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2 years ago

Find the area of the figure

Mathematics

1 answer:

Sedbober [7]2 years ago

6 0

Check the picture below.

so the figure is really a 6x21 rectangle, a 2x2 square, and a couple of quarter of a circle whose radius is 2, now, if put two quarters together, we get half, so both of those quarter circles add up to just a semi-circle with a radius of 2.

$\stackrel{\textit{\huge Areas}}{\stackrel{rectangle}{(6)(21)}~~ + ~~\stackrel{square}{(2)(2)}~~ + ~~\stackrel{two~\frac{1}{4}~circle}{\cfrac{1}{2}\pi (2)^2}}\implies 126+4+2\pi \implies 130+2\pi$

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Answer:

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Step-by-step explanation:

$(-14,1)(13,-2)$

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m is the slope and b is the y-intercept. Use the slope formula first:

$\frac{y_{2}-y_{1}}{x_{2}-x_{1}} =\frac{rise}{run}$

Rise over run is the change in the y-axis over the change in the x-axis. Plug in the coordinates:

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Simplify parentheses (two negatives makes a positive):

$\frac{-2-1}{13+14}$

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$\frac{-3}{27} =-\frac{3}{27}=-\frac{1}{9}$

The slope is $-\frac{1}{9}$ . Insert into the equation:

$y=-\frac{1}{9}x +b$

Now, to find the y-intercept, take one of the points and substitute for the x and y values of the equation:

$(13,-2)\\-2=-\frac{1}{9} (13)+b$

Solve for b, the y-intercept. Simplify parentheses:

$-\frac{1}{9}*\frac{13}{1}=-\frac{13}{9}$

$-2=-\frac{13}{9} +b$

Add $-\frac{13}{9}$ to both sides:

$-2+\frac{13}{9}=-\frac{13}{9} +\frac{13}{9} +b\\\\-2+\frac{13}{9} =b$

Simplify:

$-\frac{2}{1} +\frac{13}{9}\\\\-\frac{18}{9}+\frac{13}{9}\\ \\b=-\frac{5}{9}$

The y-intercept is $-\frac{5}{9}$ . Insert this into the equation:

$y=-\frac{1}{9} x-\frac{5}{9}$

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