I believe if that says three minus four times three minus five or three plus negative four times three plus negative five (same thing) then the answers should be positive two (2.)

4 0

2 years ago

Find the perimeter of the quadrilateral that has vertices at (8,13), (5,-18), (-6,-6), and (-2,-5).

kodGreya [7K]

Answer:

72.14 units

Step-by-step explanation:

Let A(8,13), B(5,-18),C(-6,-6) and D(-2,-5)

Apply the distance formula as;

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Finding the lengths of the segments

For AB

$AB=\sqrt{(5-8)^2+(-18-13)^2} =\sqrt{(-3^2)+(-31^2)} =\sqrt{970} =31.14$

For BC

$BC=\sqrt{(-6-5)^2+(-6--18)^2} =\sqrt{(-11^2)+12^2} =\sqrt{265} =16.28$

For CD

$CD=\sqrt{(-2--6)^2+(-5--6)^2} =\sqrt{4^2+1^2} =\sqrt{16+1} =\sqrt{17} =4.12$

For DA

$DA=\sqrt{(8--2)^2+(13--5)^2} =\sqrt{10^2+18^2} =\sqrt{424} =20.6$

Finding the perimeter will be;

31.14+16.28+4.12+20.6=72.14 units

3 0

3 years ago

Find BD???????????????

Trava [24]

See the attached picture:

5 0

4 years ago

Read 2 more answers