Calculus Preparation & Placement Evaluation, 2009 Solutions to Sample Evaluation Simplifying Expressions

Calculus Preparation & Placement Evaluation, 2009 Solutions to Sample Evaluation Simplifying Expressions

Calculus Preparation & Placement Evaluation, 2009
Department of Mathematics
Wilfrid Laurier University
Solutions to Sample Evaluation
Simplifying Expressions
Question 1
50
25
1
5
2
+9
( 1) = 54
************************************************************************
Recall order of operations:
50
1
5
25
= 50
= 50
= 50
= 54
B
E
D
M
A
S
brackets (work from inside out)
exponents (which include roots)
division/multiplication (in order, left to right)
addition/subtraction (in order, left to right)
2
+9
25 15 +9
| {z }
( 1)
| {z }
1
5+9 1
| {z }
5+9
************************************************************************
************************************************************************
Question 2
3 4
2 +
4 5
3
127
=
8
40
************************************************************************
3 4
2 +
4 5
3
8
=
11 4
+
4
5
=
110 32
+
40
40
=
127
40
3
8
(changing the mixed fraction to an improper fraction)
15
40
( nding a common denominator for all terms)
************************************************************************
************************************************************************
1
Question 3
p
3 is an example of a number from which set? irrational numbers
************************************************************************
natural numbers ( N ) : f1; 2; 3; 4; 5; :::g
integers ( Z ) : f:::; 4; 3; 2; 1; 0; 1; 2; 3; 4; :::g
na
o
rational numbers ( Q ) :
j a; b 2 Z, b 6= 0
b
! any value which can be represented as a terminating or repeating decimal
irrational numbers: all non-terminating, non-repeating decimal values
reals ( R ) : union of the rationals and irrationals
p
3 = 1:732 050 808 ::: a non-terminating,
non-repeating decimal value;
p
thus, 3 is an irrational number.
************************************************************************
************************************************************************
Question 4
4 2 1
=
3+3 1
9
32
************************************************************************
4 2 1
3+3 1
=
1
42
3+
=
=
=
=
=
1
1
3
1
16
9
+
3
15
16
10
3
15
16
3
16
9
32
16
16
1
3
3
10
3
2
************************************************************************
************************************************************************
2
Question 5
2 64x1=2 y 9
2=3
=
1
8x1=3 y 6
************************************************************************
2 64x1=2 y 9
2=3
2
=
2=3
64x1=2 y 9
2
=
2=3
x1=2
(64)
2=3
(y 9 )
2
=
=
2=3
p
3
2
64
x1=3 (y 6 )
2
2
(4)
x1=3
(y 6 )
2
16x1=3 y 6
1
= 1=3 6
8x y
=
************************************************************************
************************************************************************
Question 6
If x 6= 1, then
2 (x + 3)
x
4 (x
1
1)
=
10
x
2x
1
************************************************************************
2 (x + 3)
x
4 (x
1
1)
2x + 6 4x + 4
x 1
2x + 10
=
x 1
10 2x
=
x 1
=
************************************************************************
************************************************************************
Question 7
If x > 3, then
2 (x
1=2
3)
x (x
x 3
3)
1=2
x 6
=q
3
(x 3)
************************************************************************
3
1=2
2 (x
3)
x (x
x 3
3)
1=2
=
=
=
=
(x
3)
1=2
x
(x
3)
1=2
x
1=2
3)
x 6
(x
3)
3=2
3)
[2x
3
6
6
(x
1
x
(x
[2 (x
3
x]
x]
3)
x 6
=q
3
(x 3)
************************************************************************
************************************************************************
Question 8
If x 6=
1,
2, then
x+2
x4 16
x+1
2
=
2x 4
(x + 1) (x2 + 4)
************************************************************************
x+2
x4 16
x+1
= 2
2x 4
(x
=
=
=
=
x+2
4) (x2 + 4)
x+1
2 (x 2)
(x
x+2
2) (x + 2) (x2 + 4)
x+1
2 (x 2)
(x
(x + 2)
2) (x + 2) (x2 + 4)
2 (x 2)
(x + 1)
(x2
1
+ 4)
2
(x + 1)
2
(x + 1) (x2 + 4)
************************************************************************
************************************************************************
Question 9
If x 6= 3, then
4+x
x x2
=
4x + 3
1 x
12
x2
************************************************************************
12
x2
x2 + x 12
x2 4x + 3
(x + 4) (x 3)
=
(x 1) (x 3)
x+4
=
x 1
x+4
=
1 x
x x2
=
4x + 3
************************************************************************
************************************************************************
4
Question 10
The remainder, when 9 + 3x + 4x3
2x4 is divided by x
2, is: 15
************************************************************************
Using the Factor/Remainder Theorem, we substitute x = 2 into the expression:
9 + 3 (2) + 4 (2)
3
2 (2)
4
= 9 + 32) + 4 (8)
= 9 + 6 + 32
2 (16)
32
= 15
OR Using long division of polynomials:
x
Thus,
2
p
2x3 + 0x2 + 0x + 3
2x4 + 4x3 + 0x2 + 3x + 9
2x4 + 4x3
0x3 + 0x2 + 3x + 9
(3x 6)
15
remainder
2x4 + 4x3 + 3x + 9
=
x 2
2x3 + 3 +
15
x
2
.
