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Maslowich
2 years ago
10

I’m honestly confused on how to solve for k. please help.

Mathematics
1 answer:
stellarik [79]2 years ago
4 0

Answer:

k=9

Step-by-step explanation:

slope =  \frac{y2 - y1 }{x2 - x1}  \\  \frac{ - 5}{2}  =  \frac{ - 11 - k}{6 - ( - 2)}  \\  -  ( \frac{5}{2} ) =  - ( \frac{11 + k}{8} ) \\  \frac{5}{2}  =  \frac{11 + k}{8}  \\ 5  \times 8 = 2(11 + k) \\ 40 = 22 + 2k \\ k =  \frac{40 - 22}{2}  \\ k = 9

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\bf \cfrac{(2hx+h^2+3h)~~\begin{matrix} +x^2-x^2+3x-3x+9-9 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{h}\implies \cfrac{\stackrel{\textit{common factor}}{~~\begin{matrix} h \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~(2x+h+3)}}{~~\begin{matrix} h \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill 2x+h+3~\hfill

7 0
4 years ago
A trapezoid has base lengths of 12 centimeters and 13 centimeters. The other sides have lengths of 5 centimeters and 10 centimet
r-ruslan [8.4K]

Answer:

The area of the composite figure is 392.12cm^{2}.

Step-by-step explanation:

The area of the composite figure = area of trapezoid + area of rectangle

Area of trapezium = \frac{1}{2} ( a +b)h

Where: a is the length of the first base, b the length of the second base and h is the height of the trapzium.

Applying Pythagoras theorem, the height, h, is;

h = \sqrt{5^{2} - 1^{2}  }

  = \sqrt{24}

h  = 2\sqrt{6}

Area of trapezium = \frac{1}{2} ( a +b)h

                              = \frac{1}{2} (13 + 12) × 2\sqrt{6}

                              = 156\sqrt{6}

                              = 382.12cm^{2}

Area of trapezium is 382.12cm^{2}

Area of rectangle = length × width

                             = 5 × 2

                             = 10 cm^{2}

Area of rectangle = 10 cm^{2}

Therefore,

area of the composite figure = 382.12 + 10

                                               = 392.12cm^{2}

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Step-by-step explanation:

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I hope this helps you

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