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
1.) True. Their scales are the same, they only vary in sizes.
2.) False. Similarities between circles can be proven by reflections(1), transformation(2), dilation(3), and rotations(4) when they map the figures.
3.) True. Refer to #2.
The answer is 70
you can just divide 63 by 0.9 (90%)
you can use proportions to do that
Answer: B, D, E
Step-by-step explanation:
Statement 1 cannot be true because f(1)<0.
Statement 2 might be true.
Statement 3 cannot be true because if f is continuous, then for some value between -1 and 0, f(x) is positive by the Intermediate Value Theorem.
Statement 4 might be true.
Statement 5 might be true