************************************************************************
************************************************************************
Question 11
1
x+h+1
h
1
x+1 =
1
(x + 1) (x + h + 1)
************************************************************************
1
x+h+1
h
1
x+1
=
1
1
h x+h+1
1
x+1
=
1
1 (x + 1)
h (x + h + 1) (x + 1)
=
1 (x + 1) (x + h + 1)
h (x + 1) (x + h + 1)
=
1 x+1 x h 1
h (x + 1) (x + h + 1)
=
1
h
h (x + 1) (x + h + 1)
=
1
(x + 1) (x + h + 1)
1 (x + h + 1)
(x + 1) (x + h + 1)
************************************************************************
************************************************************************
5
Question 12
9
, then
2
3
If x >
2x
p
=
2x + 9
3+
p
2x + 9
************************************************************************
p
p
2x 3 + 2x + 9
3 + 2x + 9
2x
p
p
=
9 (2x + 9)
3
2x + 9 3 + 2x + 9
p
2x 3 + 2x + 9
=
9 2x 9
p
2x 3 + 2x + 9
=
2x
p
=
3 + 2x + 9
************************************************************************
************************************************************************
Question 13
Which of the following statements are true?
b
=
b
Answer: I only
I.
a
1+
a
b
II.
m
=0
0
III.
p
x2
9=x
3
2
IV. (q + 2) = q 2 + 4
************************************************************************
I.
a
b
=
a
b
b
a
=
b
b
1=
1+
b
m
m
II.
is unde ned [
6= 0 ]
0
0
p
III. x2 9 cannot be simpli ed
[
a
X
b
p
x2
9 6= x
3]
2
2
IV. (q + 2) = (q + 2) (q + 2) = q 2 + 2q + 2q + 4 = q 2 + 4q + 4 [ (q + 2) 6= q 2 + 4 ]
************************************************************************
************************************************************************
Question 14
The expression x2
6x + 2, upon completing the square, is the same as: (x
2
3)
7
************************************************************************
x2
6x + 2 = x2
|
= (x
= (x
6x + 9 9 + 2
{z
}
2
9+2
2
7
3)
3)
************************************************************************
************************************************************************
6
Question 15
5
In the expansion of (2x + 1) , the coefficient of the x3 term is: 80
************************************************************************
Binomial expansion:
n
(x + y) = xn +
n n
x
1
1
y+
n n
x
2
2 2
y +
+
n n
x
k
k k
y +
+
n
n
1
xy n
1
+ yn
5
Thus, for (2x + 1) , the coef cient of the x3 term corresponds to n = 5, k = 2:
5
5
(2x)
2
2
2
(1)
5!
3
2
(2x) (1)
2! (5 2)!
120
=
8x3 (1)
(2) (6)
=
[Recall: m! = m (m
1) (m
= 10 8x3
! thus, the required coef cient is 80.
= 80x3
************************************************************************
************************************************************************
Solving Equations and Inequalities
Question 16
The equation 9 (x + 8) = 2x
has solution: x =
(4
x)
38
3
************************************************************************
9x
9 (x + 8)
= 2x
(4
9x + 72
= 2x
4+x
x
=
4
72
6x
=
76
x
=
2x
=
x)
76
6
38
3
************************************************************************
************************************************************************
7
2)
(2) (1) ]
Question 17
The possible values of x for which
(2x + 1) (x
7 are:
4) =
x=
1
,3
2
************************************************************************
(2x + 1) (x
2x2
2
2x
7
4
=
7
8x + x
4+7
=0
2x2
7x + 3
=0
2
( 7)
( 7)
=
=
8x + x
Using the quadraticq
formula:
x
4)
Or using factoring techniques:
4 (2) (3)
2x2
2 (2)
p
(2x
49 24
4
p
25
7
=
4
7 5 7+5
,
=
4
4
1
= , 3
2
=
7
2x
7x + 3
1) (x
1
x
3)
=0
1
=
2
=0
=0
OR
x
3
=0
x
=3
************************************************************************
************************************************************************
Question 18
For which value(s) of m would the equation x2
6x + m = 0 have no real solutions?
