The coordinates of the vertices of A HIJ are H(-2,2), I (8,-2), and J(-4,-3). Identify the perimeter of A HIJ to the nearest tenth.

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66answer:

step-by-step explanation:

to find the perimeter of the rectangle this is the expression:

x-9+x-9+3x-4+3x-4=138

if we simplify it:

8x-26=138

remember, 138 is what it all adds up to

the length is 11.5cm

the width is 57.5cm

the area is 661.25cm^2

the answers might not be correct use at own risk

e^-kt = n/n0 = 0.79  

-kt = ln 0.79  

t = - ln 0.79/0.0001  

= 2357 years

step-by-step explanation:

hey there! : )

4y < -4

we need to isolate y. we can do this by simply dividing both sides by 4.

4y ÷ 4 > -4 ÷ 4

simplify.

y > -1

5 - y > 0

once again, we need to isolate y, so we can start off by subtracting 5 from both sides.

5 - y - 5 > 0 - 5

simplify.

-y > -5

divide both sides by -1.

** remember that when dividing any side by a negative, you must switch the sign! **

-y ÷ -1 < -5 ÷ -1

simplify.

y < 5

~hope i ! ~