<p>test 2 section 2 of the PR book #17 and #20</p>
<h1>17</h1>
<p>(c+1)^2=-b which is true</p>
<p>I. c>0
II. c=0
III. c<0</p>
<p>a.none
b.I only
c.III only
d.I and II
e. all three</p>
<h1>20</h1>
<p>(there's a diagram) rectangle with sides x and 8 and a semi circle with diameter as x. Find area of shaded region (rectangle without semicircle).</p>
<p>for #17 i got A but the book had E as the correct choice and #20 i got 8x - ((x^2)(Pi))/8 which was not one of the answer choices.</p>
<p>Q1) There is no limitation on the domain of c.
However, if they asked for b, then b has to be equal to or less than 0.</p>
<p>After this it’s just an ambiguity of the question. Since there is no limitation on the value of c, you could either say that no statement is individually correct, but the better option would be to say all three are possible (c>0, or c =0, or c<0 )So the correct answer is E.</p>
<p>Q2) Shaded region’s area = Area of Rectangle - Area of semi circle
= 8x - (1/2)[pi.(x/2)^2]
= 8x - (pi.x^2 / 8)
use and match with correct option.
which could also be:
(64x - pi.x^2) / 8</p>
<p>how can #17 be e? i thought that any number raise to the second power can never be negative.</p>
<h1>20 i got the same answer as you so i guess the answer in the book is wrong.</h1>
<p>Dude, it is (c + 1)^2 = -b
so b can be -1, -2 and so on</p>
<p>In short, (-b) must be a non-negative value.
so b must be a non - positive value.
But that has no effect on the domain of c, c can be any value.</p>
<p>oh i see, thanks for the help.</p>
<p>Wait, (sorry to hijack your thread) but then how could (-b) be <0, and, hence, how could (b) be >0?</p>
<p>^ It can’t. There’s a restriction on the domain of b, that it can only exist when b<0.</p>