1. Originally Posted by topsquark
The binomial theorem says:
$(a + b) = a^n + [ _nC_1 ]a^{n - 1}b^1 + [ _nC_2 ]a^{n-2}b^2 + ... + [ _nC_{n-1} ]a^1b^{n-1} + b^n$
Where $_nC_r = \frac{n!}{r!(n - r)!}$
(Where n is a positive integer.)

Here a = cos(x), b = 3, and n = 5 so the third term is:
$_5C_2 a^3b^2 = \frac{5!}{2!3!} \cdot cos^3(x) \cdot 3^2 = 90cos^3(x)$

-Dan
for this question can you do it on the calculator?

2. Originally Posted by topsquark
18. Which of the following has imaginary roots?

I'm going to post this one anyway, even though Jhevon has given you a good answer.

The roots of a quadratic equation have an imaginary component if the discriminant $D = b^2 - 4ac$ is negative.

So let's put all of these into standard form and see which ones have negative discriminants:
$x(5 + x) = 8$ ==> $x^2 + 5x - 8 = 0$ ==> $D = 5^2 - 4 \cdot 1 \cdot (-8) = 57$

$x(5 - x) = -3$ ==> $x^2 - 5x - 3 = 0$ ==> $D = (-5)^2 - 4 \cdot 1 \cdot (-3) = 37$

$x(x + 6) = -10$ ==> x^2 + 6x + 10 = 0[/tex] ==> $D = 6^2 - 4 \cdot 1 \cdot 10 = -4$

$(2x + 1)(x - 3) = 7$ ==> 2x^2 - 5x - 7 = 0[/tex] ==> $D = (-5)^2 - 4 \cdot 2 \cdot (-7) = 81$

So we see that #3 has a negative discriminant and thus has an imaginary component. (As Jhevon already told you.)

-Dan
i like this solution better, thanks Dan. it has less to do with intuition and more to do with cold, hard math

3. Originally Posted by Jhevon
i like this solution better, thanks Dan. it has less to do with intuition and more to do with cold, hard math
lol i dont like cold/hard math

4. Originally Posted by shanegoeswapow
for this question can you do it on the calculator?
Yes, on your calculator there should be a $_nC_r$ button. To find $_5C_2$, enter 5 in your calculator, press the $_nC_r$ 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

5. Originally Posted by shanegoeswapow
for this question can you do it on the calculator?
If your calculator can expand products then yes. But I'd recommend some facility with the method as well.

-Dan

6. [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

7. 22. Solve for m: $3^{m + 1} - 5 = 22$

$3^{m + 1} = 27 = 3^3$

Thus
m + 1 = 3

m = 4

-Dan

PS Gotta go. Good luck on the exam!

8. Originally Posted by Jhevon
you should. it makes life easier for you.

which would you prefer? someone showing you how to do a math problem using eqautions and formulas, or someone saying, "this is the answer, i can feel it. i can't explain to you why this is the answer, but i know it's right! TRUST ME"
(chuckles) Yeah, trust him on this!

-Dan

9. 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

10. Originally Posted by topsquark
22. Solve for m: $3^{m + 1} - 5 = 22$

$3^{m + 1} = 27 = 3^3$

Thus
m + 1 = 3

m = 4

-Dan

PS Gotta go. Good luck on the exam!
i just noticed an error from this
m+1=3
m=2
you subtract 1 (i feel smart lol)

11. 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

12. Originally Posted by shanegoeswapow
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
$\sqrt {3q + 7} = q + 3$ ........square both sides

$\Rightarrow 3q + 7 = (q + 3)^2 = q^2 + 6q + 9$

$\Rightarrow q^2 + 3q + 2 = 0$

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?

Originally Posted by shanegoeswapow
29 im dont know the differences b/w at least, at most and those
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?

13. Originally Posted by Jhevon
$\sqrt {3q + 7} = q + 3$ ........square both sides

$\Rightarrow 3q + 7 = (q + 3)^2 = q^2 + 6q + 9$

$\Rightarrow q^2 + 3q + 2 = 0$

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?
so i just need to factor it out then?

14. Originally Posted by Jhevon
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?
more help if you dont mind? im not good with these problemos ><

15. Originally Posted by shanegoeswapow
30 im just not good with sin and cos stuff
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) $\cos 2 \theta = \cos^2 \theta - \sin^2 \theta = 1 - 2 \sin^2 \theta$

$3 \cos 2 \theta + \sin \theta - 1 = 0$

$\Rightarrow 3 \left( 1 - 2 \sin^2 \theta \right) + \sin \theta - 1 = 0$

$\Rightarrow 3 - 6 \sin^2 \theta + \sin \theta - 1 = 0$

$\Rightarrow -6 \sin^2 \theta + \sin \theta + 2 = 0$

$\Rightarrow 6 \sin^2 \theta - \sin \theta -2 = 0$

Note that this is quadratic in $\sin \theta$, you can see this explicity if you were to say let $\sin \theta = x$ then you would get $6x^2 - x - 2 = 0$. 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 $\theta$

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