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3 years ago

What is the square root of 879?

Mathematics

2 answers:

Ivahew [28]3 years ago

3 0

The answer would be 29.6479341607
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konstantin123 [22]3 years ago

3 0

About 29.6479 is your answer :)

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It is possible to test if a quadrilateral is a square by performing calculations on its diagonals, in this case, AC and BD. a qu

Alja [10]

Answer:

The figure is not a square, because:

The diagonals DO NOT intersect at their midpoints.

The diagonals are NOT of the same length.

The diagonals are NOT perpendicular.

Step-by-step explanation:

✍️If two diagonals intersect at their midpoints, the coordinates of their midpoints will be the same.

Find the midpoints of diagonal AC and BD using the midpoint formula, $M(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$ .

Midpoint (M) of AC, for A(-4, -6) and C(6, -18):

$M(\frac{-4 + 6}{2}, \frac{-6 + (-18)}{2})$

$M(\frac{2}{2}, \frac{24}{2})$

$M(1, 12)$

Midpoint of diagonal AC = (1, 12)

Midpoint (M) of BD, for B(-12, -12) and D(13, -1):

$M(\frac{-12 + 13}{2}, \frac{-12 +(-1)}{2})$

$M(\frac{1}{2}, \frac{-13}{2})$

Midpoint of diagonal BD = $M(\frac{1}{2}, \frac{-13}{2})$

The coordinates of the midpoint of diagonal AC and diagonal BD are not the same, therefore, the diagonals do not intersect at their midpoints.

✍️Use distance formula to calculate the length of each diagonal to determine whether they are of the same length.

Distance between A(-4, -6) and C(6, -18):

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

$AC = \sqrt{(6 -(-4))^2 + (-18 -(-6))^2}$

$AC = \sqrt{(10)^2 + (-12)^2}$

$AC = \sqrt{100 + 144} = \sqrt{244}$

$AC = 15.6$ (nearest tenth)

Distance between B(-12, -12) and D(13, -1):

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

$BD = \sqrt{(13 - (-12))^2 + (-1 -(-12))^2}$

$BD = \sqrt{(25)^2 + (11)^2}$

$BD = \sqrt{625 + 121} = \sqrt{746}$

$BD = 27.3$ (nearest tenth)

Diagonal AC and BD are not of the same length.

✍️If the diagonals are perpendicular, the product of their slope would equal -1.

Slope of diagonal AC:

A(-4, -6) and C(6, -18)

$slope = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-18 -(-6)}{6 -(-4)} = \frac{-12}{10} = -\frac{6}{5}$

Slope of diagonal BD:

B(-12, -12) and D(13, -1)

$slope = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 -(-12)}{13 - (-12)} = \frac{11}{25}$

Product of their slope:

$-\frac{6}{5}*\frac{11}{25} = \frac{66}{125}$

The product of their slope doesn't equal -1. Therefore, diagonal AC and BD are not perpendicular to each other.

3 0

3 years ago