Answer: m > 9
************************************************************************
A quadratic [ ax2 + bx + c ] will have two real roots when its discriminant [ b2 4ac ]
is greater than zero; one real root when its discriminant is equal to zero; and no real roots
when its less than zero.
Thus, solve
2
( 6)
4 (1) (m)
<0
36
4m
<0
4m
<
36
m
>
36
4
m
>9
************************************************************************
************************************************************************
8
Question 19
The possible value(s) of x for which
p
x are: x =
x2 + 9 = 1
4
************************************************************************
p
x2 + 9
=1
Checking:
x
x2 + 9
= (1
x2 + 9
=1
2
x)
p
LS
=
RS
=1
16 + 9 =
p
25 = 5
( 4) = 1 + 4 = 5 = LS X
2x + x2
2x
=
8
x
=
4
************************************************************************
************************************************************************
Question 20
Solve the system of equations:
x + 5y
5x + 20y
Answer: x =
=4
= 13
3, y =
7
5
************************************************************************
Using the method of elimination:
x + 5y
5x + 20y
1
=4
= 13
2
(5) !
5x + 25y
= 20
3
5x + 20y
= 13
2
0x + 5y
=7
subtract
y
Substitute y =
7
into 1 :
5
x+5
7
5
=4
x+7
=4
x
=
Thus the system has the unique solution (x; y) =
=
4
7
5
.
3
3;
7
.
5
************************************************************************
************************************************************************
9
Question 21
A rectangle has a perimeter of 40 cm. If the length is 2 cm longer than twice the width,
nd the area of the rectangle.
Answer: 84 cm2
************************************************************************
Let w be the width of the rectangle and l its length.
) l=
P = 40 = 2w + 2l
40
2w
2
= 20
w
l+w
= 20
1
2w
3w
w
=2
= 18
=6
2
and l = 2w + 2
Using the method of elimination:
l
subtract
Substitute w = 6 into 1 :
= 20 .
l+6
l
= 14
Thus the rectangle has width of 6 cm, length of 14 cm and thus an area of 6
14 = 84 cm2 .
************************************************************************
************************************************************************
Question 22
The set fx 2 R j x
0 or
6<x<
2g
can be expressed in interval notation as:
( 6; 2) [ [0; 1)
************************************************************************
When using interval notation, a round bracket means that the
endpoint is not included in the set while a square bracket
indicates that the endpoint is included.
fx 2 R j x
x
6<x<
0 or
6<x<
2g can be represented on a real number line as:
———–>
0
2
1
————
6
2
0
1
Thus, the set can be represented by ( 6; 2) [ [0; 1).
************************************************************************
************************************************************************
10
Question 23
The solution of x2
2x
0 is the set of all x such that: x
8
2 or x
4
************************************************************************
x2
(x
(x
2x
8
0
4) (x + 2)
0
+
4) (x + 2)
1
(x
4) (x + 2)
2
0
2x
1
4
———–>
<———–
1
Therefore, x2
+
2
0 when x
8
1
4
2 or x
4.
! using interval notation, the solution set is ( 1; 2] [ [4; 1)
************************************************************************
************************************************************************
Question 24
Find all possible solutions of the equation j5x
2j = 4x.
2
Answer: x = , 2
9
************************************************************************
Note: If jaj = b then either a = b or a =
Case 1:
Case 2:
5x
5x
2
= 4x
x
=2
2
=
9x
x
Check:
b.
LS = j5 (2)
2j = j10
RS = 4 (2) = 8 = LS X
4x
Check:
=2
2
=
9
Therefore, the solution set is
LS = 5
2
9
RS = 4
2
9
2 =
=
2j = j8j = 8
10
9
18
=
9
8
= LS X
9
2
;2 .
9
************************************************************************
************************************************************************
11
8
8
=
9
9
Question 25
If jaj < jbj, then which of the following is true?
Answer:
b < a < b or b < a <
b
************************************************************************
If b > 0, then jbj = b and jaj < b means
If b < 0, then jbj =
b and jaj <
b < a < b.
b means b < a <
b.
