tricky Math 2C Problems from Barrons

Test Preparation
1

<p>pg 149 #43
If Vector v=(1, Squareroot of 3) and vector u = (3, -2), find the value of |3v-u|</p>

<p>Well 3v = (3,3sqr3)
u = (3,-2)
3v-u=(3-3, 3sqr3 + 2)= (0, 3sqr3 + 2)
so I would say that the value of |3v-u|= 3sqr3+2</p>

<p>their choices are:</p>

<p>a) 52
b) 2 + 3sqr3
c) 6
d) .2 + 3sqr3
e) 7</p>

<p>they say that the answer is e) 7 and their explanation is that 2+3sqr3 rounds to 7.... ?????</p>

<h1>44 A sector of a circle AOB with a central angle of 2pi/5 and a radius of 5 is bent to form a cone with vertex at O. What is the volume of the cone that is formed?</h1>

<p>a) 8.17
b) 6.04
c) 4.97
d) 5.13
e) 12.31</p>

<p>Answer is D... not sure how they got that, I know I'm wrong on this one, my answer isnt even a choice (19.2)</p>

<p>Thanks</p>

2

<h1>44. The sector length will form the circumference of the base of the cone. Sector length = (circle circumference)(sector angle)/2 pi</h1>

<p>= <a href="2%20pi/5">2 pi (5)</a> / (2 pi)
= 2 pi
This equals the circumference of the cone base = 2 pi (cone base radius)
So 2 pi = 2 pi (cone base radius), and the cone base therefore has a radius = 1.</p>

<p>Cone volume = (1/3) (base area) (height)
= (1/3) (pi (1)^2) (5)
= 5 pi /3
= 5.236
(I don't see how they they got 5.13).</p>

<h1>43. Your logic looks fine to me, I don't see why they had to round off.</h1>