Answer:

Step-by-step explanation:
Use the addition property of equality to get w by itself.

Answer:
i believe its D
Step-by-step explanation:
if i am wrong i am sorry i failed you
Answer:
1, 2, and 4.
Step-by-step explanation:
Have a nice day! :-)
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.
Answer:
y = -2x - 14
Step-by-step explanation:
Slope intercept form: y = mx + b
2x + y = - 14
y = -2x - 14