SAT Subject Test Mathematics Level 2

(Marvins-Underground-K-12) #1
from    P   to  base    OB  divides the base    in  half    and the x-coordinate    for point   P   is   .  Plug
into the equation y = 6 − x^2 to find the height:

So  if  the base    is   and    the height  is  3:

48 . B


Use the relationship    cos 2x  =   1   −   2sin^2 x    to  get everything  in  terms   of  sine.   And be  sure    your
calculator is in radian mode.

Of  those   solutions,  only    0.52    is  listed  in  the answer  choices.

49 . C
The entire surface area you’re looking for consists of the lateral areas of the outside cylinder
and the inside cylinder, plus the areas of the larger top and bottom circles, minus the areas
of the smaller top and bottom circles. The lateral area of the outside cylinder is 2 πrh = 2π(2)
(6) = 24π. The lateral area of the inside cylinder is 2 πrh = 2π(1)(6) = 12π. The areas of the larger
top and bottom circles are each πr^2 = π(2^2 ) = 4π. And the areas of the smaller top and bottom
circles are each πr^2 = π(1^2 ) = π. The total surface area, then, is


24 π    +   12π +   2(4π)   −   2(π)    =   42π
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