If b = 0, then jbj = 0 and jaj < 0 has no solution.
Therefore, if jaj < jbj, then
b < a < b or b < a <
b.
************************************************************************
************************************************************************
Functions and Graphing
Question 26
If f (x) = x2 + 3x then f (h + 6) = h2 + 15h + 54
************************************************************************
2
f (h + 6) = (h + 6) + 3 (h + 6)
= h2 + 12h + 36 + 3h + 18
= h2 + 15h + 54
************************************************************************
************************************************************************
Question 27
Which of the following represents a function?
Answer: f( 9
5) , ( 3; 5) , (0; 5) , (3; 5)g
************************************************************************
A function maps each element of its domain to a unique range value.
f( 9; 4) ; ( 9; 8) ; ( 3; 3) ; (0; 5)g ! f ( 9) = 4 and f ( 9) =
) this is not a function
8
f( 9; 5) ; ( 3; 5) ; (0; 5) ; (3; 5)g ! represents a function X
f( 3; 9) ; ( 3; 3) ; ( 3; 0) ; ( 3; 6)g ! not a function, as the domain value 3
is mapped to 4 different range values
f( 9; 4) ; (0; 5) ; (0; 5) ; (9; 4)g ! not a function as f (0) = 5 and f (0) =
5
************************************************************************
************************************************************************
12
Question 28
1
is the
4 x2
the set of all x such that:
2<x<2
The domain of f (x) = p
************************************************************************
We cannot divide by zero nor take the square root of a negative value.
Therefore, the domain of f is the set of x such that:
x2
>0
x) (2 + x)
>0
4
(2
(2
–
x) (2 + x)
(2
2
1
x2 > 0 when
1
2
————–
2
2
x) (2 + x) > 0
Therefore, 4
–
+
1
1
2 < x < 2.
************************************************************************
************************************************************************
Question 29
The function f (x) = x2 1 can be defined by:
(
x2 1 if x
1 or x 1
f (x) =
1 x2 if
1<x<1
************************************************************************
f (x) =
(
x2
(x
if x2
if x2
x2 1
x2 1
1
= (x
1) (x + 1)
1) (x + 1)
–
+
1
Therefore, f (x) =
1 0
1<0
(
1
+
1
1
x2 1 if x
1 or x
1 x2 if
1<x<1
1
Note, the equality can be added to either piece of(the function, but not both;
x2 1 if x < 1 or x > 1
i.e., a valid answer would also be f (x) =
1 x2 if
1 x 1
************************************************************************
************************************************************************
13
Question 30
The function h (x) = x2 + 2x
f (x) = x2=3
2=3
can be thought of as the composition f
g, where:
g (x) = x2 + 2x
************************************************************************
(f
g) (x) = f (g (x))
Thus, taking f (x) = x2=3 and g (x) = x2 + 2x,
(f
g) (x)
= f x2 + 2x
= x2 + 2x
2=3
= h (x)
NOTE:
If f (x) = x2 + 2x, g (x) = x2=3
then: (f
g) (x)
If f (x) = x2 + 2x
then: (f
= f x2=3
2=3 2
=x
=f
+ 2 x2=3
= x
4=3
g) (x)
g (x) =
+ 2x
=
If f (x) = x4=3
then: (f
2
3
=
2=3
2=3
2=3
= f (2x)
2=3
= (2x)
2
2=3
+ 2 (2x)
= 24=3 x4=3 + 25=3 x2=3
************************************************************************
************************************************************************
Question 31
3
The inverse function of y = (x + 4) + 8 is:
p
3
y= x 8 4
************************************************************************
3
original function: y
= (x + 4) + 8
inverse function: x
= (y + 4) + 8
p
3
3
3
x
8
= (y + 4)
x
8
=y+4
p
3
= x 8
y
2
+2
4 4
16
+ =
9 3
9
g (x) = (2x)
g) (x)
2
3
2
3
4
************************************************************************
************************************************************************
14
2
3
Question 32
Which of the following is not a polynomial?
Answer: f (x) = x
2
+4
************************************************************************
A polynomial has the form:
an xn + an
n 1
1x
+ an
n 2
2x
where n 2 N and ai 2 R for 0
i
+ a2 x2 + a1 x1 + a0
+
n.
Thus:
A.
x2
B.
