Please help me :(:(:(:(:(:(

#9=-8, #10= -18, #11=-20

Step-by-step explanation:

For these, you need to remember that a positive*a positive is always positive, a negative * a negative is always positive, and a negative * a positive is negative.

So just do 2*4, 6*3, and 4*5 and add a negative in front of them.  

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

is only a line? i think so