
Solutions with explanations to compass math test practice questions in sample 1.
A  Numerical Skills/PreAlgebra

Solution
Use order of
operation to evaluate multiplication and division first from left to
right 
9 ÷ 3 · 2 = 6 
Insert the result in
the whole expression 
72  6 + 2 
Evaluate addition and subtraction from left to right to obtain final value of expression 
72  6 + 2 = 68 

In scientific notation, 3.0 x 10^{5} + 0.0000022 =
Solution
We first rewrite 0.0000022 using 10^{5}
as follows 
0.0000022 = 0.22 ×10^{5} 
We now add 3.0 x 105 and 0.22
×10^{5
}as follows 
3.0 x 105 + 0.22 ×10^{5 }
= 10^{5 }(3.0 + 0.22)
= 3.22 x 105 

If
.
is calculated and the fraction obtained reduced to simplest form, then what is its numerator?
Solution
Reduce all fractions to the least
common denominator 12, add, subtract and reduce the final result. 
2 / 3 + (1 / 2  1 / 6) + (1 / 3  3 / 4)
= 8 / 12 + (6 / 12  2 / 12) + (4 / 12  9 / 12)
= (8 + 6  2 + 4  9) / 12
= 7 / 12 
The numerator after reduction is 
7 

.
Solution
Evaluate expressions
inside brackets 
4 / 3 × 3 / 5 = 4 / 5
1 / 4 ÷
4 / 5 = 1 / 4 ×
5 / 4 = 5 / 16 
Substitute
expressions inside brackets by their values found above 
1 / 4 + (4 / 3 × 3 / 5)  (1 / 4
÷ 4 / 5) = 1 / 4 + 4 / 5  5
/ 16 
Reduce to the least
common denominator 80 and evaluate from left to right to obtain
the final answer. 
= 20 / 80 + 64 / 80  25 / 80 = 59 / 80 

A shop owner increased the selling price of a shirt from $20 to $26. By what percentage was the price increased?
Solution
Original price 
20 
Selling price 
26 
Difference / change 
26  20 = 6 
Relative change 
change / original price = 6 / 20 
Change fraction into
percent 
6 / 20 = 30 / 100 = 30 % 

3/4+0.85+20% =
Solution
Given expression

3 / 4 + 0.85 + 20 % 
Change fraction into
decimal number (divide) 
3 / 4 = 0.75 
Change 20% to decimal 
20% = 20 / 100 = 0.2 
Evaluate the given
express using decimal forms 
3 / 4 + 0.85 + 20 % = 0.75 + 0.85 + 0.2
= 1.8 

Tom worked for 6 hours and was paid $5.50 per hour. He then went shopping and bought two magazines at $9.50 each and a pen at $8.25. How much money does Tom have left?
Solution
Tom worked 6 hours at
$5.50 per hour; he
earned 
6 × $5.50 = $33 
Shopping 2 magazines
at $9.50 each 
2 × $9.50 = $19 
Shopping a pen at
$8.25 
$8.25 
Total Shopping (he spent) 
$19 + $8.25 = $27.25 
Tom has left = what he earned  what he spent 
$33  $27.25 = $5.75 

Bill bought 1(3/4) pounds of Cheddar cheese, 3.75 pounds of blue cheese and 4(1/2) pounds of goat cheese. How many pounds of cheese did Bill buy?
Solution
Change the mixed
number 1 (3/4) into
decimal number 
1 (3/4) = 1.75 
Change the mixed number 4 (1/2) into decimal number 
4 (1/2) = 4.5 
Total amount of
cheese bought 
1 (3/4) + 3.75 + 4 (1/2)
= 1.75 + 3.75 + 4.5
= 10 pounds 

In a certain school, 40% of all students are 8 years old or younger. The number of the remaining students is 120. How many students are 8 years old or younger?
Solution
Let x be the total
number of students 
x 
Number of students
that are 8 years old or
younger is given by 
40% x 
Remaining students 
120 students
or 100%  40% = 60% of total
or 60% x 
Hence 
60% x = 120 or x = 120 / 60% 
Number of students
that are 8 years old or
younger is given by 
40% x = 40% (120 / 60%) = 80 

