Subtract 6(x^2-xy) from 3x(x-y)

Mathematics

2 answers:

den301095 [7]2 years ago

7 0

$3 {x}(x - y) - 6 ({x}^{2} - xy) \\ = 3 {x}^{2} - 3xy - 6 {x}^{2} + 6xy \\ = 3xy - 3 {x}^{2}$

=3x(y-x)

rusak2 [61]2 years ago

4 0

Answer:

-3x ( x - y )

Step-by-step explanation:

3x ( x - y ) - 6 ( x² - x y )

Solve the brackets.

3x² - 3 x y - 6x² + 6 x y

Combine like terms.

3x² - 6x² - 3 x y + 6 x y

-3x² + 3 x y

Factorize the answer.

-3x ( x - y )

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Let n be a positive integer and define [n] to be the set of the first n positive integers. That is, [n] = {1, 2, 3, . . . , n}.

yaroslaw [1]

Answer: There are $2^{n-1}$ ways of doing this

Hi!

To solve this problem we can think in term of binary numbers. Let's start with an example:

n=5, A = {1, 2 ,3}, B = {4,5}

We can think of A as 11100, number 1 meaning "this element is in A" and number 0 meaning "this element is not in A"

And we can think of B as 00011.

Thinking like this, the empty set is 00000, and [n] =11111 (this is the case A=empty set, B=[n])

This representation is a 5 digit binary number. There are $2^5$ of these numbers. Each one of this is a possible selection of A and B. But there are repetitions: 11100 is the same selection as 00011. So we have to divide by two. The total number of ways of selecting A and B is the $2^{5-1} = 2^4$ .