The document contains a math review with examples of different techniques for counting combinations and permutations, as well as examples involving polynomials, solid geometry, and other math concepts. It provides the steps and work for 14 different math problems, showing how to use techniques like multiplication, combinations, and permutations to calculate things like the number of combinations for choosing food items or essays. It also includes examples of multiplying polynomials, finding volumes of solids, and solving geometry problems involving parallel lines and transversals.
APEX INSTITUTE was conceptualized in May 2008, keeping in view the dreams of young students by the vision & toil of Er. Shahid Iqbal. We had a very humble beginning as an institute for IIT-JEE / Medical, with a vision to provide an ideal launch pad for serious JEE students . We actually started to make a difference in the way students think and approach problems.
APEX INSTITUTE was conceptualized in May 2008, keeping in view the dreams of young students by the vision & toil of Er. Shahid Iqbal. We had a very humble beginning as an institute for IIT-JEE / Medical, with a vision to provide an ideal launch pad for serious JEE students . We actually started to make a difference in the way students think and approach problems.
B.Sc (Pass) Nautical & Engineering Model Question 2 Mathematics Second Paper
(Differential Calculus, Integral Calculus, Two-dimensional & Three- dimensional Geometry)
APEX INSTITUTE was conceptualized in May 2008, keeping in view the dreams of young students by the vision & toil of Er. Shahid Iqbal. We had a very humble beginning as an institute for IIT-JEE / Medical, with a vision to provide an ideal launch pad for serious JEE students . We actually started to make a difference in the way students think and approach problems.
APEX INSTITUTE was conceptualized in May 2008, keeping in view the dreams of young students by the vision & toil of Er. Shahid Iqbal. We had a very humble beginning as an institute for IIT-JEE / Medical, with a vision to provide an ideal launch pad for serious JEE students . We actually started to make a difference in the way students think and approach problems.
B.Sc (Pass) Nautical & Engineering Model Question 2 Mathematics Second Paper
(Differential Calculus, Integral Calculus, Two-dimensional & Three- dimensional Geometry)
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Proton Training Solution (PTS) is a trusted Training Institute in Pune for MBA Entrance Exams (CAT | IIFT | XAT | SNAP | NMAT | CMAT | TISS | MH CET | MAT | ATMA and many others), BBA Entrance Exams (IPM-AT | DUJAT | SET | NPAT | BMCC | MIT and many others). Proton is also associated with many institutes as knowledge partner & provides Aptitude Training for Placement Preparation.
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good luck guys and girls...simple and short ans also sweet formulas..
Directions Please show all of your work for each problem. If app.docxduketjoy27252
Directions: Please show all of your work for each problem. If applicable, you may find Microsoft Word’s equation editor helpful in creating mathematical expressions in Word. The option of hand writing your work and scanning it is acceptable.
1. List all the factors of 88.
2. List all the prime numbers between 25 and 60.
3. Find the GCF for 16 and 17.
4. Find the LCM for 13 and 39.
5. Write the fraction in simplest form.
6. Multiply. Be sure to simplify the product.
7. Divide. Write the result in simplest form.
8. Add.
9. Perform the indicated operation. Write the result in simplest form. –
10. Perform the indicated operation. Write the result in simplest form. ÷
11. Find the decimal equivalent of rounded to the hundredths place.
12. Write 0.12 as a fraction and simplify.
13. Perform the indicated operation. 8.50 – 1.72
14. Divide.
15. Write 255% as a decimal.
16. Write 0.037 as a percent.
17. Evaluate. 56 ÷ 7 – 28 ÷ 7
18. Evaluate. 9 42
19. Multiply: (-1/4)(8/13)
20. Translate to an algebraic expression: Twice x, plus 5, is the same as -14.
21. Identify the property that is illustrated by the following statement. 5 + 15 = 15 + 5
22. Identify the property that is illustrated by the following statement.
(6 · 13) 10 = 6 · (13 · 10)
23. Identify the property that is illustrated by the following statement.
10 (3 + 11) = 10 3 + 10 11
24. Use the distributive property to remove the parentheses in the following expression. Then simplify your result where possible. 3.1(3 + 7)
25. Add. 14 + (–6)
26. Subtract. –17 – 6
27. Evaluate. 3 – (–3) – 13 – (–5)
28. Multiply.
29. Divide.
30. Evaluate. (–6)2 – 52
31. Evaluate. (–9)(0) + 13
32. A man lost 36 pounds (lb) while dieting. If he lost 3 pounds each week, how long has he been dieting?
33. Write the following phrase using symbols: 2 times the sum of v and p
34. Write the following phrase using symbols. Use the variable x to represent the number: The quotient of a number and 4
35. Dora puts 50 cents in her piggy bank every night before she goes to bed. If M represents the money (in dollars) in her piggy bank this morning, how much money (in dollars) is in her piggy bank when she goes to bed tonight?
