12. ∆ABD is similar to ∆PQD, so
QD/z = BD/x
Likewise ∆CDB is similar to ∆PQB, so
QB/z = BD/y
Since QD + QB = BD, you have
QD/z + QB/z = BD/x + BD/y
BD/z = BD/x + BD/y
Dividing by BD gives the desired result:
1/x + 1/y = 1/z
13. Triangles ABD, BCD, and ACB are all similar. This means
AB/AC = AD/AB
AB² = AC×AD = (4 cm)×(9 cm)
AB² = 36 cm²
AB = 6 cm
and
BD/CD = AD/BD
BD² = CD×AD = (5 cm)×(4 cm)
BD² = 20 cm²
BD = 2√5 cm
Answer:
= 5.E14 im a genius mark me as brainliest
Answer: $140.50
Step-by-step explanation: Just multiply $13.50 by 3 then add the $100 initial fee.
Answer:
0.3 =Three tenths
24.20 = Twenty four and twenty hundredths
24.20 - 0.3 = 23.9
23.9 is the answer..
HOPE THIS HELPS..