1/5 of students in a school have no brothers or sisters. Of the remaining students, 40% have 1 brother or sister. What percent of all students have more than 1 brother or sister?
Solution
Let x be the total
number of students 
x 
Number of students
with no brothers or sisters 
1/5 of x = x(1/5) 
Number of students
with brothers or sisters 
(1  1/5) = (4/5) x 
40% of those with
brothers or sisters have only one brother or sister 
40% of (4 / 5) x = 40% (4/5) x 
Students with more
than one brother or sister 
(100%  40%) of (4 / 5) x = 60% (4 / 5) x = 48 %
x 
Percent of all
students with more than one brother or sister is 
48 %


A car hire company offers two plans A and B. In plan A, a customer pays a flat fee of $20 for the day plus 5 cents for each kilometer. In plan B, a customer pays a flat fee of $15 for the day plus 7 cents for each kilometer. For how many kilometers would a customer pay the same amount for the two plans?
Solution
Let x be the number
of km 
x 
The cost CA for plan
A is given by 
CA = 20 + 0.05 x 
The cost CB for plan
B is given by 
CB = 15 + 0.07 x 
Same amount paid 
CA = CB 20 + 0.05 x =
15 + 0.07 x 
Solve the above
equation to find x which is the number of km so that a customer would
pay the same amount. 
20  15 = 0.07 x  0.05 x
x = 250 

Joe worked part time for x hours and earned y dollars. How much does he earn if he works z hours?
Solution
Earn y dollars in x
hours, earn per hour 
y / x 
Earn in z hours 
z (y / x) = z y / x 

In a science test, a class of 30 students had an average of 80. The 20 girls in the class had an average of 85. What is the average of the 10 boys in the class?
Solution
Let T be the total of
marks for the 30 students. If average is 80, then 
T / 30 = 80 T = 30 *
80 = 2400 
Let TG be the total
marks for the 20 girls. If average is 85, then 
TG / 20 = 85 TG = 20
* 85 = 1700 
Let TB be the total
marks for the 10 boys, then 
TB = 2400  1700 = 700 
Average for boys is
given by 
700 / 10 = 70 

Find y if y / 5 = 10 / 25.
Solution
Given equation 
y / 5 = 10 / 25 
Cross product gives 
25 y = 5 * 10 
Solve for y 
y = 2 

√(3^{2} + 4^{2}) =
Solution
Given expression 
√(3^{2} + 4^{2}) 
Simplify expression
inside radical 
3^{2} + 4^{2 }= 9 + 16 = 25 
Simplify given expression by taking square root 
√(3^{2} + 4^{2}) = √25
= 5 
B  Algebra

If a = 2 and b = 2, what is the value of (a^{3}  1) / (b  1)
Solution
Given expression 
(a^{3}  1) / (b  1) 
Substitute a by 2
and b by 2 in given expression 
((2)^{3}  1) / ((2)  1) 
Simplify expression
above 
(8  1) / (2  1) = 3 

If 2(x + 9) = 20, then  4x =
Solution
Given equation 
2(x + 9) = 20 
Solve for x 
2x  18 = 20 2x =
38
x = 19 
Substitute x in  4x
by  19 
 4 x =  4 (19) = 76 

x is a variable such that if 20% of it is added to its fifth the result is equal to 12 subtracted from seven tenths of x. Find x.
Solution
x is a variable 
x 
20% of it 
20% x 
its fifth 
(1 / 5) x 
x is a variable such that
if 20% of it is added to its fifth 
20% x + (1 / 5) x 
seven tenths of x 
(7 / 10) x = 0.7 x 
the result is equal to 12
subtracted from seven tenths of x 
20% x + (1 / 5) x = 0.7x  12 
Change % and fractions to
decimal numbers 
0.2 x + 0.2 x = 0.7x  12 
Solve for x 
12 = 0.3 x x = 40 

