Answer:
This is proved by ASA congruent rule.
Step-by-step explanation:
Given KLMN is a parallelogram, and that the bisectors of ∠K and ∠L meet at A. we have to prove that A is equidistant from LM and KN i.e we have to prove that AP=AQ
we know that the diagonals of parallelogram bisect each other therefore the the bisectors of ∠K and ∠L must be the diagonals.
In ΔAPN and ΔAQL
∠PNA=∠ALQ (∵alternate angles)
AN=AL (∵diagonals of parallelogram bisect each other)
∠PAN=∠LAQ (∵vertically opposite angles)
∴ By ASA rule ΔAPN ≅ ΔAQL
Hence, by CPCT i.e Corresponding parts of congruent triangles PA=AQ
Hence, A is equidistant from LM and KN.
Just do normal division, just like ignore the decimal point. for a moment once you got your answer bring back the decimal point back up to your answer. You would get 0.27 as your answer.
Answer:
28% I'm pretty sure
Step-by-step explanation:
I hope this helps you
x=4 q (4)=8.4-7
q (4)=32-7
q (4)=25