Solve the equation -8=-(×+4)
2 answers:
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Check the picture below.
so the perimeter of the polygon is the sum of all its sides, namely, AB + BC + CD + DA.
now, let's check how long each side is,
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &A&(~{{ -6}} &,&{{ -4}}~) % (c,d) &B&(~{{ -3}} &,&{{ 6}}~) \end{array} \\\\\\ d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}\\\\ -------------------------------\\\\ AB=\sqrt{[-3-(-6)]^2+[6-(-4)]^2} \\\\\\ AB=\sqrt{(-3+6)^2+(6+4)^2} \\\\\\ AB=\sqrt{3^2+10^2}\implies \boxed{AB=\sqrt{109}}\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26A%26%28~%7B%7B%20-6%7D%7D%20%26%2C%26%7B%7B%20-4%7D%7D~%29%20%0A%25%20%20%28c%2Cd%29%0A%26B%26%28~%7B%7B%20-3%7D%7D%20%26%2C%26%7B%7B%206%7D%7D~%29%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0AAB%3D%5Csqrt%7B%5B-3-%28-6%29%5D%5E2%2B%5B6-%28-4%29%5D%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AAB%3D%5Csqrt%7B%28-3%2B6%29%5E2%2B%286%2B4%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AAB%3D%5Csqrt%7B3%5E2%2B10%5E2%7D%5Cimplies%20%5Cboxed%7BAB%3D%5Csqrt%7B109%7D%7D%5C%5C%5C%5C%0A-------------------------------)
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &B&(~{{ -3}} &,&{{6}}~) % (c,d) &C&(~{{ 4}} &,&{{ 0}}~) \end{array} \\\\ -------------------------------\\\\ BC=\sqrt{[4-(-3)]^2+[0-6]^2}\implies BC=\sqrt{(4+3)^2+(0-6)^2} \\\\\\ BC=\sqrt{7^2+(-6)^2}\implies \boxed{BC=\sqrt{85}}\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26B%26%28~%7B%7B%20-3%7D%7D%20%26%2C%26%7B%7B6%7D%7D~%29%20%0A%25%20%20%28c%2Cd%29%0A%26C%26%28~%7B%7B%204%7D%7D%20%26%2C%26%7B%7B%200%7D%7D~%29%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0ABC%3D%5Csqrt%7B%5B4-%28-3%29%5D%5E2%2B%5B0-6%5D%5E2%7D%5Cimplies%20BC%3D%5Csqrt%7B%284%2B3%29%5E2%2B%280-6%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0ABC%3D%5Csqrt%7B7%5E2%2B%28-6%29%5E2%7D%5Cimplies%20%5Cboxed%7BBC%3D%5Csqrt%7B85%7D%7D%5C%5C%5C%5C%0A-------------------------------)
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &D(~{{ 2}} &,&{{-1}}~) % (c,d) &A&(~{{ -6}} &,&{{ -4}}~) \end{array}\\\\ -------------------------------\\\\ DA=\sqrt{[-6-2]^2+[-4-(-1)]^2}\\\\\\ DA=\sqrt{(-6-2)^2+(-4+1)^2} \\\\\\ DA=\sqrt{(-8)^2+(-3)^2}\implies \boxed{DA=\sqrt{73}}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26D%28~%7B%7B%202%7D%7D%20%26%2C%26%7B%7B-1%7D%7D~%29%20%0A%25%20%20%28c%2Cd%29%0A%26A%26%28~%7B%7B%20-6%7D%7D%20%26%2C%26%7B%7B%20-4%7D%7D~%29%0A%5Cend%7Barray%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0ADA%3D%5Csqrt%7B%5B-6-2%5D%5E2%2B%5B-4-%28-1%29%5D%5E2%7D%5C%5C%5C%5C%5C%5C%20DA%3D%5Csqrt%7B%28-6-2%29%5E2%2B%28-4%2B1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0ADA%3D%5Csqrt%7B%28-8%29%5E2%2B%28-3%29%5E2%7D%5Cimplies%20%5Cboxed%7BDA%3D%5Csqrt%7B73%7D%7D)
sum those sides up, and that's the perimeter of the polygon.
so the perimeter of the polygon is the sum of all its sides, namely, AB + BC + CD + DA.
now, let's check how long each side is,
sum those sides up, and that's the perimeter of the polygon.
Answer:
Step-by-step explanation:
The range of a projectile is given by :
Put R = 20 m, u = 15 m/s and finding the value of angle of projection
So,
So, the required angle of projection is equal to .