The area A of a trapezoid is given by the formula A = 0.5 (b + B) h, where b and B are the sizes of the bases and h the size of the height of the trapezoid. Express B in terms of A, b and h.
Solution
Given formula 
A = 0.5 (b + B) h 
Divide both sides of
the formula by 0.5 h and simplify 
(A / 0.5 h) = b + B 
Solve for B 
B = (A / 0.5 h)  b 
Note that 1
/ 0.5 = 2, hence 
B = 2 A / h  b 

Tom, Linda and Alex have $120 dollars. Alex has the third of what Tom has and Linda has twice as much as Alex. How much money, in dollars, does Linda have?
Solution
Let x be the amount
of money in dollars that Tom has 
x 
Alex has the third of
Tom 
(1 / 3) x 
Linda has twice as
much as Alex 
2 (1 / 3) x 
All together
they have 120 dollars, hence 
x + x / 3 + 2 x / 3 = 120 
Multiply all terms of
the above equation, simplify and solve for x 
3 x + x + 2 x = 360
x = 60 
Linda has 
2 (1 / 3) x = (2 / 3) 60 = $40 

Which of the following is equivalent to 6x^{2}  11x  2?
Solution
Given 
6x^{2}  11x  2 
The linear terms that will make the factors has coefficients 6 and 1 (since 6*1 = 6) or 2 and 3 (since 2*3=6) 
(6x )(x
) (2x )(3x ) 
The constant terms
should 1 and 2 or 1 and 2 
(6x  1)(x + 2) 
If (6x  1)(x
+ 2) is expanded, it does not give 6x^{2}  11x  2. 
(6x  1)(x + 2) = 6x^{2 }+ 11x  2^{ }

We now try (6x + 1)(x  2) which when
expanded gives 6x^{2 } 11x  2^{ } 
(6x + 1)(x  2) = 6x^{2 } 11x  2^{ }


Which of the following is a factor of x^{2}  7x  8?
Solution
Given 
x^{2}  7x  8 
Factor 
(x + 1)(x  8) 
One of the factors is

x + 1 

(2xy^{2}  3x^{2}y)  (2x^{2}y^{2}  4x^{2}y) =
Solution
Given 
(2xy^{2}  3x^{2}y)  (2x^{2}y^{2}  4x^{2}y) 
Eliminate brackets
taking signs into consideration 
= 2xy^{2}  3x^{2}y  2x^{2}y^{2} + 4x^{2}y 
Group like terms  3x^{2}y and 4x^{2}y 
= 2xy^{2}  2x^{2}y^{2} +
x^{2}y 

During the same journey, Stuart drove x miles for 2 hours, and 200 miles for 3 hours. Find x if the average speed for the entire journey is 70 miles per hour.
Solution
Average speed =

Total distance / total time = 70 miles/hour 
Total distance =

x + 200 miles 
Total time = 
2 + 3 = 5 hours 
Substitute in formula
above 
(x + 200) / 5 = 70 
Solve above equation
for x 
x + 200 = 350 x = 150
miles 

Given the equations of the lines
(I) 2y + 3x = 3
(II) 3y  2x = 5
(III) 6y + 4x = 9,
(IV) 2y + 6x = 9
which two lines are perpendicular?
Solution
Rewrite all 4
equations in slope intercept form 
(I) y = ( 3 / 2) x + 3 / 2
(II) y = (2 / 3) x  5 / 3
(III) y = (2 / 3) x  3 / 2
(IV) y =  3x + 9 / 2 
The slopes of all 4
lines are 
(I) 3/2
(II) 2/3
(III) 2/3
(IV) 3 
For two lines with
slopes m and n to be perpendicular, the product of their slopes must
equal 1 or m*n = 1. The product of the slopes of (I) and (III) gives
1; and hence (I) and (III) are perpendicular. 
(3 / 2)*(2 / 3) = 1 

