# How do you find the coefficient of #x^3y^2# in the expansion of #(x-3y)^5#?

The coefficient of

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To find the coefficient of (x^3y^2) in the expansion of ((x-3y)^5), you can use the binomial theorem or Pascal's triangle. The coefficient can be calculated using the formula:

[\binom{n}{k} \cdot a^{n-k} \cdot b^k]

where (n) is the power to which the binomial is raised, (k) is the term number, (a) and (b) are the terms in the binomial, and (\binom{n}{k}) is the binomial coefficient.

In this case, (n = 5), (a = x), (b = -3y), and (k) is determined by the power of (x) and (y) in the term, which is (3) for (x^3) and (2) for (y^2).

Substituting these values into the formula:

[\binom{5}{3} \cdot (x)^{5-3} \cdot (-3y)^{3}]

[= \binom{5}{3} \cdot x^2 \cdot (-3)^3y^3]

[= 10 \cdot x^2 \cdot (-27)y^3]

[= -270x^2y^3]

So, the coefficient of (x^3y^2) in the expansion of ((x-3y)^5) is (-270).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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