The area of a sector = 0.5 * radius^2 * central angle measure (in radians)
So that:
sq in
So that:
Perputation PLZ HELP ME 20 pionts A college questionnaire was distributed among high school students asking each student to rank
2 answers:
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Well, let's take a peek at the graph of that line, hmmm let's pick two points, heck, those tow on the extremes anyway, and those are (-5,2) and (5,-1), alrite.. now, let's do some checking on that.
![\bf \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-2=-\cfrac{3}{10}[x-(-5)] \\\\\\ y-2=-\cfrac{3}{10}(x+5)\implies y-2=-\cfrac{3}{10}x-\cfrac{3}{2}\implies y=-\cfrac{3}{10}x-\cfrac{3}{2}+2 \\\\\\ y=-\cfrac{3}{10}x+\cfrac{1}{2}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bpoint-slope%20form%7D%7D%7By-%7B%7B%20y_1%7D%7D%3D%7B%7B%20m%7D%7D%28x-%7B%7B%20x_1%7D%7D%29%7D%5Cimplies%20y-2%3D-%5Ccfrac%7B3%7D%7B10%7D%5Bx-%28-5%29%5D%0A%5C%5C%5C%5C%5C%5C%0Ay-2%3D-%5Ccfrac%7B3%7D%7B10%7D%28x%2B5%29%5Cimplies%20y-2%3D-%5Ccfrac%7B3%7D%7B10%7Dx-%5Ccfrac%7B3%7D%7B2%7D%5Cimplies%20y%3D-%5Ccfrac%7B3%7D%7B10%7Dx-%5Ccfrac%7B3%7D%7B2%7D%2B2%0A%5C%5C%5C%5C%5C%5C%0Ay%3D-%5Ccfrac%7B3%7D%7B10%7Dx%2B%5Ccfrac%7B1%7D%7B2%7D)
If a2 -2ab + b2 = 9 and a
then
[tex]\begin{gathered} (a-b)^2=9 \\ \sqrt{(a-b)^2}=\sqrt{9} \\ a-b=\pm3 \\ \\ aa-b=-3