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vivado [14]
2 years ago
7

Can someone please help me ASAP

Mathematics
2 answers:
masha68 [24]2 years ago
6 0

Answer:

<h2><u>I want to say it is Graph A because it is more alined compared to any other graph. And not a single one of the other graphs touched the dotted line. So that brings me to my conclusion picking Graph A.</u></h2><h2><u /></h2><h2><u>Sorry If I'm Wrong.</u></h2>

Mashcka [7]2 years ago
3 0

Answer:

GRAPH A

Step-by-step explanation:

<h2><u>LINE OF BEST FIT</u></h2>

<em>notes</em> : the line of best fit is the line that passes close to majority of the points. it can either pass through most of the points or none of the points but it has to be close to them.

hence the most appropriate answer in relation to my explanation is:

<h3><u>GRAPH A </u></h3>
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Please find the exact length of the midsegment of trapezoid JKLM with vertices J(6, 10), K(10, 6), L(8, 2), and M(2, 2). Thank y
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Answer:

the exact length of the midsegment of trapezoid JKLM  = \mathbf{ = 3 \sqrt{5} } i.e 6.708 units on the graph

Step-by-step explanation:

From the diagram attached below; we can see a graphical representation showing the mid-segment of the trapezoid JKLM. The mid-segment is located at the line parallel to the sides of the trapezoid. However; these mid-segments are X and Y found on the line JK and LM respectively from the graph.

Using the expression for midpoints between two points to determine the exact length of the mid-segment ; we have:

\mathbf{ YX = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }

\mathbf{ YX = \sqrt{(8-5)^2+(8-2)^2} }

\mathbf{ YX = \sqrt{(3)^2+(6)^2} }

\mathbf{ YX = \sqrt{9+36} }

\mathbf{ YX = \sqrt{45} }

\mathbf{ YX = \sqrt{9*5} }

\mathbf{ YX = 3 \sqrt{5} }

Thus; the exact length of the midsegment of trapezoid JKLM  = \mathbf{ = 3 \sqrt{5} } i.e 6.708 units on the graph

8 0
2 years ago
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