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mash [69]
3 years ago
13

Please help!!!

Mathematics
1 answer:
Lerok [7]3 years ago
5 0

The sum of x^2−3xy−y^2 and 2x^2+5xy−4y^2:

x^2−3xy−y^2 + 2x^2+5xy−4y^2

= 3x^2 + 2xy - 5y^2


From 6x^2−7xy+8y^2 subtract the sum of x^2−3xy−y^2 and 2x^2+5xy−4y^2:

6x^2−7xy+8y^2 - (3x^2 + 2xy - 5y^2)

= 6x^2−7xy+8y^2 - 3x^2 - 2xy + 5y^2

= 3x^2−9xy + 13y^2


Answer

3x^2−9xy + 13y^2

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Answer:

The area in factored form is A=(x+2)(x-5).

The area in standard form is A=x^2-3x-10.

Step-by-step explanation:

The area of a rectangle is length times width.

So the area here is (x+2)(x-5).

They are probably not looking for A=(x+2)(x-5) because it requires too little work.

They probably want A in standard form instead of factored form.

Let's use foil:

First x(x)=x^2

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Inner: 2(x)=2x

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---------------------Adding together:  x^2-3x-10.

The area in factored form is A=(x+2)(x-5).

The area in standard form is A=x^2-3x-10.

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