Question

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

Answer 1

hello

Answer 2

ummmmmm

Answer 3

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

Answer 4

@mumustar78612

Answer 5

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

Answer 6

I got 7.7 idk if that's right

Answer 7

I think you need to check your answer

Answer 8

Is it 9.4?

Answer 9

\[WX = \sqrt{(6 -2)^2 + (3 - (-1))^2}\]

\[YZ = \sqrt{(7 - 5)^2 + (0 - 2)^2}\]

find those 2 values..

Answer 10

I got 4

Answer 11

Well 32/8

Answer 12

I calculated everything

Answer 13

\[WX = \sqrt{32} = 4\sqrt{2}~~~and~~~YZ = \sqrt{8} = 2\sqrt{2}\]

then the area is

\[A = \frac{1}{2} \times WX \times YZ\]

which is

\[A = \frac{1}{2} \times 4 \sqrt{2} \times 2 \sqrt{2}\]

Answer 14

12??

Answer 15

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