10. The correct answer is (D). To handle this item, just ask yourself how to obtain the minimum and
maximum values for xy. Since x > 4 and y > 6, the lower limit for xy is 24. And since x < 12 and y < 8,
the upper limit for xy is 96.
11. The correct answer is (B). Since the triangles are equilateral, the unmarked angles with vertices at
the common point have degree measures of 60. Then, the sum of the degree measures of all six of the
angles with vertices at the common point (those marked x, y, and z, and those not marked) is 360.
Therefore:
x + 60 + y + 60 + z + 60 = 360
x + y + z + 180 = 360
x + y + z = 180
12. The correct answer is (E). This is a “defined operation” problem. Don’t let the Greek letter throw
you. Just substitute –2 for x and –1 for y into the expression xy – y – x:
(–2)( –1) – (–1) – (–2) = 2 + 1 + 2 = 5
13. The correct answer is (C).
For this problem, you have to know that the degree measure of an inscribed
angle is one–half that of the intercepted arc. (Or, an inscribed angle intercepts twice the arc.) Since the
inscribed angle has a measure of 30 degrees, the minor arc PR has a measure of 60 degrees. And since the
total measure of the circle is 360 degrees, minor arc PR is
60
360
1
6
=
of the circumference.
PR
C
=
6
And since C = 2
πr
PR
r
= π
2
6
And r = 3:
PR
= π = π = π
2 3
6
6
6
Lesson 8
210
w w w . p e t e r s o n s . c o m / a r c o
ARCO
■ SAT II Subject Tests
Do'stlaringiz bilan baham: |
