just add a zero
There are only two possible outcomes in finding out whether a statement is true or false.In testing a conjecture, even one contradiction is sufficient to disprove it. However, it can never be proven. All you can do is add support to the likelihood that the conjecture is true. But there remains a possibility that some other test will prove it false.Furthermore, in view of Godel's incompleteness theorem, some conjectures cannot be proven to be true even if you can prove that their negation is false.
This is a very broad question, so I'll try to cover all bases. Scientists rely on math when they see patterns. For example, if I drop my pen, it always falls in the same manner with the same acceleration. By using math scientists have found a way to give concrete numbers that they can analyze and use to make further conjectures. In and of itself, math does not support scientific principles. Math is just a way for scientists to express what happens. If some event shows a mathematical relationship between two or more variables, then scientists will tend to use whatever equations they find to fit what has happened. Sometimes these equations are completely wrong, or only work for a certain set of circumstances. (For example, Newton proposed some ideas about gravity that work fine on earth but break down in large scale.) Basically, math is just a tool for scientists to show consistancy and patterns.
Math is a subject that the Americans teach in school.This is from Wikipedia.!Mathematics is the study of quantity, structure, space, change, and related topics of pattern and form. Mathematicians seek out patterns whether found in numbers, space, natural science, computers, imaginary abstractions, or elsewhere.[2][3] Mathematicians formulate new conjectures and establish their truth by rigorous deduction from appropriately chosen axioms and definitions.[4]There is debate over whether mathematical objects exist objectively by nature of their logical purity, or whether they are manmade and detached from reality. The mathematician Benjamin Peirce called mathematics "the science that draws necessary conclusions".[5] Albert Einstein, on the other hand, stated that "as far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality."[6]Through the use of abstraction and logical reasoning, mathematics evolved from counting, calculation, measurement, and the systematic study of the shapes and motions of physical objects. Knowledge and use of basic mathematics have always been an inherent and integral part of individual and group life. Refinements of the basic ideas are visible in mathematical texts originating in the ancient Egyptian, Mesopotamian, Indian, Chinese, Greek and Islamic worlds. Rigorous arguments first appeared in Greek mathematics, most notably in Euclid's Elements. The development continued in fitful bursts until the Renaissance period of the 16th century, when mathematical innovations interacted with new scientific discoveries, leading to an acceleration in research that continues to the present day.[7]Today, mathematics is used throughout the world as an essential tool in many fields, including natural science, engineering, medicine, and the social sciences such as economics and psychology. Applied mathematics, the branch of mathematics concerned with application of mathematical knowledge to other fields, inspires and makes use of new mathematical discoveries and sometimes leads to the development of entirely new disciplines. Mathematicians also engage in pure mathematics, or mathematics for its own sake, without having any application in mind, although practical applications for what began as pure mathematics are often discovered later.[8]
