3 years ago

What is the area of a quadrilateral with vertices at (-2, - 6),

Mathematics

1 answer:

svet-max [94.6K]3 years ago

4 0

Answer:

18 sq. unit

you can draw it on no. line

actually it's rectangle with l=3 and b=6

You might be interested in

Please help me....I really need help!!

scoundrel [369]

$\dfrac{c^2-4}{6c^4+15c^3}\div\dfrac{c^2+4c+4}{12c^3+30c^2}=\dfrac{c^2-2^2}{3c^3(2c+5)}\div\dfrac{c^2+2(c)(2)+2^2}{6c^2(2c+5)}\\\\_{\text{use}\ a^2-b^2=(a-b)(a+b)\ \text{and}\ (a+b)^2=a^2+2ab+b^2}\\\\=\dfrac{(c-2)(c+2)}{3c^3(2c+5)}\cdot\dfrac{6c^2(2c+5)}{(c+2)^2}=\dfrac{(c-2)}{c}\cdot\dfrac{2}{(c+2)}=\boxed{\dfrac{2(c-2)}{c(c+2)}}$