Mathematics, 25.07.2019 03:30 juniorvalencia4

The coordinates of a quadrilateral are (1,3), (7,-3), (1,-9), and (-5,-3). which shape is formed select all that apply -kite -rectangle -rhombus -square -trapezoid

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Answer from: riahhrene

rectangle , rhombus , and square

Step-by-step explanation:

First we find the slope of the line between each pair of points. We will first name each point:

A(1, 3); B(7, -3); C(1, -9); and D(-5, -3).

The formula for slope is

$m=\frac{y_2-y_1}{x_2-x_1}$

This means the slope for AB is

m = (3--3)/(1-7) = (3+3)/(-6) = 6/-6 = -1

The slope for BC is

m = (-3--9)/(7-1) = (-3+9)/6 = 6/6 = 1

The slope for CD is

m = (-9--3)/(1--5) = (-9+3)/(1+5) = (-6)/6 = -1

The slope for DA is

m = (-3-3)/(-5-1) = (-6)/(-6) = 1

If two sides are parallel, then the slopes of their lines are the same. The slopes of AB and CD are the same; this means they are parallel. The slopes of BC and DA are the same; this means they are parallel. This makes this figure a parallelogram.

If two sides form a right angle, then their slopes are negative reciprocals. The slopes of AB and BC are negative reciprocals, so they form a right angle. The slopes of BC and CD are negative reciprocals, so they form a right angle. The slopes of CD and AB are negative reciprocals, so they form a right angle. This means the fourth angle must be a right angle as well. This makes the figure a rectangle.

We next use the distance formula to find the length of each side:

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

The length of AB is

$d=\sqrt{(3--3)^2+(1-7)^2}=\sqrt{6^2+(-6)^2}=\sqrt{36+36}=\sqrt{72}$

The length of BC is

$d=\sqrt{(-3--9)^2+(7-1)^2}=\sqrt{6^2+6^2}=\sqrt{36+36}=\sqrt{72}$

The length of CD is

$d=\sqrt{(-9--3)^2+(1--5)^2}=\sqrt{(-6)^2+6^2}=\sqrt{36+36}=\sqrt{72}$

The length of DA is

$d=\sqrt{(-3-3)^2+(-5-1)^2}=\sqrt{(-6)^2+(-6)^2}=\sqrt{36+36}=\sqrt{72}$

Since all four sides have the same length, the figure is a square.

Answer from: Quest

answer: the answer is one of

$f=\dfrac{1}{2},~\dfrac{1}{4},~\dfrac{3}{4},~\dfrac{1}{6},~\dfrac{5}{6},~\dfrac{1}{8},~\dfrac{3}{8},~\dfrac{5}{8},~\dfrac{7}{8}.$

step-by-step explanation: we are given to write a proper fraction where the whole is divided into an even number of equal parts less than ten, and an odd number of those parts exist.

an even number less than 10 may be one of 2, 4, 6 or 8. so, the whole may be one of these four evens.

now, an odd number of these parts must exist, so the numerator must be an odd number, one of 1, 3, 5 or 7. it cannot be 9 because then the fraction will become improper.

thus, our fraction may be one of the following

$f=\dfrac{1}{2},~\dfrac{1}{4},~\dfrac{3}{4},~\dfrac{1}{6},~\dfrac{5}{6},~\dfrac{1}{8},~\dfrac{3}{8},~\dfrac{5}{8},~\dfrac{7}{8}.$

Answer from: Quest

there is no question content here, but you for the free points! : d

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