"What is the formula in finding the surface area of a regular hexagonal prism with side s units and height h units?"by rcdalivaCHAPTER IINRODUCTIONAccording to Doris Kearns Goodwin, the past is not simply the past, but a prism which the subject filters his own changing self - image. In relation to this quote, the students like us should not forget the past because it was always perpendicular to ones life like a prism. Prism which means a polyhedron with two congruent parallel faces known as the bases, the other faces are called lateral faces are parallelograms and the height of a prism is the perpendicular distance between the planes of the bases (Soledad, Jose-Dilao Ed. D and Julieta G. Bernabe, 2009).There are formulas in finding the surface areas which means the sum of all areas faces of the prism. Perimeter is the outer boundary of a body or figure, or the sum of all the sides. Geometry is a branch of mathematics which investigates the relations, properties, and measurement of solids, surfaces, lines, and angles; the science which treats of the properties and relations of magnitudes; the science of the relations of space. This subject is being taught in the third year students.When one of the researchers was playing footing with his friends, one of the third year students approached and asked him about their assignment on the surface area of hexagonal prism whose side and height were given. In the very start, the researcher thinks deeply and approached some of his classmates to solve the problem. By this instance, we as the fourth year researchers were challenged to find out the solution for the third year assignment.The problem drove the researchers to investigate and that problem was: "What is the formula in finding the surface area of a regular hexagonal prism, with side s units and height h units?"This investigation was challenging and likewise essential. It is important to the academe because the result of this investigation might be the bases of further discoveries pertaining to the formula in finding the surface area of a hexagonal prism. This is also beneficial to the Department of Education because it will give the administrators or the teachers the idea in formulating formulas for other kinds of prisms. And it is so very significant to the students and researchers like us because the conjectures discovered in this study will give them the simple, easy and practical formulas or approaches in solving the problems involving the surface area of prisms.However, this investigation was limited only to the following objectives:1. To answer the question of the third year students;2. To derive the formula of the surface area of hexagonal prism; and3. To enrich the students mathematical skills in discovering the formula.In view of the researchers desire to share their discoveries, their conjectures, they wanted to invite the readers and the other students' researchers to read, comment and react if possible to this investigation.CHAPTER IISTATEMENT OF THE PROBLEMThe main problem of this investigation was: "What is the formula in finding the surface area of a regular hexagonal prism with side s units and height h units?"shSpecifically, the researchers would like to answer the following questions:1. What are the formulas in finding the areas of the regular hexagon and of the rectangle as lateral faces?2. What is the formula in finding the surface area of the regular hexagonal prism?CHAPTER IIIFORMULATING CONJECTURESBased on the thorough investigation of the researchers, the tables and conjectures discovered and formulated were as follows:Table 1. Perimeter of a Regular Hexagon sHEXAGON WITH SIDE (s) in cmPERIMETER (P) in cm162123184245306367428489541060s6sTable 1 showed the perimeter of a regular hexagon. It revealed that the perimeter of the said polygon was 6 times its side. Thus, the conjecture formed was:CONJECTURE 1:The perimeter of a regular hexagon with side s is equal to 6 times the side s. In symbols P = 6s.Table 2. The Apothem of the Base of the Hexagonal PrismHexagonal prism with side(s) in cmMeasure of the apothem (a)in cm1½ √32√333√3242√355√3263√377√3284√399√32105√3s√3 s2sTable 2 showed that the measure of the apothem is one-half the measure of its side times √3.CONJECURE 2The measure of the apothem of the base of the regular hexagonal prism is ½ the side s times √3.In symbols: ½ √3 s or √3 s.2Table 3. The Area of the Bases of Regular Hexagonal PrismSIDE (cm)APOTHEM (cm)PERIMETER (cm)AREA OF THE BASES (cm²)11√3263√32√31212√333√321827√342√32448√355√323075√363√336108√377√3242147√384√348192√399√3254243√3105√360300√3s√3s26s3√3 s²Table 3 revealed that the area of the base of regular hexagonal prism was 3√3 times the square of its side.CONJECTURE 3The total areas of the two bases of the regular hexagonal prism is 3√3 times