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kaheart [24]
3 years ago
14

Add the following polynomails -3x^2-2xy+y^2,-4x^2-xy+6y^2, x^2+6xy-y^2

Mathematics
1 answer:
Tomtit [17]3 years ago
8 0

Answer:

When we add the given polynomials the result is

(-3x^2-2xy+y^2)+(-4x^2-xy+6y^2)+(x^2+6xy-y^2)=3(2(y^2-x^2)+xy)

Step-by-step explanation:

Given polynomials are

-3x^2-2xy+y^2 ,-4x^2-xy+6y^2 , x^2+6xy-y^2

Now addding the polynomials we get

(-3x^2-2xy+y^2)+(-4x^2-xy+6y^2)+(x^2+6xy-y^2)=-3x^2-2xy+y^2+-4x^2-xy+6y^2+x^2+6xy-y^2

=-6x^2+3xy+6y^2  (adding the like terms on RHS)

=3(-2x^2+xy+2y^2)  (taking 3 outside on RHS)

=3(2(-x^2+y^2)+xy)

Rewritting the above equation we get,

=3(2(y^2-x^2)+xy)

Therefore (-3x^2-2xy+y^2)+(-4x^2-xy+6y^2)+(x^2+6xy-y^2)=3(2(y^2-x^2)+xy)

When we add the given polynomials the result is

(-3x^2-2xy+y^2)+(-4x^2-xy+6y^2)+(x^2+6xy-y^2)=3(2(y^2-x^2)+xy)

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Question 14, Part (i)

Focus on quadrilateral ABCD. The interior angles add to 360 (this is true for any quadrilateral), so,

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==================================================

Question 14, Part (ii)

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We have the two equations x+y = 180 and y+z = 180 to form this system of equations

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