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A cyclist starts a journey from town A. He rides 10km north, then 5km east and finally 10km on a bearing of 045.
(a) How far east is the cyclist's destination from town A?
(b) How far north is the cyclist's destination from A?
(C) Find the distance and bearing of the cyclist's destination from town A.
(Correct your answers to nearest kilometers and degree)
Is hard for me using vector approach
I started by saying
(10km, 000°) (5km, 000°) (10km, 045°) I used Zero degrees for the unknown degrees.
(10cos000° and 10sin000°) =(10*cos90 and 10*sin 90) = (0 And 10) I did the same for the rest.
Thank you. I will put a bearing question here which I think cannot be solved by vector appraoch. As far as I remember, Bob bundy made it clear to me that not all bearing questions could be solved using vector approach, I wanted to know this truth from different perspective. Thanks once more!
Pece kololo!
Hi,
Please, I wanted to know if all bearing problems could be solved by using vector approach method. Please confirm
Many thanks!
Hi;
I am having problems getting the answer;
An arc substends angle of 60° of the centre of a circle of radius 14cm.
(a) Find the perimeter of major sector
I had 76.67 as the answer but not sure - please confirm.
Many thanks!
Thanks Agnashom!
One question left and that is;
When one crosses diagonal lines within a rectangle which triangle is formed?
Hi; I have some few questions to ask;
(1) Are the internal angles of rhombus, square, rectangle and parallelograms equals 360?
(2) When one makes diagonal angles within rhombus is the triangle formed be an isosceles or equilateral?
(3) When one crosses diagonal lines within a square, is the triangle formed be equilateral or isosceles?
Many thanks
I have made the amendment- thank you
I have bracket
Hi;
I am given it a try concerning the conversion;
Log x = 1/2log x
so, it will be
log (x) = 1/2log (x)
Is that correct?
Please, help.
I don't understand the question.
Show that
Log x = 1/2logx and hence solve log x + 1/2 = logx
In both cases the LHS is in base 4 and the RHS is in base 2
Thanks
This is it;
log(x^2 + 1) - 2logx = 1
Ask the tutor to plug and check his answer.
Yes - have seen the tutor - the tutor gave me a calculator which I plugged 1/3 into the equation and I had 1(one), so it seems his answer is correct. What do you say Bobbym?
I have plug it in and the calculator gave me "math error". But I want to add that the base is five [5] for all.
Ok. Let me put this
Log(x-3) - logx = 2.
Is negative one (-1) the correct answer?
Good!
Thanks bobym - I will post problems.
Big thanks!
Again;
x = 10/log16
correct?
like this;
xlog16 = 10
No, the original is log16^x.
xlog16 = log10
xlog16 = 1
x = 1/log16
Correct?
Thanks! Bobbym
I am now getting logarithm better.
See;
solve for x in, log16^x.
log16^x = 10
xlog16 = log10
x = log10/log16
correct?
I think your methods are good concerning the simplification of like terms in logs. Thanks for answering.
Look it if it is good
log16^-x = log2 + log7
-xlog16 = log14
x= -log14/log16
Is it good?
Still I haven't come across the tutor - but I will definitely meet him.
My one question is - when do one take antilogs of both sides of an equation?
I am away from the tutor now - but I shall tell him later
The book says - if the base is not written then it is base 10. And with this problem the base is not written - so it used base 10