the square of its side s. In symbols, A=3√3 s².Table 4. The Total Areas of the 6 Rectangular Faces of the Hexagonal PrismSIDE (cm)HEIGHT (cm)TOTAL AREA (cm²)11622243354449655150662167729488384994861010600sh6shBased on table 4, the total areas of the 6 rectangular faces of the regular hexagonal prism with side s units and height h units was 6 times the product of its side s and height h.CONJECTRE 4The total areas of the six rectangular faces of a regular hexagonal prism with side s units and height h units is equal to 6 times the product of its side s and height h. In symbols, A = 6sh.Table 5. The Surface Area of the Regular Hexagonal PrismSIDE (cm)HEIGHT (cm)AREA OF THE BASES (cm²)AREA OF THE 6 FACES (cm²)SURFAE AREA (cm²)113√363√3+62212√32412√3+243327√35427√3+544448√39648√3+965575√315075√3+15066108√3216108√3+21677147√3294147√3+29488192√3384192√3+38499243√3486243√3+4861010300√3600300√3+600sh3√3 s²6sh3√3s²+6shTable 5 showed the surface area of the regular hexagonal prism and based from the data, the surface area of a regular hexagonal prism with side s units and height h units was the sum of the areas of the bases and the areas of the 6 faces.CONJECTURE 5The surface area of the regular hexagonal prism with side s units and height h units is equal to the sum of the areas of the two bases and the 6 faces. In symbols, SA =3√3s²+6sh.CHAPTER IVTESTING AND VERIFYING CONJECTURESA. Testing of ConjecturesCONJECTURE 1:The perimeter of a regular hexagon with side s is equal to 6 times the side s. In symbols P = 6s.To test the conjecture 1, the investigators applied the said conjecture in finding the perimeter of the base of the following regular hexagonal prisms and regular hexagons. 5.5 cm1. 10 cm 2. 3.11 cm4. 5.12 cm20mSolutions:1. P = 6s 2. P = 6s 3. P = 6s 4. P = 6s= 6 (10cm) = 6 (5.5 cm) = 6 (11 cm) = 6 (12 cm)= 60 cm = 33 cm = 66 cm = 72 cm5. P = 6s= 6 (20 cm)= 120 cmCONJECURE 2The measure of the apothem of the base of the regular hexagonal prism is ½ the side s times √3.In symbols: ½ √3 s or √3 s.2The investigators applied this conjecture to the problem below to test its accuracy and practicality.Problem: Find the apothem of the base of each of the regular hexagonal prism in the figures under the conjecture 1.Solutions:1. a = √3 s 2. a = √3 s 3. a = √3 s 4. a = √3 s2 2 2 2= √3 (10 cm) = √3 (5.5 cm) = √3 (11 cm) = √3 (12 cm)2 2 2 2= √3 (5 cm) = √3 (2.75 cm) = √3 (5.5 cm) = √3 (6 cm)= 5√3 cm = 2.75 √3 cm = 5.5√3 cm = 6√3 cm5. a = √3 s2= √3 (12 cm)2= √3 (6 cm)= 6√3 cmCONJECTURE 3The total areas of the two bases of the regular hexagonal prism is 3√3 times the square of its side s. In symbols, A=3√3 s².To test this conjecture, the investigators applied its efficiency in the problem, "Find the total area of the bases of each regular hexagonal prism in figures 1, 2 and 3 under the testing of conjecture 1".Solutions:A= 3√3 s² 2. A= 3√3 s² 3.A= 3√3 s²= 3√3 (10cm) ² = 3√3 (5.5 cm)² = 3√3 (11cm)²= 3√3 (100cm²) = 3√3 (30.25) cm² = 3√3 (121 cm²)= 300 √3 cm² = 90.75 √3 cm² = 363 √3 cm²CONJECTRE 4The total areas of the six rectangular faces of a regular hexagonal prism with side s units and height h units is equal to 6 times the product of its side s and height h. In symbols, A = 6sh.This conjecture can be applied in finding the total areas of the faces of regular hexagonal prism like the problems below.a. Find the total areas of the faces of a regular hexagonal prism whose figure is8 cmSolution: A= 6sh= 6 (8cm) (20cm) 20 cm= 960 cm2b. What is the total areas of the bases of the regular hexagonal prism whose side is 15 cm and height 15cm.Solution: A = 6 sh= 6 (15 cm) (15 cm)= 1,350 cm 2CONJECTURE 5The surface area of the regular hexagonal prism with side s units and height h units is equal to the sum of the areas of the two bases and the 6 faces. In symbols, SA =3√3s²+6sh.The investigators tested this conjecture by solving the following problems:How much material will be needed to make a regular hexagonal prism whose side equals 25cm and height 50cm?Solution:SA= 3√3 s² + 6sh= 3√3 (25cm) ² + 6 (25cm) (25cm)= 3√3 (625cm²) + 3750 cm²SA = 1,875 √3 + 3750 cm2Find the surface area of the solid at the right.28 cmSolution:SA= 3√3 s²+ 6sh 18 cm= 3√3 (18cm) 2 + 6 (18cm)(28cm)= 3√3 (324cm²) + 3024 cm2SA = 972 √3 cm² + 