Melody

I would assume that they would if they requested to be notified in the first place. 

They may not have made this request.

(Plus it is an assumption on my part)

If sin(x) = 2/3 and sec(y) = 5/4 , where x and y lie between 0 and π/2, evaluate sin(x + y).

Thanks guest,

It is really good to see you give the question a go.

Unfortunately, your answer cannot be fully correct because sine is a ratio.  It is not going to be in degrees or radians.

----------------

x and y are both acute angles.

Consider a right angled triangle with one acue angle equal to x

opp=2

hyp=3

so   adj=sqrt(9-4) = sqrt 5

sinx=2/3        cosx=sqrt5 / 3

Consider a right angled triangle with one acue angle equal to x

cosy=4/5

adj=4

hyp=5

so   opp=sqrt(16) = 4

\(sin(x+y)=sinx cosy+cosxsiny\\ sin(x+y)=\frac{2}{3} \cdot\frac{4}{5}+\frac{\sqrt5}{3}\cdot\frac{4}{5}\\ sin(x+y)=\frac{20-4\sqrt5}{15}\\ sin(x+y)=\frac{4}{15}(5-\sqrt5)\)

4/15(5-sqrt(5)) approx = 0.7370485393333894    This is a ratio, it is not going to be in degrees or radians.

Maybe you are sometimes blocked for a short period of time until a mod unblocks you...

Even I am blocked if I post a web link. 

Once you have been unblocked there is no trail left behind (that i can see) to indicate that you were ever blocked in the first place.

I see no evidence that you are commonly blocked. 

\(\beta=3+2i\\ \alpha+\beta=k \qquad k\in R\\ so\\ \alpha=n-2i\\ \alpha-2\beta=wi \qquad w\in R\\ (n-2i)-2(3+2i)=wi\\ n-6-2i-4i=wi\\ (n-6)-6i=wi\\ n-6=0\\ n=6\\ \alpha=6-2i \)

I think I made that more complicated than necessary.

Your account has not been deleted because I still have all your old messages :)

Hi DS it is really good to see you here again!

There might be a problem but I think it is more likely that you have forgotten your password.

Maybe it is different from what you remember???

Tom..riddle asked me to reopen this one. 

I think she would like you to explain more how you got your answer Chris.

(I have not looked at it at all as yet)

2000 potatoes at 84 euro each  is 90000 euros         (my goodness they are expensive potatoes!)

X  potatoes at     45 euros is             45X     euros

Now we have  2000+X  potatoes.  And we want them to have an average price of  69 euros

So the total cost of all the potatoes will have to be     69(2000+X)    euros.

so we have

\(2000 \times 84  + 45x = 69(2000+x)\\ 2000 \times 84  + 45x = 69\times 2000+69x\\ 2000 \times 84  -69\times 2000 = 69X-45x\\ 2000(84-69)=(69-45)x\\ x=\frac{2000(84-69)}{(69-45)}\\ x=\frac{30000}{24}\\ x=1250 potatoes\\ \)

check

(2000*84+1250*45)/3250 = 69   Good :)

So what do we have 

\( x=\frac{2000(84-69)}{(69-45)}\\ \)

Let the original number of units be N     So  N=2000 potatoes

Let the original highher price be      H    so  H=84 euros

Let the new low price be       L                so L = 45 euros

Let the desired average price be A        so A= 69 euros

x is the additional number we need to buy to reduce the average price to the one that is desired.

and we have:

\(x=\frac{N(H-A)}{A-L}\)