If f(x) = (x + 1)^{2}, then f(t + 2) =
Solution
Given 
f (x) = (x + 1)^{2} 
To find f (t + 2), we
substitute x by t + 2 in f (x); hence 
f (t + 2) = ((t + 2)+ 1)^{2}
=
(t + 3)^{2}
= t ^{2 }+ 6t + 9 

For x > 0 and y > 0,
(√x + √y) (√x  √y)  (√x  √y)^{2} =
Solution
Given 
(√x + √y) (√x  √y)  (√x  √y)^{2} 
Expand (√x + √y) (√x
 √y) 
(√x + √y) (√x  √y) = x 
y 
Expand (√x  √y)^{2} 
(√x  √y)^{2} = x + y  2√x √y 
Hence 
(√x + √y) (√x  √y)  (√x  √y)^{2}
= (x  y)  (x + y  2√x √y)
=  2 y + 2√x √y 

What is the slope of the line whose equation is given by
x / 2  y / 4 = 7
Solution
Equation of line given
in standard form 
x / 2  y / 4 = 7 
To determine the slope,
we need to write in slope intercept form. Multiply all terms by 4 and
simplify and solve for y. 
2 x  y = 38 y = 2x 
38 
Slope is the
coefficient of x 
2 

For x > 3,
(x / (x  3) + 1 / 2)(2/(x  1)) =
Solution
Given 
(x / (x  3) + 1 / 2) (2 / (x  1)) 
We first simplify, if
possible, the expression on the left side. Reduce to the lowest common denominator. 
( x / (x  3) + 1 / 2)
= 2x / ( 2 (x  3) ) + ( x 3) / (2 (x  3))
= (3x  3) / (2(x  3)) 
Hence 
x / (x  3) + 1 / 2) (2 / (x  1)) = (3x  3) (2/(x  1))) / (2(x
 3))
= 3(x  1)(2 / (x  1)) / (2(x 3)) 
Simplify 
= 3 / (x  3) 

In a standard rectangular system of axes, point A has the coordinates (2 , 1). What must be the coordinates of point B if M(3 , 2) is the midpoint of the segment AB?
Solution
Let a and b be the x
and y coordinates of point B,
hence 
M is the midpoint of A and B 
Since the midpoint
M(3,2) is known and point A(2,1) is also known, we can write 
(3,2) = ( (2 + a)/2 , (1 + b)/2 ) 
Two equations may be
written as follows 
3 = (2 + a) / 2 and 2 = (1 + b) / 2 
Solve for a and b 
a = 4 and b = 3 
C  College Algebra

What is the value of X in the following geometric sequence
X , 64 , ___ , ___ , 8
Solution
Let r be the common
ratio of the geometric sequence. Hence the terms of the sequence may be
written as 
X , 64 , __ , ___ , 8 or
X , r X , r^{ 2} X , r^{ 3} X , r^{ 4} X 
Two terms are known,
the second and the last. Hence 
r X = 64 and r^{ 4} X = 8 
Divide the left hand
side and the right hand side of the two equations in order to eliminate
X, simplify and solve for r. 
r^{ 4} X / r X = 8 / 64 r^{ 3} = 1 / 8
r = 1 / 2 
The second term is
equal to 64 and also given by r X; write the equation and solve for X. 
(1/2) X = 64 X = 128 

For what value of K will the system of equations given below have no solution?
2x + 5y = 9
3x  K y = 4
The given system of
equations will not have solutions if its determinant is equal to 0.
Hence 
2(K)  5 (3) = 0 
Solve the above for K 
K = 15 / 2 

If f(x) and g(x) are two functions defined by
f(x) = √(x + 1)
g(x) =  x  1 ,
then what is the value of f(g(9))?
Solution
f and g given 
f (x) = √(x + 1) and g (x) =  x  1  
Hence 
f (g(9)) = f(9  1)
= f(8)
= f(8)
= √(8 + 1) = 3 