C.
x
p
D.
x3
4 + 6x is a polynomial with n = 2, a2 = 1, a1 = 6, a0 =
2
4
+ 4 is not a polynomial [ n =
22
= N]
p
3 is a polynomial with n = 0, a0 = 3
4 is a polynomial with n = 3, a3 = 1, a2 = a1 = 0, a0 =
4
************************************************************************
************************************************************************
Question 33
If the graph of y = f (x) is obtained from the graph of y = x4 by shifting it one unit down
and two units to the right, then the formula for the function y = f (x) is:
4
y = (x
2)
1
************************************************************************
Transformations and Translations: Consider the graph of the function f (x). Then, if c > 0,
f (x)
c
f (x
c)
is the graph of f shifted up [+] or down [ ] by c units
is the graph of f shifted to the left [+] or to the right [ ] by c units
cf (x)
is the graph of f stretched [ if c > 1 ]
or compressed [ if c < 1 ] vertically by a factor of c
f (cx)
is the graph of f compressed [ if c > 1 )]
or stretched [ if c < 1 ] horizontally by a factor of c
f (x)
is the graph of f reflected in the x-axis
f ( x)
is the graph of f reflected in the y-axis
! f (x) shifted one unit down is represented by f (x) 1
! f (x) 1 shifted two units to the right is represented by f (x
Thus, y = (x
4
2)
2)
1
1
************************************************************************
************************************************************************
15
Question 34
The relationship between the graphs of y = f (x) and y =
f (x) is:
a re ection in the x-axis
************************************************************************
See solution for Question 33.
************************************************************************
************************************************************************
Question 35
Graph the region represented by the solution of 6x
7y
42.
Answer:
************************************************************************
First, we consider the line 6x
(i) 2 points on the line
7y =
42. We either need:
! x-intercept: 6x
! y-intercept: 6 (0)
or
(ii) 1 point and the slope of the line
10
42 ,
7y =
42 , x =
7
42 , y = 6
6
6x + 42
= x+6
7
7
6
! y-intercept: y = 6 and slope: m =
7
! 6x
LS = 6 (0)
10
-10
7y =
42 , 6x =
7y =
42 , y =
Then determine which side of the line satis es the inequality
by using a test point, say (0; 0) :
Graph the line:
-10
7 (0) =
7 (0) = 0
?
42 = RS
This is a true statement, thus the point (0; 0) is part of the
solution and we shade the region below the line to represent
all solutions.
************************************************************************
************************************************************************
16
Geometry
Question 36
The distance between the points P (3; 7) and Q (6; 9) is:
p
13
************************************************************************
The distance between the points (x1 ; y1 ) and (x2 ; y2 ) is given by:
q
2
2
d = (x2 x1 ) + (y2 y1 )
Thus, the distance between the points P (3; 7) and Q (6; 9):
q
p
p
2
2
d = (6 3) + (9 7) = 9 + 4 = 13
************************************************************************
************************************************************************
Question 37
An equation of the line with a slope of 3 and
passing through the point (2; 1) is: 3x
y
7=0
************************************************************************
A line with slope m passing through point (x0 ; y0 )
will have equation:
y
x
y0
=m
x0
Thus, a line with slope 3 passing through point (2; 1)
has equation:
y+1
x 2
=3
y+1
= 3 (x
0
2)
= 3x
6
= 3x
y
7
************************************************************************
************************************************************************
17
Question 38
The vertex of the parabola given by f (x) = 4x2
8x
2 is:
(1; 6)
************************************************************************
We “complete the square” to write the equation of the parabola
in standard form:
f (x)
= 4x2
= 4 x2
h
= 4 (x
= 4 (x
8x
2
2x
2
1)
2
1)
2
i
1
2
6
Thus, the parabola has its vertex located at (1; 6).
************************************************************************
************************************************************************
Question 39
Which of the following represent lines:
I. y = 3
II. xy = 3
III. x2 + y 2 = 3
IV. x + y = 3
Answer: I and IV
************************************************************************
Horizontal lines (slope = 0) have general equation: y = c, c 2 R.
Vertical lines (slope is unde ned) have general equation: x = k, k 2 R.
Lines with slope = m have general equation: y = mx + b, m; b 2 R.
I. y = 3 ! is a horizontal line X
II. xy = 3 ! not a line
III. x2 + y 2 = 3 ! not a line (actually this represents a circle)
IV. x + y = 3 [or y =
x + 3] ! a line with slope m =
1 X
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Question 40
A circle centered at (0; 0) with radius 1 has equation:
x2 + y 2 = 1
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The general equation of a circle centered at (p; q) with radius r
2
has equation: (x
2
q) = r2 .
p) + (y
Thus, a circle centered at (0; 0) with radius 1
has equation:
2
(x
0) + (y
2
0)
= (1)
x2 + y 2
2
=1
Note: A circle centred at the origin, with radius 1, is referred to as the unit circle.
