Math Is Fun Forum

At the back of the book the answer given was 21.

It says we should simplify it. But I don't know if it could be factorized. I tried doing it but the six has given me a tough time, it cannot be reduced to have  3 in order to have the same base as the others.

I used zeros for the unprovided bearings.  And got every thing wrong. I don't know any method well, apart from the vector approach. I don't seem to understand your workings, but see how I did mine:

From A to north;  vector AN = (10km 000)
From north to east; vector NE = (5km, 000)
From east to the end let say P, therefore: vector EP = (10km, 045).

I used zeros for the unknown bearings, am I right?

27^n+2 - 6*3^3n+3/3^n9^n+2
That is it.

Thanks.

between the six and the three is multiplication sign. that is only problem.

Please produce them here.
Thanks.

Please help me solve the following, I can solve,  but the 6 has become my headache, I can't figure out how to go about the 6.

b)

Please, the

Thanks in advance.

Son.

Thanks bobbym, I get you.

Thank you very much indeed Sir!, in fact you're darling!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

This is it:

A cyclist starts a journey from town A . He rides 10 km north, then 5 km east and finally ten 10 km on a bearing of 045 degrees

a) How far east is the cyclist's destination from A?

b) How far north is the cyclist's destination from town A? 

c) Find the distance and the bearing of the  cyclist's destination from A?. ( correct  your answer to the nearest km and degree).

This is the problem.

Thanks. God bless.

I must factorise the denominator, and here I go:

x^3 - 2x + x = 0

= x^3 - x^2 - x^2 + x = 0
= x^2(x - 1) - x(x - 1) = 0

Therefore:

(x^2 - x)(x- 1)

Numerator looks like difference of two square. There I go:

(x - 1)(x+ 1)/(x^2 - x)(x- 1) = x + 1/x^2 - x) = x + 1/ x(x - 1) . I hope I am correct.

With the denominator I had: x^3 - 2x^2 + x.