The range of
f(x) = x^{2} 2x + 1
is given by the interval
Solution
Given 
f (x) = x^{2} 2x + 1 
f is a quadratic function and its graph is a parabola that opens up since its leading coefficient is positive and therefore has a minimum
point which is also the vertex. The coordinates of the vertex V of the graph of is given by 
V (h , k) = (b/2a , f(b/2a))
= (1 , 0) 
Since (1 , 0) is a
minimum point, the range of values of f(x) is given by the interval 
[ 0 , + infinity) 

If f(x) is a linear function such that f(1) = 11 and f(2) = 5,
then f(0) =
Solution
f(x) is a linear
function which may be written as 
f (x) = m x + b , where m and b are real numbers 
f(1) = 11 and f(2) = 5
give the following equations 
 m + b = 11 2m + b = 5 
We need to find f(0)
= b. Hence solve the above system of equations for b. Multiply all terms
in the equation  m + b = 11 by 2 and rewrite
the system as follows 
2m + 2 b = 22 2m + b
= 5 
Add the sides of the
two equations, solve and evaluate f(0). 
3 b = 27
b = 9
f(0) = 9 

Which of these functions is(are) always increasing?
(I) f(x) = x^{2} + 2x + 1
(II) f(x) = 2x  1000
(III) f(x) = x  4 + 255
(IV) f(x) = e^{x},
Solution
Given 
(I) f(x) = x^{2} + 2x + 1
(II) f(x) = 2x  1000
(III) f(x) = x  4 + 255
(IV) f(x) =  e^{ x},

Function (I) is a
quadratic one and therefore is not always increasing.
Function (II) is a linear function with positive slope and therefore always increasing.
Function (III) contains an absolute value and its graph is V shaped and therefore is not always
increasing.
Function (IV): e^{ x} is an exponential function with base
equal to 1/e and therefore less than 1 and hence always decreasing but
when multiplied by 1 it becomes always increasing. Hence f(x) =  e^{ x}
is always increasing.

Of all the functions given the ones in (II) and (IV)
are always increasing. 

i is the imaginary unit such that i = √1. What is the value if i^{99}?
Solution
Note that 
i^{ 4} = 1 , where i is the imaginary unit

We need to evaluate i^{ 99}. Divide 99 by 4 and
write 99 as follows 
99 = 4 * 24 + 3 
Hence 
i ^{99 }= i ^{(4*24 + 3)}
= (i ^{4})^{ 24} i^{3}
= 1 * i^{3} =  i 

If
.
Find a, b and c.
Solution
Multiply row [3 , 2]
in the first matrix by column [0 , b] in the second matrix and equate it
to the corresponding term in the third matrix. 
3*0 + 2*b =  4 
Multiply row [3 , 2] in the first
matrix by column [a , c] in the second matrix and equate it to the
corresponding term in the third matrix. 
3*a + 2*c = 9 
Multiply row [1 ,  2] in the first
matrix by column [a , c] in the second matrix and equate it to the
corresponding term in the third matrix. 
 a  2 c =  7 
Solve the first equation to find b
and solve simultaneously the second and third equations to find a and c.
Hence 
a = 1, b = 2 and c = 3 

