4 years ago

PLEASE HELP MATH GOOD POINTS

1 answer:

4 0

I'd say it's option C, (-1, 3)

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s344n2d4d5 [400]

True got it right in edge 2020

3 0

3 years ago

WINSTONCH [101]

In general the binomial expansion is

$(a+b)^n = {n \choose 0} a^0 b^n + {n \choose 1} a^1 b^{n-1} + {n \choose 2} a^2 b^{n-2} + ... + {n \choose n} a^n b^0$

So in our case, because we want ascending powers of x we'll write,

$(-3x + 1)^{11} = {11 \choose 0} (-3x)^0 1^{11} + {11 \choose 1} (-3x)^{1} 1^{10} + {11 \choose 2} (-3x)^{2} 1^9 + {11 \choose 3 } (-3x)^3 1^8 + ...$

We need to calculate the binomial coefficients:

${11 \choose 0} = 1$

${11 \choose 1} = 11$

${11 \choose 2} = \dfrac{11 \times 10}{2} = 55$

${11 \choose 3} = \dfrac{11 \times 10 \times 9}{3 \times 2} = 165$

$(-3x+1)^{11} = 1 (-3x)^0 1^{11} + 11(-3x)^{1} 1^{10} + 55 (-3x)^2 1^{9} + 165 (-3x)^3 1^8 + ...$

$(1-3x)^{11} = 1 -33 x + 495 3x^2 - 4455 x^3+ ...$

6 0

3 years ago

kramer

The correct answer is 57

4 0

3 years ago

Dmitrij [34]

First distribute
combine like terms
subtract 6x from both sides
then you get no solutions which is a special case

8 0

3 years ago

Bogdan [553]

The first option is how you wrote your statement in math.

7 0

3 years ago