Answer:
A???
Step-by-step explanation:
For this case, the first thing you should know is that the opposite sides of the parallelogram are the same.
Therefore, we have the following equations:

From equation 1 we have:


From equation 2 we have:


Answer:
the value of the variables are:

Answer: Yes, He bicycled fewer in this week than last week.
Step-by-step explanation:
Let he bicycled x hours in the last week,
Then, According to the question,




Hence, He bicycled fewer in this week than last week.
Answer:
CE = 17
Step-by-step explanation:
∵ m∠D = 90
∵ DK ⊥ CE
∴ m∠KDE = m∠KCD⇒Complement angles to angle CDK
In the two Δ KDE and KCD:
∵ m∠KDE = m∠KCD
∵ m∠DKE = m∠CKD
∵ DK is a common side
∴ Δ KDE is similar to ΔKCD
∴ 
∵ DE : CD = 5 : 3
∴ 
∴ KD = 5/3 KC
∵ KE = KC + 8
∵ 
∴ 
∴ 
∴ 
∴ 
∴ KC = (8 × 9) ÷ 16 = 4.5
∴ KE = 8 + 4.5 = 12.5
∴ CE = 12.5 + 4.5 = 17