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anonymous
 one year ago
CAN SOMEONE PLEASE HELP!! Just need to finish just 1 and I'm done!!
anonymous
 one year ago
CAN SOMEONE PLEASE HELP!! Just need to finish just 1 and I'm done!!

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Calculate the area of triangle WXY with altitude YZ, given W (2,1), X (6,3), Y (7,0) and Z (5,2) 8 square units 9.4 square units 7.7, square units 12 square units

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.0well I'd say you need the distance formula to find the base length and altitude the base would be WX, so find the distance from (2, 1) tp (6,3) and the altitude YZ is the distance from (7,0) to (5,2) then use the normal area of a triangle A =1/2bh

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got 7.7 idk if that's right

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.0I think you need to check your answer

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.0\[WX = \sqrt{(6 2)^2 + (3  (1))^2}\] \[YZ = \sqrt{(7  5)^2 + (0  2)^2}\] find those 2 values..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I calculated everything

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.0\[WX = \sqrt{32} = 4\sqrt{2}~~~and~~~YZ = \sqrt{8} = 2\sqrt{2}\] then the area is \[A = \frac{1}{2} \times WX \times YZ\] which is \[A = \frac{1}{2} \times 4 \sqrt{2} \times 2 \sqrt{2}\]

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.0so you have \[A = \frac{1}{2} \times 4 \sqrt{2} \times 2\sqrt{2} = 2 \times 2 \times \sqrt{2} \times \sqrt{2}\] and you need to know \[\sqrt{a} \times \sqrt{a} = a\] hope that helps
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