Mathematics, 15.01.2020 22:31 vcasela

Hijk is a parallelogram because the midpoint of both diagonals is which means the diagonals bisect each other.

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HIJK is a parallelogram because the midpoint of both diagonals is unique and intersect each other, which means the diagonals bisect each other.

Step-by-step explanation:

We are given the following in the question.

HIJK is a parallelogram.

So it satisfies the property of parallelogram.

The diagonals of a parallelogram bisect each other.That is the intersection of the diagonals of the parallelogram is unique and divide the diagonals in two equal parts each.

Thus,

HIJK is a parallelogram because the midpoint of both diagonals is unique and intersect, which means the diagonals bisect each other.

Answer from: kaygirlnelson6258

HIJK is the parallelogram

The mid point of both the diagonals is unique and divides both parts of same diagonal in equal ratios , which means the diagonals bisect each other.

A parallelogram is a quadrilateral with opposite sides equal and parallel to each other. The adjacent angles of parallelogram are not equal to 90° rather they have one pair of equal and opposite acute angles and one pair of equal and opposite obtuse angles.

The figure is attached

From the figure we can write

OH = OJ

OK = OI

According to the property of parallelogram the diagonals of a parallelogram bisect each other.

So we can see that for a parallelogram HIJK

The mid point of both the diagonals is unique and divides both parts of same diagonal in equal ratios , which means the diagonals bisect each other.

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Hijk is a parallelogram because the midpoint of both diagonals is which means the diagonals bisect

Answer from: Quest

What is the question

Answer from: Quest

x= -3 and $x=\frac{-3+/-3i\sqrt{3} }{2}$

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

$(x^3-a^3)=$ $(x-a)(x^2+ax+a^2)$

where a is the cube root of $a^3$ .

here this is 27 and the cube root of 27 is 3. so we write using the form:

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

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

$x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

where a = 1, b=3, and c=9

$x=\frac{-3+/-\sqrt{3^2-4(1)(9)} }{2(1)} \\x=\frac{-3+/-\sqrt{9-36)} }{2} \\x=\frac{-3+/-\sqrt{-27)} }{2}\\x=\frac{-3+/-3i\sqrt{3} }{2}$

this cubic has two complex roots.

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