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Free_Kalibri [48]
4 years ago
14

A quadrilateral has vertices (2, 0), (0, –2), (–2, 4), and (–4, 2). Which special quadrilateral is formed by connecting the midp

oints of the sides?
a.kite
b.rectangle
c.trapezoid
d.rhombus
Mathematics
1 answer:
HACTEHA [7]4 years ago
8 0
In order to know what the polygon is, you have to plot the coordinates. After plotting, it is obviously shown a form of a rectangle. After connecting the midpoints of the sides, it formed a (D) rhombus, not a kite.
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Step-by-step explanation:

\because \: time \times speed  = distance \\  \therefore \: 4(t + 3) \times 10(t + 3) = 810 \\ \therefore \: 4(t + 3)^{2} = 81 \\ \therefore \:  \{2(t + 3) \}^{2} =  {9}^{2}  \\  \therefore \:  \{2(t + 3) \} =  {9}   \\  \therefore \: t + 3 =  \frac{9}{2}  \\ \therefore \: t  =  \frac{9}{2}  - 3 \\  \therefore \: t  =  \frac{9 - 6}{2}  \\  \therefore \: t  =  \frac{3}{2}  \\ \therefore \: t  =  1.5 \: hrs  \\

\because \: time \times speed  = distance \\  \therefore \: 4(t + 3) \times 10(t + 3) = 810 \\ \therefore \: 4(t + 3)^{2} = 81 \\ \therefore \:  \{2(t + 3) \}^{2} =  {9}^{2}  \\  \therefore \:  \{2(t + 3) \} =  {9}   \\  \therefore \: t + 3 =  \frac{9}{2}  \\ \therefore \: t  =  \frac{9}{2}  - 3 \\  \therefore \: t  =  \frac{9 - 6}{2}  \\  \therefore \: t  =  \frac{3}{2}  \\ \therefore \: t  =  1.5 \: hrs  \\ speed \: of \: car = 10(t + 3) \\ \hspace{60 pt} = 10(1.5 + 3) \\   \hspace{60 pt}= 10 \times 4.5 \\ \red{ \boxed{ \bold{ \therefore \: speed \: of \: car  = 45 \:km/ h}}} \\

3 0
3 years ago
Pls help ill give branliest​
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Answer:

looks to me like it could be (A.)

Step-by-step explanation:

i might be wrong if i am im sorry

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