CC H. Geometry Answer Key

Honors Geometry Summer Math Packet Answers
1.
2. 10π‘₯ ! βˆ’ 3π‘₯ βˆ’ 18 or its equivalent
3.
4. -5 and 3 2
5 5
5. n = ⎧⎨ ,βˆ’ ⎫⎬
⎩ 3
3 ⎭
6. -4.2 to -4
7. a) c = 5
b) b = 12
c) a = 6
d) c =
98 or approximately 9.9
Honors Geometry Summer Math Packet Answers
8. 9
9. d =
20 or approximately 4.47
10. d =
234 or approximately 15.3
11. d =
78.25 or approximately 8.85
12. d =
( J βˆ’ P) 2 + ( K βˆ’ Q) 2
13. a) 17 + 2𝑀 13 + 2𝑀 = 396 or its equivalent
b) w =
5
or equivalent values
2
14. D
15. 2𝑑 + 3𝑑 = 24 or an equivalent equation
16. Sample Response:
I would buy the ticket from Airline P. Both airlines are likely to have an ontime arrival since they both have median values at 0. However, Airline Q has
a much greater range in arrival times. Airline Q could arrive anywhere from
35 minutes early to 60 minutes late. For Airline P, the flights arrived within
10 minutes on either side of the scheduled arrival time about 2/3 of the time,
and for Airline Q, that number was only about 1/2. For these reasons, I think
Airline P is the better choice.
Honors Geometry Summer Math Packet Answers
17. Sample response:
Tina is incorrect because some orders of basic transformations do not
produce the same results. Suppose we move triangle A 2 units to the right
first. The point (4, 3) is then (6, 3). Then, we take the reflection across the
x-axis, which makes that point (6, -3). A reflection of (6, -3) across the yaxis gives us (-6, -3), which is not one of the vertices of triangle A’.
Therefore, the basic transformations done in any order do not produce the
same result.
18. B
19.
20. (π‘₯ + 3)(π‘₯ + 1)
21. 5(π‘₯ ! + 5)
22. (2π‘₯ + 5)(π‘₯ βˆ’ 2)
23. (3π‘₯ βˆ’ 4)(π‘₯ + 2)
24. (π‘₯ βˆ’ 6)(π‘₯ + 2)
Honors Geometry Summer Math Packet Answers
25. -2(2π‘₯ + 5)(2π‘₯ βˆ’ 5)
26. βˆ’4π‘₯(3π‘₯ βˆ’ 1)
27. (π‘₯ + 5)(π‘₯ βˆ’ 6)
28. (π‘₯ + 9)(π‘₯ βˆ’ 2)
29. π‘₯ βˆ’ 9 π‘₯ βˆ’ 1 = 0
π‘₯ = 1 π‘œπ‘Ÿ π‘₯ = 9
30. 3π‘₯ 2π‘₯ βˆ’ 5 = 0
π‘₯ = 0 π‘œπ‘Ÿ π‘₯ =
!
!
31. π‘₯ βˆ’ 5 π‘₯ + 7 = 0
π‘₯ = βˆ’7 π‘œπ‘Ÿ π‘₯ = 5
32. π‘₯ βˆ’ 6 π‘₯ βˆ’ 6 = 0
π‘₯=6
33. π‘₯ βˆ’ 2 π‘₯ + 8 = 0
π‘₯ = βˆ’8 π‘œπ‘Ÿ π‘₯ = 2
34. 2π‘₯ + 7 π‘₯ + 1 = 0
!
π‘₯ = βˆ’ π‘œπ‘Ÿ π‘₯ = βˆ’1
!