What types of lines are coplanar and intersect at a right angle?

Mathematics

2 answers:

lisabon 2012 [21]3 years ago

6 0

Answer:

Parallel lines are lines in a plane that are always the same distance apart. Parallel lines never intersect. Perpendicular lines are lines that intersect at a right (90 degrees) angle.

Step-by-step explanation:

pantera1 [17]3 years ago

4 0

Parallel lines are lines in a plane that are always the same distance apart. Parallel lines never intersect. Perpendicular lines are lines that intersect at a right (90 degrees) angle. Answer from goggle

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Which products result in a difference of squares? Select three options. (x minus y)(y minus x) (6 minus y)(6 minus y) (3 + x z)(

ValentinkaMS [17]

Answer:

Third option: $(3 + xz)(-3 + xz)$

Fourth option: $(y^2 - xy)(y^2 + xy)$

Sixth option: $(64y^2 + x^2)(-x^2 + 64y^2)$

Step-by-step explanation:

By definition, we can factor the Difference between two squares:

$a^2 - b^2 = (a+b)(a-b)$

In order to find which products results in a difference of squares, we need to check each option:

$1.\ (x - y)(y - x)=(x)(y)+(x)(-x)-(y)(y)+(-y)(-x)=xy-x^2-y^2+xy=-x^2-y^2+2xy$

The product does not result in a difference of squares.

$2.\ (6 - y)(6 - y)$

Since the signs are equal ( $-$ ), the product will not result in a difference of squares.

Using Distributive Property for the other options, we get:

$3.\ (3 + xz)(-3 + xz)=(3)(-3)+(3)(xz)+(xz)(-3)+(xz)(xz)=\\\\=-9+3xz-3xz+x^2z^2=x^2z^2-9=(xz)^2-(3)^2$

The product results in a difference of squares.

$4.\ (y^2 - xy)(y^2 + xy)=(y^2)(y^2)+(y^2)(xy)-(xy)(y^2)-(xy)(xy)=\\\\=y^4+xy^3-xy^3-x^2y^2=y^4-x^2y^2=(y^2)^2-(xy)^2$

The product results in a difference of squares.

$5.\ (25x - 7y)(-7y + 25x)=(25x - 7y)(25x - 7y)$

Since the signs are equal ( $-$ ), the product will not result in a difference of squares.

$6.\ (64y^2 + x^2)(-x^2 + 64y^2)=(64y^2 + x^2)(64y^2 - x^2)=\\\\=(64y^2)(-x^2)+(64y^2)(64y^2)+(x^2)(-x^2)+(x^2)(64y^2)=\\\\=-64y^2x^2+4096y^4-x^4+64y^2x^2=\\\\=4096y^4-x^4=(64y^2)^2-(x^2)^2$