36. Write the following geometric expression using the given symbols.
times the Area of the base (A) times the height(h)
37. Evaluate if x = 12, y = , and z = .
38. A formula that relates Fahrenheit and Celsius temperature is . If the current temperature is 59°F, what is the Celsius temperature?
39. If the circumference of a circle whose radius is r is given by C = 2πr, in which π ≈ 3.14, find the circumference when r = 15 meters (m).
40. Combine like terms: 9v + 6w + 4v
41. A rectangle has sides of 3x – 4 and 7x + 10. Provide a simplified expression for its perimeter.
42. Subtract 4ab3 from the sum of 10ab3 and 2ab3.
43. Use the distributive property to remove the p.
APEX INSTITUTE was conceptualized in May 2008, keeping in view the dreams of young students by the vision & toil of Er. Shahid Iqbal. We had a very humble beginning as an institute for IIT-JEE / Medical, with a vision to provide an ideal launch pad for serious JEE students . We actually started to make a difference in the way students think and approach problems.
Final Exam Name___________________________________Si.docxcharlottej5
Final Exam Name___________________________________
Silva Math 96 Spring 2020
YOU MUST SHOW ALL WORK AND BOX YOUR ANSWERS FOR CREDIT. WORK ALONE.
Solve the absolute value inequality. Write your answer
in interval notation.
1) |2x - 12 |> 2
Solve the compound inequality. Graph the solution set.
Write your answer in interval notation.
2) -4x > -8 and x + 4 > 3
Solve the three-part inequality. Write your answer in
interval notation.
3) -1 < 3x + 2 < 14
Solve the absolute value equation.
4) 4x + 9 = 2x + 7
Solve the compound inequality.
5) 3( x + 4 ) ≥ 0 or 4 ( x + 4 ) ≤ 4
Solve the inequality. Graph the solution set and write
your answer in interval notation.
6) |5k + 8| > -6
Solve the inequality graphically. Write your answer in
interval notation .
7) x + 3 ≥ 1
x-8 -6 -4 -2 2
y
8
6
4
2
x-8 -6 -4 -2 2
y
8
6
4
2
1
Graph the system of inequalities.
8) 2x + 8y ≥ -4
y < - 3
2
x + 6
x-10 -8 -6 -4 -2 2 4 6 8 10
y
10
8
6
4
2
-2
-4
-6
-8
-10
x-10 -8 -6 -4 -2 2 4 6 8 10
y
10
8
6
4
2
-2
-4
-6
-8
-10
Find the determinant of the given matrix.
9) 10 5
0 -4
Use Cramer's rule to solve the system of linear
equations.
10) 6x + 5y = -12
2x - 2y = -4
Write a system that models the situation. Then solve the
system using any method. Must show work for credit.
11)A vendor sells hot dogs, bags of potato chips,
and soft drinks. A customer buys 3 hot dogs,
4 bags of potato chips, and 5 soft drinks for
$14.00. The price of a hot dog is $0.25 more
than the price of a bag of potato chips. The
cost of a soft drink is $1.25 less than the price
of two hot dogs. Find the cost of each item.
Use row reduced echelon form to solve the system.
12) x + y + z = 3
x - y + 4z = 11
5x + y + z = -9
2
Find the domain of f. Write your answer in interval
notation.
13) f(x) = 13 - 9x
If possible, simplify the expression. If any variables
exist, assume that they are positive.
14) 2x + 6 32x + 6 8x
Match to the equivalent expression.
15) 100-1/2
A) 1
1000
B) 1
10
C) 1
100
D) 1
10
Write the expression in standard form.
16) (5 + 8i) - (-3 + i)
Simplify the expression. Assume that all variables are
positive.
17) 5 t
5
z10
Solve the equation.
18) 3x + 1 = 3 + x - 4
Write the expression in standard form.
19) 3 + 3i
5 + 3i
3
Write the equation in vertex form.
20) y = x2 + 5x + 2
The graph of ax2 + bx + c is given. Use this graph to solve
ax2 + bx + c = 0, if possible.
21)
x-5 5 10
y
50
40
30
20
10
-10
-20
-30
-40
-50
x-5 5 10
y
50
40
30
20
10
-10
-20
-30
-40
-50
Solve the equation. Write complex solutions in standard
form.