In an arithmetic series, the 3rd term is equal to 114 and the last term is equal to 27. The sum of all terms in the series is equal to 2325. What are the last three terms in the series?
Solution
Let a be the first
term of the series, d be the common difference. The third term given by
a + 2d is equal to 114. Hence 
(I) a + 2 d = 114 
Let n be the number of terms in
the series. The last term is given by a + (n  1)d and is equal to  27.
Hence 
(II) a + (n  1) d = 27 
The sum of the n terms is given by (n
/ 2)(a  27) and is equal to 2325. Hence 
(III) (n / 2)(a  27) = 2325 
Equation (I) gives 
a = 114  2d 
Substitute a by 114  2d in equations
II and III 
(II) 114  2d + (n  1)d = 27
(III) (n / 2)(114  2d  27) = 2325 
Group multiply all terms of equation
(III) group like terms and simplify to obtain 
(II) n d  3d = 141
(III) 87n  2 n d = 4650 
Solve equation (II) for d 
d = 141 / (n  3) 
Substitute d by 141 / (n  3) in
equation (III) 
87n  2n (141 / (n  3)) = 4650 
Expand and simplify and set the
numerator equal to zero to obtain the quadratic equation. 
87n^{2}  4629 n + 13950 = 0 
Solve the above equation for integer
n to find 
n = 50 and n = 93 / 29 
Only n = 50 is a solution since it is
an integer. Find d using the above equation 
d = 141 / (n  3) = 141 / (50  3) =  3 
Last term is equal to 27 and hence
the last three terms are equal to 
27
27  d = 24
24  d = 21 
D  Geometry

In the figure below, line T1 is parallel to line T2 and line L1 is parallel to line L2. The pair of lines T1 , T2 are not perpendicular to L1,L2. Which of the set of angles below has all 4 angles equal in measurement?
.
A) {a,b,c,d}
B) {e,f,g,h}
C) {e,f,m,h}
D) {j,k,f,g}
E) {a,b,f,g}
Solution
Angles j and k are verticcal angles and therefore are equal in size. Angles f and g are also vertical angles and equal in size. Angles j and f are corresponding angles to the same angle n and are therofore equal in size. Hence all angles in the set {j,k,f,g} in D) are equal in size.

A cube with edge length x and a right cylinder with circular base have the same volume. Find x ,in cm, if the radius of the cylinder is 5 cm and its height is 5 cm.
Solution
Volume V of cylinder is given by
V = height*area of base = 5 * (pi*5^{2}) = 125 pi
Volume W of cube with edge x is given by
W = x^{3}
The two volumes are equal
x^{3} = 125 Pi
Solve above for x
x = 5 (Pi)^{1/3}

Find, in square cm, the area of an equilateral triangle of side 20 cm.
Solution
Draw the altitude from one vertex and make a right triangle of legs h and x/2 and hypotenuse x. The altitude h of the triangle
h = √ (x^{2}  (x/2)^{2}) = (x/2) √3
The area A of the triangle is given by
A = (1/2) x (x/2) √3 = 100 √3

The two circles shown below have diameters AB and BD and are tangent at B. ABC and BDE are isosceles triangles inscribed inside the circles. What is the measurement, in degrees, of angle CBE?
.
Solution
Since AB is the diameter of the circle, then ABC is a right triangle. Since it is isosceles the size of angle CBA is 45 degrees. Since BD is the diameter of the circle, then BDE is a right triangle. Since it is isosceles, the size of angle EBD is 45 degrees.
Since the circles are tangent at B, the points A, B and D are collinear and therefore the sum of the sizes of angles CBA, CBE and EBD is equal to 180 degrees which gives the size of angle CBE of 90 degrees.

In the figure below, EB and FC are both perpendicular to AC and x is the length of AB. If the length of AC is 11, what is x?
.
Solution
Since EB and FC are perpendicular to the same line, they are parallel and therefore triangles ABE and ACF are similar and the proportionality of the corresponding sides gives
x / 11 = 4 / 9
solve for x: x = 44 / 9

ABC is a right triangle such that the length of BC is 5 units. The legs have length x and 2x. Find x.
.
Solution
Since the triangle is right we use Pythagora's theorem
x^{2} + (2x)^{2} = 5^{2}
5 x^{2} = 5^{2}
solve for x: x = √5

