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3 years ago

6

(p+q)2-(5-5) p=1 q=1

2 answers:

IgorC [24]3 years ago

7 0

Answer: 4

Step-by-step explanation:

Since p=1 and q=1

Step 1: substitute the values of p and q into the equation to give

(1+1) 2- (5-5)

Step 2- apply BODMAS, open bracket first

(2) 2- (0)

(2) 2

2 x 2 =4

valkas [14]3 years ago

6 0

Answer:

The answer is 4

Step-by-step explanation:

Solve:

p = 1

q = 1

= (1+1)2 - (5-5)

= 2+2 -5 +5

= 4

Make sure to give me brainliest answer!

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3 years ago

How do i do 3 part a ?

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:

${11 \choose 0} = 1$

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

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

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

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