Yes, on your calculator there should be a $\displaystyle _nC_r$ button. To find $\displaystyle _5C_2$, enter 5 in your calculator, press the $\displaystyle _nC_r$ button, then press 2. and finally, press enter, or ans or whatever. but this one is simple enough to do without a calculator. expecially if you do it using Pascal's triangle. the binomial theorem way is no exception however. you need to learn how to cancel between factorials and break down the answer
[quote=Jhevon;55955]what are you talking about? 27 has nothing to do with exponents
lol sorry i meant number 22 instead of 27
buut 27 wouldve been my next question, im trying to reveiw this regents before mine on thursday and im so confused
but you guys are helping alot
thanks you very much for all the help
i have a few more to finish off this regents
28 i have no idea how to do
29 im dont know the differences b/w at least, at most and those
30 im just not good with sin and cos stuff
31 is just confusing
32 i get since i can put it on a calculator
and 33 and 34 are just plain hard
if you guys can answer it thank you so much and if not still thank you so much for all your help
im okayish with part I and alittle for part II but the part IIIs and IVs are hard
sorry im using up all your time asking you guys these questions but your answers are helping me alot
sorry for all the questions but 2 more questions i forgotten to ask
for number 25 would i use area of a triangle? if so would the setup for it be K=1/2(11 x 13) sin(70)? if not how do you do it, if so say yes? thanks
and nuber 26 how would i do that also?
thanks alot for all the help
$\displaystyle \sqrt {3q + 7} = q + 3$ ........square both sides
$\displaystyle \Rightarrow 3q + 7 = (q + 3)^2 = q^2 + 6q + 9$
$\displaystyle \Rightarrow q^2 + 3q + 2 = 0 $
This is a quadratic eqaution. solve it, then plug in the values for q that you get to see which solutions are extraneous (we often get extraneous, that is, solutions that don't work, when we square both sides of an equation).
Can you take it from here?
at least means greater than or equal to. so at least 5 means 5 or more.
at most means less than or equal to, so at most 5 means 5 or less
i don't suppose you have a problem with phrases like "greater than," "less than," or "exactly"
you need to use Bernoulli Trials to do this problem. Now that you know what at least means, can you do it?
well, you need to get good with trig. there are a lot of problems that use trig, as you can see. for this problem you need to know two things:
(1) How to solve a quadratic equation
(2) $\displaystyle \cos 2 \theta = \cos^2 \theta - \sin^2 \theta = 1 - 2 \sin^2 \theta$
$\displaystyle 3 \cos 2 \theta + \sin \theta - 1 = 0$
$\displaystyle \Rightarrow 3 \left( 1 - 2 \sin^2 \theta \right) + \sin \theta - 1 = 0$
$\displaystyle \Rightarrow 3 - 6 \sin^2 \theta + \sin \theta - 1 = 0$
$\displaystyle \Rightarrow -6 \sin^2 \theta + \sin \theta + 2 = 0$
$\displaystyle \Rightarrow 6 \sin^2 \theta - \sin \theta -2 = 0$
Note that this is quadratic in $\displaystyle \sin \theta$, you can see this explicity if you were to say let $\displaystyle \sin \theta = x$ then you would get $\displaystyle 6x^2 - x - 2 = 0$. But don't bother with that, it's usually a waste of time. Just solve this like a regular quadratic and tell me the answers you get for $\displaystyle \theta$