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GREYUIT [131]
11 months ago
11

A rectangular garden has a walkway around it. The area of the garden is 6(6.5x + 2.5). The combined area of the garden and the w

alkway is 6.5(8x + 6). Find the area of the walkway around the garden as the sum of two terms.
Mathematics
1 answer:
Rasek [7]11 months ago
7 0

According to the information given in the exercise, the following expression represents the area of the rectangular garden:

6\mleft(6.5x+2.5\mright)

And the following expression represents the combined area of the walkway around the rectangular garden and the area of the garden:

6.5\mleft(8x+6\mright)

You can identify that the word "combined" indicates that that expression was obtained by adding both areas.

Knowing the above, you can set up the following equation:

6(6.5x+2.5)+A=6.5(8x+6)

Where "A" is the area of the walkway around the rectangular garden.

Solving for "A", you get the following expression:

\begin{gathered} A=6.5(8x+6)-6(6.5x+2.5) \\ A=52x+39-39x-15 \\ A=13x+24 \end{gathered}

The answer is:

13x+24

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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} }

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