 

Download Fun-Angles.mw

In you expression

test := subs(a = 2, x^2/2 + x*a + a^2*ln(x - a))

ln(x-a) becomes complex with x<a.

I dont know is this is the Clifford package you mean. I found this on last friday. Index of /12/eng/files/st_files/kyrchei/kyr_files/Maple Package

You can set the number of digits displayed on the screen using for eample

interface(displayprecision=5)

This does not affect the accuracy of calculation set by Digits

I use 4k Video Downloader. There is a free version of it. AFAIK if you are not redestributing the videos so purely for you own use it is not a problem. 

There are  a couple you tube chanels I use. Insights into Mathematics and Wild Egg Maths. I downloaded most of the videos. About 3 weeks ago both chanels were hacked and all the content was gone.So far only Wild egg maths has been restored.

Basically if you value it get you own copy.

I tried setting transparency to 0 but that didn't work. I used plottools:-point as a work around.

L:=plottools:-line([0,0],[0,3],color="blue"):
c1:=plottools:-disk([0,1],.1):
c2:=plottools:-disk([0,2],.1,color="green",transparency=0):
c3:=plottools:-point([0,1],symbol=solidcircle,symbolsize=45,color="red");
plots:-display([L,c1,c2,c3],scaling=constrained,axes=none)

 

I answered something similar in 2016. I don't have the time right now to investicate you example.

How to take derivative of sum? - MaplePrimes

10:55-10:10 =45
11:58-10:55=63
9 divides 45 and 63

so 9

Use align and \n (within the text) as an option in textplot or textplot3d. This is just a quick example.

with(plots);with(plottools)
A:=[0,0,0];C[1]:=[1,1,1]
plt1:=point([A,C[1]],symbol=solidcircle,symbolsize=16,color=blue)
plt2:=textplot3d([ [A[],"A",align={above,right}] , [C[1][],typeset("C[1]=\n",C[1]),align={below,left}] ]):

display(plt1,plt2)

One way of doing it.

restart

local D;

BD:=4;
AB:=6;

D

 

4

 

6

(1)

eq1:=BD*sin(4*theta)=AB*sin(2*theta)

4*sin(4*theta) = 6*sin(2*theta)

(2)

sol:=solve(eq1,[theta])

[[theta = 0], [theta = (1/2)*Pi], [theta = (1/2)*arctan((1/3)*7^(1/2))], [theta = -(1/2)*arctan((1/3)*7^(1/2))]]

(3)

assign(sol[3]);
theta:=simplify(theta)

(1/4)*arctan(3*7^(1/2))

(4)

eq2:=AB*sin(2*theta)=BC*sin(theta)

6*sin((1/2)*arctan(3*7^(1/2))) = BC*sin((1/4)*arctan(3*7^(1/2)))

(5)

#eq3:=eval(eq2,(sol[3]))

 

simplify(solve(eq2,[BC])[])

[BC = 12*cos((1/4)*arctan(3*7^(1/2)))]

(6)

assign(%)

BC

12*cos((1/4)*arctan(3*7^(1/2)))

(7)

DC:=BD*cos(Pi-4*theta)+BC*cos(theta)

-1/2+12*cos((1/4)*arctan(3*7^(1/2)))^2

(8)

simplify( (8) );

10

(9)
 

 

Download 2024-09-18_A_Find_DC_geometry.mw

180-2m+60+k = 2n+40+k  implies m+n=100  so m=100-n

m+40+n+x=180

100-n+40+n+x=180

therefore x=40

Looking at it.ABC is half the area and ABE is quater the area. Draw a vertical line from E intersects the diagonal midpoint at say G. Triangle ABF and EGF are similar. with a 4;1 area ratio. So the horizontal coordinate of F is 2/3 AE. also ABF+EGF is quater of the total area.

The triangle EGC area.  is 1/8 of the total area so 3cm^2.

ABF+2* EGF would be 6 (quater of the total area) 

EGF would be 1/6 of that =1. So  total area of 4.

Could you put the zip file on One Drive Google Docs or Dropbox for example and have a link to it it in your post.

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