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Exponential and Logarithmic Functions
Question 41
The expression log3
1
27
evaluates to:
3
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loga b represents “the exponent which is applied to base a to obtain b”
Letting x = log3
1
27
is equivalent to the equation 3x =
By inspection, we nd that x =
Thus, log3
1
27
=
3 as 3
3
=
1
3
3
=
1
.
27
1
.
27
3.
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19
Question 42
Considering the domain x 2 ( 3; 0) [ (3; 1), the expression log
log (x + 3) + log jx
3j
x2 9
x3
is equivalent to:
3 log jxj
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Properties of logarithms include:
loga (b c)
loga
= loga b + loga c
b
c
= loga b
loga (bc )
log
loga c
= c loga b
x2 9
x3
= log
(x
3) (x + 3)
x3
= log (x
3) + log jx + 3j
log x3
= log (x
3) + log jx + 3j
3 log jxj
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Question 43
The equation log5 (2x
3) = 0 has solution: x = 2
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First note that logarithms are only de ned for positive values. Thus, for the given equation
3
.
2
We keep this in mind while solving the equation.
2x
3 > 0 () 2x > 3 () x >
log5 (2x
2x
3)
3
2x
x
=0
= 50 = 1
=4
= 2 which is >
3
.
2
Thus, the equation has one solution, x = 2.
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20
Question 44
The equation 4x = 8x
has solution: x = 15
5
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4x
= 8x
OR
5
x
= 23
x 5
22x
= 23x
15
2x
= 3x
22
x
=
x
= 15
4x
ln (4x )
x ln 4
15
15
= 8x
5
= ln 8x
5
x ln 4
= (x
x ln 8
=
x
=
5 ln 8
ln 4 ln 8
=
5 ln 23
ln 22 ln 23
=
15 ln 2
2 ln 2 3 ln 2
=
5) ln 8
5 ln 8
15 ln 2
ln 2
= 15
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************************************************************************
Question 45
The function f (x) = 2x+3
f
1
1 has an inverse function given by:
(x) = log2 (x + 1)
3
************************************************************************
original function: y
= 2x+3
1
inverse function: x
= 2y+3
1
x+1
= 2y+3
log2 (x + 1)
y
=y+3
= log2 (x + 1)
3
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************************************************************************
21
Trigonometry
Question 46
3
radians
4
Convert 135o to radian measure:
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1 radian =
180
o
1o =
OR
Therefore, 135o = 135
180
=
radians
180
135
180
=
27
3
=
radians
36
4
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Question 47
5
. The angle is in radians.
6
p
3
Answer:
2
Evaluate cos
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The angle
5
6
radians falls within the second quadrant,
so the CAST Rule tells us that cos
Thus, cos
5
6
=
cos
6
=
5
6
will be negative.
p
3
.
2
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22
Question 48
Find the value of c in the given right triangle
if B = radians and a = 6 cm.
3
Answer: c = 12 cm
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In relation to angle B, c is the length of the hypotenuse side and a the length of the adjacent side.
Thus, we have
cos B
cos
3
c
adjacent
a
=
hypotenuse
c
6
=
c
6
=
cos 3
=
=
6
1
2
= 12
Therefore, the length of the hypotenuse is 12 cm.
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************************************************************************
Question 49
Find the value of Angle C in the given triangle
p
if B = radians, b = 3 cm and c = 3 2 cm.
6
Answer: C =
4
radians
************************************************************************
Note that this is not a right triangle.
Applying the sine law:
sin C
c
sin C
C
sin B
b
c sin B
=
b
p
3 2 sin
6
=
3
p
1
1
= 2
=p
2
2
=
=
4
[as sin
1
=p ]
4
2
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************************************************************************
23
Question 50
The function given by f ( ) = 3 sin (2 ) + 4 has period:
************************************************************************
Considering the general equation y = a sin (b ( + c)) + d:
a = the amplitude of the sine wave
2
= the period of the sine wave
b
c = the phase shift of the sine wave
d = a shift upwards [ d > 0 ] or downwards [ d < 0 ] of the wave.
The period of f is then given by
2
= .
2
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************************************************************************
24