anonymous
  • anonymous
CAN SOMEONE PLEASE HELP!! Just need to finish just 1 and I'm done!!
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
hello
anonymous
  • anonymous
ummmmmm
anonymous
  • anonymous
Calculate 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
campbell_st
  • campbell_st
well 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
  • anonymous
I got 7.7 idk if that's right
campbell_st
  • campbell_st
I think you need to check your answer
anonymous
  • anonymous
Is it 9.4?
campbell_st
  • campbell_st
\[WX = \sqrt{(6 -2)^2 + (3 - (-1))^2}\] \[YZ = \sqrt{(7 - 5)^2 + (0 - 2)^2}\] find those 2 values..
anonymous
  • anonymous
I got 4
anonymous
  • anonymous
Well 32/8
anonymous
  • anonymous
I calculated everything
campbell_st
  • campbell_st
\[WX = \sqrt{32} = 4\sqrt{2}~~~and~~~YZ = \sqrt{8} = 2\sqrt{2}\] then the area is \[A = \frac{1}{2} \times WX \times YZ\] which is \[A = \frac{1}{2} \times 4 \sqrt{2} \times 2 \sqrt{2}\]
anonymous
  • anonymous
12??
campbell_st
  • campbell_st
so 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.