AB is tangent to the circle with center O and radius 10 cm. The length of OA is 15 cm and the length of OB is 20 cm. Find the area, in square cm, of triangle ABO.
.
Solution
If the altitude from vertex o to the base AB is drawn, we end up with two right triangles of hypotenuses 15 and 20 cm. Let T being the point of tangency of AB and the circle. The remaining legs of the two triangles may be found using Pythagora's theorem.
AT^{2} = 15^{2}  10^{2} = 125
BT^{2} = 20^{2}  10^{2} = 300
AT = 5 √ 5
BT = 10 √ 3
The area of triangle ABO is given by
(1/2)*OT*(AT + BT) = (1/2)*10*(5 √ 5 + 10 √ 3)
= 25 √ 5 + 50 √ 3

The area of the trapezoid shown below is 128 square cm. Find x in cm.
.
Solution
If h is the altitude of the trapezoid, then the area is given by
(1/2)*h*(10 + 22) = 128
solve for h: h = 8
If we draw an altitude so that side x, the altitude and part of the lower base form a right triangle, then the length of that part of the lower base is equal to
(22  10) / 2 = 6
We now use Pythagora's theorem to find x
x^{2} = 8^{2} + 6^{2}
solve for x: x = 10

C1, C2 and C3 are three tangent circles with their centers on the same line. If R is the radius of the large circle and the diameter of circle C1 is twice the diameter of circle C3, then what is the area of the shaded region?
.
Solution
The area of the largest circle is given by.
Pi R^{2}
The sum of the diameters of C1 and C3 is equal to the diameter of the largest circle which is 2R. The diameter of C1 is twice that of C3. Hence the diameter of C3 is given by
2 R / 3 and its radius is 2 R / 6
and the diameter of C1 is given by
4 R / 3 and its radius is 4 R / 6
Area of C1 is given by
Pi (4 R / 6)^{2}
Area of C3 is given by
Pi (2 R / 6)^{2}
Area of shaded region is given by area of largest circle minus area of C1 and area of C3
Pi R^{2}  Pi (4 R / 6)^{2}  Pi (2 R / 6)^{2}
= (4 / 9)R^{2}Pi
E  Trigonometry

If x is a positive angle smaller than 90° and cos(x) = 1/3, then sin(x) =
Solution
Use the identity sin^{2}x + cos^{2}x = 1 to find sin x as follows
sin^{2}x = 1  cos^{2}x
Solve for sin x
sin x = + or  √(1  cos^{2}x)
Since x is positive and less than 90 degrees, sin x is positive and hence
sin x = + √(1  cos^{2}x)
= √( 1  (1/3)^{2} ) = 2 √2 / 3

Convert 135° into radians.
Solution
135° = (135° * Pi) / 180° = 3 Pi / 4

The measurement of an angle of a right triangle is 40° and the leg opposite to this angle is 5 cm long. What is the length, in cm, of the other leg?
Solution
Let us use the definition of the tangent
tan(40°) = opposite side / adjacent side = 5 / adjacent side
solve to find adjacent side
adjacent side (second leg) = 5 / tan(40°)

Which of these graphs is the graph of y = cos(2x)?
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Solution
Graphs C and E do not satisfy the fact that y = 1 at x = 0. The period of y = cos(2x) is given by
2 Pi / 2 = Pi
Graphs B has a period equal to Pi and is the graph of y = cos(2x).

What is the smallest positive value of x for which y = cos(3x) has a minimum value?
Solution
The smallest positive value of A that makes cos(A) minimum is Pi. If instead of cos(A) we have cos(3x), then the smallest value of 3x that makes cos(3x) minimum is also Pi. Hence
3x = Pi
x = Pi / 3 is the smallest positive value of x that makes cos(3x) minimum.

If you put 6 equilateral triangles together as shown below, you can make a regular hexagon. What is the area, in square units, of a regular hexagon of side lenght 6 units?
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Solution
The altitude h of an equilateral triangle of side 6 is given by
h = √(6^{2}  3^{2}) = 3 √ 3
The area of an equilateral triangle of side 6 is given by
(12)*6*3 √ 3 = 9 √ 3
The area of the hexagon is given by
6*9 √ 3 = 54 √ 3
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