22) 4x2 + 5x + 5 = 0
Graph the quadratic function by its properties.
23) f(x) = 1
3
x2 - 2x + 3
x
y
x
y
Solve the equation. Find all real solutions.
24) 2(x - 1)2 + 11(x - 1) + 12 = 0
Solve the problem.
25) The length of a table is 12 inches more than its
width. If the area of the table is 2668 square
inches, what is its length?
4
Solve the equation..
1. Colegio San Patricio
2nd Period Math Review´s ANSWERS
School Year 2009 – 2010
Counting Techniques
5) The school’s cafeteria has new options. Now they have: 3 types of salads, 3 types of
hamburgers, 5 types of fruit beverages, and 5 dessert options. How many different
combinations you to choose from?
MULTIPLICATION Technique (you have 4 groups)
Salads = 3
Hamburgers = 3
Fruit Beverages = 5
Desserts = 5
3*3*5*5 = 225 different combinations
6) How many ways are there for choosing any 4 books from the library’s desk from the 12
books they are on the top of the desk?
COMBINATION Technique (you have 1 group = books & order in which you choose the books
does NOT matters.)
nCr = n! (until r) = 12C4 = 12*11*10*9 = 495 combinations
r! 4*3*2*1
7) In how many ways can a jury choose the best 3 essays from the top 20 essays they
received? (The first chosen person will have the 3rd
. place; the second one will have the 2nd
.
place, and the last person will be the 1st
. place.)
PERMUTATION Technique (you have 1 group = essays & order in which the jury selects the
essays DOES matters.)
nPr = n! (until r) = 20P3 = 20*19*18 = 6840 permutations
Grouping and Analyzing Data
8) The following data shows the grades of different students.
74 95 96 85 77 65
83 88 93 12 81 92
100 20 40 35 98 87
a) Create a class interval table (interval, tally, frequency). The first interval is 0-20
b) Graph the information using a frequency polygon graph.
2. a) Interval TableTable:
Intervals Tallies Frequency
0 - 20 II 2
21 - 40 II 2
41 - 60 0 0
61 - 80 III 3
81 - 100 IIIII IIIII II 11
b)Graph
Fre que nc y Po ly g o n Gra ph
0
2
4
6
8
10
12
0-20 21-40 41-60 61-80 81-100
Inte r v al s
Fr ecuency
Multiplying Polynomials
9) (x – 5) (3x + 7) 10) (3a - 2)(a2
-5 a + 5)
3x2
+ 7x – 15x – 35 3a3
– 15a2
+ 15a – 2a2
+ 10a - 10
3x2
– 8x – 35 3a3
– 17a2
+ 25a - 10
11) (6x – 5y)2
12) (5x + 6y) (5x – 6y)
36x2
– 60xy + 25y2
25x2
– 36y2
13) Find an expression for the area of a trapezoid with short base length x + 3, long base 2x +
9, and height is 10.
* Remember: Area = (B1 + b2)H = (x + 3 + 2x + 9) 10 = (3x + 12)10 = (3x + 12)5 = 15x + 60
of trapezoid 2 2 2
14) What is the area of the rectangle? ___b___
a) 7x² + 5xz + 2xy +yz
b) 10x2
+ 5xz + 2xy + yz
c) 7x² + 10xz + xy + 2yz
d) 10x² + 5xz + 2xy + 2yz²
(Remember: You need to multiply each side, because you have 2 binomials. (5x + y)(2x + z).)
Solid Geometry
5x
y
2x z
3. 15) Draw and answer the following
a) Pentagonal Prism Faces: 5 + 2 = 7
Edges: 5 * 3 = 15
Vertices: 5 * 2 = 10
b) Nonagon Pyramid Faces: 9 + 1 = 10
Edges: 9 * 2 = 18
Vertices: 9 + 1 = 10
16) Draw the different views for the following structure:
Front Left Side Right Side Top
17. Volume
Figure Formula Procedure
a) V = Ab*H
A = Pa
2
V = Ab*8cm = 30cm2
(8cm) = 240cm3
A = (10*3cm)2cm = 30cm2
2
b) V = Ab*H
A = Pa
2
V = Ab*12cm = 72cm2
(12cm) = 864cm3
A = (6*6cm)4cm = 72cm2
2
c) V = Ab*H
3
A = Pa
2
V = Ab*10cm = 37.5cm2
(10cm)= 125cm3
3 3
A = (5*5cm)3cm = 37.5cm2
2
Each side = 3cm
H = 8cm
a = 2cm
a = 4m
Each side = 6m
H = 12 m
a = 3cm
Each side = 5cm
H = 10cm
4. 18) What is the volume of a cube if the length is 6cm, width is 4cm, and it’s height is 11cm?
19) Find the volume of the following figure.
20) A cereal company decided to increase the height of its boxes by 30% and reduce the
width in order to maintain the same volume. What will be the new height and width if the
length stays the same?
Initially: Length = 20 cm
Width = 30cm
Height = 40cm
Answer:
Actual cereal’s box volume= (20cm)(30cm)(40cm) = 24,000cm3
.
NEW MEASURES =
L= 20cm (stays the same)
H= 52cm (it increased a 30% = 40 +30% = 40 + 12 = 52cm)
W= ?
* They what the SAME volume, so:
24,000cm3
=(20cm)(52cm)W =
24,000cm3
= 1040cm2
* W =
24,000cm3
/ 1040cm2
= W
W = 23.08cm.
So this is the new size of the width in order to keep the same volume of the boxes.
V = l*w*h = 8cm(5cm)(9cm) = 264cm3
V= (4cm*5cm*6cm) – (2cm*2cm*6cm)
= (120cm3
) – (24cm3
) = 96cm3
Remember: You need to “take away” the volume of the
empty space (the hole, which has the same deepness of the
complete figure). That’s why you subtract the volume of the
hole FROM the volume of the COMPLETE figure.
5. 21)Express the perimeter of the figure in simplest form.
Remember: Perimeter is the addition of ALL the sides of the figure. So, you just combine “like
terms”.
P= (2n-1)+(2n-1)+(-3n2
+n+3)+(-6n+8)+(5n2
+3n-4) = 2n2
+2n+5
22) The figure below shows two parallel lines cut by a transversal. In each exercise find the
measures of all eight angles under the given conditions.
a) m ‹ 3 = 2x + 40 and m ‹ 7 = 3x + 27
• ‹ 3 and ‹ 7 are CORRESPONDING angles, so they are EQUAL.
2x+40 = 3x+27
-27 + 40 = 3x – 2x
13 = x
So, if x= 13, the value of the angle is= 2(13) +40 = 26+40 = 66°
If ‹ 3 = 66°, then ‹ 4 = 114° (because they’re supplementary angles). Now you can fill the
rest of the angles.
b) m ‹ 4 = 3x + 40 and m ‹ 5 = 2x
• ‹ 4 and ‹ 5 are CO-INTERIOR or ALLIED angles, so they are SUPPLEMENTARY ANGLES.
3x+40 + 2x = 180°
5x + 40 = 180
5x = 140
X = 28
So, if x= 28, the values of the angles are= ‹ 4 = 3(28) +40 = 84+40 = 124° and ‹ 5= 2(28) = 56°
Now you can fill the rest of the angles.
2n-1
-6n+8
5n2
+3n-4
2n-1
-3n2
+n+3
1
2
4
3
5 8
6 7
6. 21)Express the perimeter of the figure in simplest form.
Remember: Perimeter is the addition of ALL the sides of the figure. So, you just combine “like
terms”.
P= (2n-1)+(2n-1)+(-3n2
+n+3)+(-6n+8)+(5n2
+3n-4) = 2n2
+2n+5
22) The figure below shows two parallel lines cut by a transversal. In each exercise find the
measures of all eight angles under the given conditions.
a) m ‹ 3 = 2x + 40 and m ‹ 7 = 3x + 27
• ‹ 3 and ‹ 7 are CORRESPONDING angles, so they are EQUAL.
2x+40 = 3x+27
-27 + 40 = 3x – 2x
13 = x
So, if x= 13, the value of the angle is= 2(13) +40 = 26+40 = 66°
If ‹ 3 = 66°, then ‹ 4 = 114° (because they’re supplementary angles). Now you can fill the
rest of the angles.
b) m ‹ 4 = 3x + 40 and m ‹ 5 = 2x
• ‹ 4 and ‹ 5 are CO-INTERIOR or ALLIED angles, so they are SUPPLEMENTARY ANGLES.
3x+40 + 2x = 180°
5x + 40 = 180
5x = 140
X = 28
So, if x= 28, the values of the angles are= ‹ 4 = 3(28) +40 = 84+40 = 124° and ‹ 5= 2(28) = 56°
Now you can fill the rest of the angles.
2n-1
-6n+8
5n2
+3n-4
2n-1
-3n2
+n+3
1
2
4
3
5 8
6 7