What does -3 (-5.6) equal?

Determine the most precise name for ABCD (parallelogram, rhombus, rectangle, or square). Explain how you determined your answer.

Sonja [21]

<h3>Answer: Rhombus</h3>

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Reason:

Let's find the distance from A to B. This is equivalent to finding the length of segment AB. I'll use the distance formula.

$A = (x_1,y_1) = (3,5) \text{ and } B = (x_2, y_2) = (7,6)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(3-7)^2 + (5-6)^2}\\\\d = \sqrt{(-4)^2 + (-1)^2}\\\\d = \sqrt{16 + 1}\\\\d = \sqrt{17}\\\\d \approx 4.1231\\\\$

Segment AB is exactly $\sqrt{17}$ units long, which is approximately 4.1231 units.

If you were to repeat similar steps for the other sides (BC, CD and AD) you should find that all four sides are the same length. Because of this fact, we have a rhombus.

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Let's see if this rhombus is a square or not. We'll need to see if the adjacent sides are perpendicular. For that we'll need the slope.

Let's find the slope of AB.

$A = (x_1,y_1) = (3,5) \text{ and } B = (x_2,y_2) = (7,6)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{6 - 5}{7 - 3}\\\\m = \frac{1}{4}\\\\$

Segment AB has a slope of 1/4.

Do the same for BC

$B = (x_1,y_1) = (7,6) \text{ and } C = (x_2,y_2) = (6,2)\\\\m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}\\\\m = \frac{2 - 6}{6 - 7}\\\\m = \frac{-4}{-1}\\\\m = 4\\\\$

Unfortunately the two slopes of 1/4 and 4 are not negative reciprocals of one another. One slope has to be negative while the other is positive, if we wanted perpendicular lines. Also recall that perpendicular slopes must multiply to -1.

We don't have perpendicular lines, so the interior angles are not 90 degrees each.

Therefore, this figure is not a rectangle and by extension it's not a square either.

The best description for this figure is a <u>rhombus</u>.

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1 year ago

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Find the slope between these two points: (3, -10) and (-2, -30). Show all of your work. what the answer

Harlamova29_29 [7]

$\huge\text{$m=\boxed{4}$}$

Hey there! Start with the slope formula, where $(x_1,y_1)$ and $(x_2,y_2)$ are the two known points.

$\begin{aligned}m&=\dfrac{y_2-y_1}{x_2-x_1}\\&=\frac{-30-(-10)}{-2-3}\end{aligned}$

Simplify.

$\begin{array}{c|l}\textbf{Solving}&\textbf{Reason}\\\cline{1-2}\\m=\dfrac{-30+10}{-2-3}&x-(-y)=x+y\\\\m=\dfrac{-20}{-5}&\text{Addition and subtraction}\\\\m=\dfrac{20}{5}&\text{The negatives cancel out}\\\\m=\dfrac{4}{1}&\text{Divide the numerator and denominator by $5$}\\\\m=\boxed{4}&\dfrac{x}{1}=x\end{aligned}$