PLEASE PLEASE HELP:(((

PQ= 10

Step-by-step explanation:

To find intersection of the line ,put y=0 in the equation,cuz points on x axis have ordinate equals to '0'

you have, y=2x+8 ; put y =0,to find it's intersection to x-axis

i.e, 0=2x+8

i.e, 2x= -8

i.e, x= -8/2

i.e, x= -4

Point P is (-4,0)

again, you have y=2x-12, put y=0

i.e,0= 2x-12

i.e, 2x=12

i.e, x=6

Point Q is (6,0)

now distance PQ is

√(6-(-4))^2 +(0-0)^2

i.e, √(10)^2

i.e, 10

distance btwn PQ is 10 unit!

✌️:)

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)

its an acute right triangle

step-by-step explanation: