Melody

A is easy,

Work on the LHS to prove it is equal to the RHS

Multiply it by   $$\\(1+cos\theta)/(1+cos\theta)\\$$

$$\\$you will also need to recognise that $\\sin^2\theta+cos^2\theta=1$$

I will leave you to work out the details  :))

That is easy it is 27

The question really is

If 27is divided by 128 what is the remainder.

Well 128 goes into 27  ZERO time and the remainder is 27

Thanks Bertie   

That makes god sense  :)

I share your concern.

It looks like that answer was not given until an hour after the question was posted so it most likely did not help in the actual exam.

This is why internet access should NEVER be allowed in exams!  

(unless the ability to use the internet is being tested)

$$\\x^2+y^2=29\qquad (1)\\
y=x^2+1\\
x^2=y-1 \qquad (2)\\
$Note: y-1 MUST be $ \ge 0 \;\;SO\;\; y\ge 1\\
$sub (2) into (1)$\\
y-1+y^2=29\\
y^2+y-30=0\\
(y-5)(y+6)=0\\
y=5 \;\;or\;\;y=-6\\\\
$y must be greater or equal to 1 so y=-6 is invalid in the real number system$
y=5\;\;sub\;to\;find\;x\\
x^2=5-1=4\\
x=\pm2\\\\
$So the 2 solutions are (2,5)\;\;and\;\;(-2,5)$$

Here is a really neat graph to show you what is happening 

0.89 is the square root of 0.89^2

You need to be a member to post pictures :)

Thanks Chris, I still have to think about it :))

@@ End of Day Wrap   Thurs 4/6/15   Sydney, Australia Time   1:05am  (Yes its really Friday)   ♪ ♫

Greetings,

Wonderful answers today from CPhill, Radix, Alan, DarkBlaze347, thejamesmachine, helper, mybfisbest, Brodudedoodebrodude and asinus.   Thank you  

Sorry anon but you cannot post pictures unless you join up.

It is very easy so why don't you do it :)