2math problems

[tRNA]

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ok, i don't know how to expand this (x+3)^5, do you just multiply it by itself 5 times?? topscore gives this long formula, i have attached it, but is there easier way to expand, or do i just have to memorize that formula? they also say for that problem k=3, n=5, a=x, b=3

2) when they give you a=(3,-6), b=(1,1), c=(7,6) then ask u to find the area of the tiangle? i drew it on the #lines then tried to find out the length of base and hight by counting the spaces then doing the regular 1/2bh formula for the area of tiangle but they said NO, you have to use this formula to find the area
area=1/2((x1)y2+(x2)y3+(x3)y1)-((y1)x2+(y2)x3+(y3)x1) (the #s next to x &y are subscripts)
why won't using just the regular formula for area of triagle give the answer and what if you had these same points and it asked for perimeter, is there a formula for that too or can I use the regular formula of adding the 3side lengths??

thanks alot for helping

 

[Streetwolf]

#1 - Use Binomial Theorem (think: Pascal's Triangle):

(x+y)^n = (n choose 0)(x^n)(y^0) + (n choose 1)(x^(n-1))(y^1) + (n choose 2)(x^(n-2))(y^2) + ... + (n choose (n-1))(x^1)(y^(n-1)) + (n choose n)(x^0)(y^n)

Note a few things:

(1) We start with (n choose 0) and end with (n choose n).
(2) We start with x^n and end with x^0.
(3) We start with y^0 and end with y^n.
(4) Each subsequent term subtracts 1 from the x exponent and adds 1 to the y exponent.
(5) The y exponent is the same as the number 'k' in (n choose 'k').
(6) The x exponent is the same as the number 'n-k'.
(7) As a result, x + y = (k) + (n-k) = n. Always.
(8) If, instead of x or y, you have an actual number, you just put that number into the above equation. So if y = 3, as in our example, and you were up to the term x^2 * y^3, you would instead have x^2 * (3^3) = 27x^2.
(9) Remember that (n choose k) = n!/((n-k)!k!)

(x+3)^5 = (5 choose 0)x^5*3^0 + (5 choose 1)x^4*3^1 + (5 choose 2)x^3*3^2 + (5 choose 3)x^2*3^3 + (5 choose 4)x^1*3^4 + (5 choose 5)x^0*3^5

= 1*x^5*1 + 5*x^4*3 + 10*x^3*9 + 10*x^2*27 + 5*x^1*81 + 1*1*243

= x^5 + 15x^4 + 90x^3 + 270x^2 + 405x + 243

#2 - The equation is correct after a quick glance. It uses determinants and is just simplified for your convenience. The problem with the normal formula is that you DON'T know the base and height. You can randomly choose one of the sides to be either the base or the height, but since there are no perpendicular lines, you don't know the value of the other. You have to use determinants. The formula you have takes this into consideration.

I think there's another thread with that exact problem. If so, the answer is 26.