Melody

A thread on how to upload a pic was posted very recently.

It is the first one in the STICKY TOPICS on the bottom  right of your screen

Here is where

Mathcourts24, 

How about saying thankyou to heueka for all his time consuming effort which he afforded you!

Don'r just grumble about getting it wrong!

You tried to vote your question up about 10 times but you have not even attempted to give Heureka a point.

That is incredibly rude of you!  Do you think that will encourage anyone to help you next time?

Coolstuff, could you put only one of these questions per post.

It makes it more confusing to hve 2 or more  plus there should only be one question per post anyway. :)

ok so you use    \cap = intersection    only for sets

           and      \wedge  for the same thing when it is not written in set notation.

But 

you use  \cup = union    for sets and non-sets notation...

Thanks  Rom.

Hi Rom,

I was trying to work out what your  \(\wedge \)    (\wedge)  meant ....

It intersection I think  ??       \(\cap\)              \cap   

I have never seen a wedge used before....

 For dinner, Javier will choose one of three options and one of the three side dishes. His choices for fish are salmon, trout, and halibut. His choices for side dishes are fries, cooked carrots, or coleslaw. 

          (a) Draw a tree diagram for the sample space of all dinner combinations. 

In a given plane, points A and B are 10 units apart. How many points C are there in the plane such that the perimeter of ABC is 50 units and the area of ABC is 100 units?

Without much thought I would assume the answer is infinite.

But let's take a look.

I am using the triangle below. You can see that the sides add up to 50.

Now

\(Area= 0.5*10*h=100\\ 5*h=100\\ h=20\\ \)

I just did a stack of work that i will not even bother to display.  

BECAUSE

If the perpendicular height must be 20 then the sum of the two slant heights of the triangle must be more than 40.

But this is impossible so there are no solutions.    I think that you guessed correctly!

You beat me to it StarStrike!

Good Work!

Conside the line from E to RC that is parallel to BC.     Let the point of intersection be X

EX=24

XR=24-6=18

find ER

\(ER=\sqrt{24^2+18^2}\\ ER=\sqrt{900}\\ ER=30 \)

Now PR is just the diagonal of one side of the square

\(PR=\sqrt{24^2+24^2}\\ PR=\sqrt{1152}\\ \)

Now I have all 3 sides of the triangle so I can use cosine rule to find the angle between them

the sides are      \(30, \quad30, \quad \sqrt{1152}\)

\(1152=900+900-2*900*Cos(PER)\\ 1152=1800-1800*Cos(PER)\\ -648=-1800*Cos(PER)\\ 0.36=Cos(PER)\\ \angle{PER}=cos^{-1}0.36=68.8998^\circ=68^\circ54' \)

That is assuming I've not made any stupid mistakes.