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3 years ago

Photo attached. Thanks in advance!

Mathematics

2 answers:

jonny [76]3 years ago

7 0

A zero in front of the x would make this a difference of two perfect squares. All of the other option would interfere by adding another term.

Iteru [2.4K]3 years ago

5 0

$HEY THERE !!$

CORRECT answer is C. 0

because 0 will produce difference of squares other all will not.

HOPE IT HELPED YOU.

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3 years ago

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3 years ago

Find the perimeter of the triangle with vertices A(negative 2,2,3), B(4,negative 4,3), and C(7,6,4).

gayaneshka [121]

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$P\approx 28.872\,\text{units}$

Step-by-step explanation:

We are given coordinates of each of the vertices of the triangle ABC, hence we can use the distance formula to find the lengths of each of the side.

the distance formula is generally written as:

$r = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

for the side AB: A(-2,2,3) and B(4,-4,3)

$AB = \sqrt{(-2-4)^2+(2-(-4))^2+(3-3)^2}$

$AB = \sqrt{(-6)^2+(6)^2+(3-3)^2}$

$AB = \sqrt{72}$

for the side BC: B(4,-4,3) and C(7,6,4)

$BC = \sqrt{(4-7)^2+((-4)-6)^2+(3-4)^2}$

$BC = \sqrt{110}$

for the side CA: C(7,6,4) and A(-2,2,3)

$CA = \sqrt{(7-(-2))^2+(6-2)^2+(4-3)^2}$

$CA = \sqrt{98}$

Now we all the sidelengths:

to find the perimeter we need to just sum the three side lengths:

$P = AB + BC + CA$

$P = \sqrt{72} + \sqrt{110} + \sqrt{98}$

$P\approx 28.872\,\text{units}$

this is the perimeter of triangle ABC

6 0

3 years ago