Math Help Needed!!!

[BaliBowler]

so i looked at my HW today, and realized its review, and i dont rember any of it, and i couldnt get the answers, or the way to get the answers on ask.com so could anyone help me out?

1.) if one root of the equation x^2 + kx - 15 = 0 is -3, what is the other root?
(A)-2 (B)2 (C)3 (D)5
2.) the product of 6x^a and x is...
(A)6x^a (B)6x^a+1 (C)6x^2a (D)(6x^2)^a I chose D, it seems to be correct?(FIGURED IT OUT!!, thanks)
3.) what is the sum of the coefficients in the expansion of (a+b)^5
(A)5 (B)16 (C)32 (D)40

                     3 + 5 radical 3    
4.) the experssion  --------------  is equivalent to...
                     4 - 2 radical 3
    
    (A)-9+7 radical 3  (B)21+13radical3  (C)-18+14radical3  (D) 42-26radical3
        ---------------   ---------------     --------------       -----------
                2                 2                   4                  4

5.) witch statements could be used to prove that triangle ABC and triangle A'B'C' are congruent? (they put in a square as the symbol that i will replace with @) it seems to me that all of them work...
(A) (AB @ A'B'), (BC @ B'C') and (<A @ <A')
(B) (AB @ A'B'), (<B @ <B') and (<A @ <A')
(C) (<A @ <A'), (<B @ <B') and (<C @ <C')
(D) (AC @ A'C'), (BC @ B'C') and (<A @ <A')


can anyone help me finish this before tomarrow???????
thanks

 

[Garett]

The second one is B
Number 3 is 32, you can figure that out using Pasqual's Triangle.

 

[BaliBowler]

thanks forthe help i just remberd how to do #2, but i have never learned pasquals triangle before.... math 11R, i dont know if i just never learned(skiped) or its in a more advanced class?

 

[BaliBowler]

i dont get how to do this one... i do it like this, witch must be wrong lol
5c0(a)^5(b)^0 = a^5
5c1(a)^4(b)^1 = 5a^4 b
5c2(a)^3(b)^2 ect
5c3(a)^2(b)^3 ect
5c4(a)^1(b)^4 ect
5c5(a)^0(b)^5 ect

and if i do all that i get some big exponents with the a and b, but they arent in any of the answers...

 

[Garett]

Search for Pasqual's Triangle on Google and an image will come up. Its used for many things but in this case its easily used to find coefficients of expansion.

 

[Garett]

To do it manually you go (a+b)(a+b)(a+b)(a+b)(a+b)

 

[Point44]

5.) witch statements could be used to prove that triangle ABC and triangle A'B'C' are congruent? (they put in a square as the symbol that i will replace with @) it seems to me that all of them work...
(A)AB @ A'B', BC @ B'C' and <A @ <A'
(B)AB @ A'B', <B @ <B' and <A @ <A'
(C)<A @ <A', <B @ <B' and <C @ <C'
(D)AC @ A'C', BC @ B'C' and <A @ <A'


can anyone help me finish this before tomarrow???????
thanks

I think someone needs to help him with his english as well.

 

[BaliBowler]

yea i dont even know where to start......??

 

[BaliBowler]

I think someone needs to help him with his english as well.

i typed it fast, cut me some slack lol

 

[Garett]

(a+b)^2 = (a+b)(a+b) = a^2 + 2ab + b^2
does that look familar in the triangle?

 

[BaliBowler]

um, not really, im guessing a teacher forgot to teach this, as im not even close to understanding that...sorry im not asking you to teach me(its not your job), so you dont have to worry about it, thanks for trying though

 

[BaliBowler]

fixed #1, i copied it wrong, all are correct now

 

[Garett]

No worries...you may not have been taught that, but it make it alot easier to find coefficients of large expansions.

 

[Garett]

1.) if one root of the equation x^2 + kx - 15 = 0 is -3, what is the other root?
(A)-2 (B)2 (C)3 (D)5

The other root will be 5, if you need it k=-2

(x - 5)(x + 3)=0
x^2 - 2x -15 =0

 

[BaliBowler]

The other root will be 5, if you need it k=-2

(x - 5)(x + 3)=0
x^2 - 2x -15 =0

that makes sence lol, thanks it seems rather obvious now, but until i saw it i was completely blank
thanks

 

[phil000]

1. D
2. 6x^(a+1)
3. C
4. (please retype man, I can not tell what the question is&#8230
5. No clue

Thanks for all the help you've given me (with random tricks), i hope this helps...

My apologies for bad language here, I do really actually apologize...
(and I didn't realize mods actually checked language)


P.s. I landed 3 drunken helixes in a row last night, can't seem to repeat this feat...

 

[BaliBowler]

re did #4 hope that helps understand it

P.s. I landed 3 drunken helixes in a row last night, can't seem to repeat this feat...

awesome! its always great when you get more than 1, the trick is so smooth its great, good job though, probably one of the harder moves to get down efficently
OK i have 1,2 and 3 figured out, now 4 i have an idea, multiply by the conjugate(sp) but i dont rember how to do this, man i feel pretty dumb right now... and 5 still seems like they could all be right...

 

[BaliBowler]

alright, goin to sleep in a few, i will check tomarrow to see if anyone could help on #5
thanks for all the help everyone!

 

[johnniet]

Number 3 is 32, you can figure that out using Pasqual's Triangle.

It's simpler than Pascal's triangle though. Summing the coefficients for (a+b)^n
is always 2^n. Just add up (1+1)^n.

 

[Point44]

Have a look at this link...

No problem with Q no. 5.

Eg. S=Side, A=Angle, H=Hypotenuse, L=Leg
The only combinations that work are...

SSS
SAS
ASA
AAS
HL

(A) (AB @ A'B'), (BC @ B'C') and (<A @ <A')-------->SSA
(B) (AB @ A'B'), (<B @ <B') and (<A @ <A')--------->ASA
(C) (<A @ <A'), (<B @ <B') and (<C @ <C')--------->AAA
(D) (AC @ A'C'), (BC @ B'C') and (<A @ <A')-------->SSA

Hope that helps.