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Dmitriy789 [7]
3 years ago
13

Sam can not read part of this homework problem. -7y +5(3 + ___y) Sam knows the expression simplifies to 3y + 15. What number bel

ongs on the blank line in the homework problem? Question 5 options: A. 10 B. -4 C. 4 D. 2
Mathematics
1 answer:
choli [55]3 years ago
7 0
Write and solve an equation, as follows:

-7y + 5(3+ny) = 3y + 15.  We are to find the value of 'n.'

-7y + 15 + 5ny = 3y + 15.

Subtracting 15 from both sides, we get    -7y + 5ny = 3y 

Grouping like terms, we get   5ny = 3y + 7y = 10y

Dividing both sides by 5y, we get n = 2   (answer)
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What is the coefficient of x2y3 in the expansion of (2x + y)5?
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Option C:

The coefficient of x^{2} y^{3} is 40.

Solution:

Given expression:

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$=\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}+\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1}+\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}+\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}

                                                            $+\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}+\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}

Let us solve the term one by one.

$\frac{5 !}{0 !(5-0) !}(2 x)^{5} y^{0}=32 x^{5}

$\frac{5 !}{1 !(5-1) !}(2 x)^{4} y^{1} = 80 x^{4} y

$\frac{5 !}{2 !(5-2) !}(2 x)^{3} y^{2}= 80 x^{3} y^{2}

$\frac{5 !}{3 !(5-3) !}(2 x)^{2} y^{3}= 40 x^{2} y^{3}

$\frac{5 !}{4 !(5-4) !}(2 x)^{1} y^{4}= 10 x y^{4}

$\frac{5 !}{5 !(5-5) !}(2 x)^{0} y^{5}=y^{5}

Substitute these into the above expansion.

(2x+y)^5=32 x^{5}+80 x^{4} y+80 x^{3} y^{2}+40 x^{2} y^{3}+10 x y^{4}+y^{5}

The coefficient of x^{2} y^{3} is 40.

Option C is the correct answer.

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