3024 cm²B. Verifying ConjecturesCONJECTURE 1:The perimeter of a regular hexagon with side s is equal to 6 times the side s. In symbols P = 6s.F EA DB CsProof 1.If ABCDEF is a regular hexagon with BC=s, then AB+BC+CD+DE+EF+FA= 6SStatementsReasons1. ABCDEF is a regular hexagon with BC=s.2. AB=BC=CD=DE=FA3.AB=sCD=sDE=sFA=sEF=s4.AB+BC+CD+DE+EF+FA=s+s+s+s+s+s5.AB+BC+CD+DE+EF+FA=6S1. Given2. Definition of regular hexagon3.Transitive Property4.APE5. Combining like terms.Proof 2.Sides(s)12345678910Perimeter f(s)61218243036424854606 6 6 6 6 6 6 6 6Since the first differences were equal, therefore the table showed linear function f(x) = mx+b. To derive the function, (1, 6) and (2, 12) will be used.Solve for m:m= y2-y1 Slope formulax2-x1= 12-6 Substituting y2= 12, y1=6, x2=2 and x1=1.2-1= 6 Mathematical fact1m = 6 Mathematical factSolve for b:f(x)=mx+b Slope-Intercept formula6=6(1) + b Substituting y=6, x=1, and m=6.6=6+b Identity0=b APEb=0 SymmetricThus, f(x) = 6x or f(s) = 6s or P = 6s.CONJECURE 2The measure of the apothem of the base of the regular hexagonal prism is ½ the side s times √3.In symbols: ½ √3 s or √3 s. E D2Proof I.Given: ABCDEF is a regular hexagon F CAB=saProve: a= √3 s2A G BsStatementsReasons1. ABCDEF is a regular hexagon.AB= s1. Given2.AG= ½ s2. The side opposite to 30˚ is one half the hypotenuse.3. a=(½ s)(√3)3. The side opposite to 60˚ is equal to the side opposite to 30˚ times √3.4. a= √3 s24. ClosureProof 2.Side (s)12345678910Apothem (a)F(s)√32√33√322√35√323√37√324√39√325√3√3 √3 √3 √3 √3 √3 √3 √3 √32 2 2 2 2 2 2 2 2Since the first differences were equal, therefore the table showed a linear function in the form f(x) = mx+b.Solving for m using (1, √3) and (2, √3).2m = y2-y1 Slope formulax2-x1m = √3 - √3 Substitution22-1m= √3 Mathematical fact/ Closure21m= √32Solving for b: Use (1, √3)2f(x) = mx + b Slope - intercept form√3 = (√3) (1) +b Substitution2 2√3 = √3+ b Identity2 20=b APEb=0 SymmetricThus, f(x) = √3 or f(s) = √3s or a = √3s2 2 2CONJECTURE 3The total areas of the two bases of the regular hexagonal prism is 3√3 timesB Cthe square of its side s. In symbols, A=3√3 s².Proof 1 A DGiven: ABCDEF is a regular hexagonal prism.FE = s unitsProve: AABCDEF = 3(√3)s² F s E22AABCDEF = 3√3s²StatementsReasons1. ABCDEF is a regular hexagonFE =sGiven2.a= 3√3sThe side opposite to 60 is the one half of the hypotenuse time's √3.3.A = ½bhThe area of a triangle is ½ product of its side and height4.A =(½)s(√3/2s)Substituting the b=s and h=a=√32s.5.A = (√3/4)s²Mathematical fact6.AABCDEF= 6AIn a regular hexagon, there are six congruent triangles formed7.AABCDEF= 6(√3/4s²)Substitution8.AABCDEF= 3 (√3/2) s²Mathematical fact9.2AABCDEF= 2[3 (√3/2)]s²MPE10.2AABCDEF= 3 √3 s²Multiplicative inverse / identityProof 2Based on the table, the data were as follows:Side (s)12345678910Area of the bases f(s)3 √312√327√348√375√3108√3147√3192√3243√3300√39√3 15√3 21√3 27√3 33√3 39√3 45√3 51√3 57√3First difference6√3 6√3 6√3 6√3 6√3 6√3 6√3 6√3Second differenceSince the second differences were equal, the function that the investigators could derive will be a quadratic function f(x) = ax²+bx+c.Equations were:Eq. 1 f(x) = ax²+bx+c for (1, √3)6√3 = a (1)²+ b(1)+c Substitution6√3 = a+b+c Mathematical fact / identitya+b+c=6√3 SymmetricEq. 2 f(x) = ax²+bx+c for (2, 12√3)24√3=a (2)²+b(2)+c Substitution24√3=4a+2b+c Mathematical fact4a+2b+c=24√3 SymmetricEq. 3 f(x) = ax²+bx+c for (3, 27√3)54√3=a (3)²+b(3)+c Substitution54√3=9a+3b+c Mathematical fact9a+3b+c=54√3 SymmetricTo find the values of a, b, and c, elimination method was utilized.Eliminating cEq. 2 4a+2b+c=24√3 Eq. 3 9a+3b+c=54√3- Eq. 1 a+b+c=6√3 - Eq. 2 4a+2b+c=24√3Eq. 4 3a+b = 18√3 Eq. 5 5a+b = 30√3Eliminating b and solving aEq. 5 5a+b = 30√3- Eq. 4 3a+b = 18√32a = 12√3a = 6√3 MPESolving for b if a = 6√3Eq. 5 5a+b = 30√35(6√3) +b= 30√3 Substitution30√3+b=30√3 Closureb = 0 APESolving for c if a = 6√3 and b=0Eq. 1 a + b + c=6√36√3 + 0+c =6√3 Substitution6√3 + c = 6√3 Identityc = 0 APETherefore, f(x) = 6√3x² or f(s) = 6√3s² or A= 6√3s²CONJECTURE 4The total areas of the six rectangular faces of a regular hexagonal prism with side s units and height h units is equal to 6 times the product of its side s and height h. In symbols, A = 6sh.AProofGiven: ABCD is a rectangle. BAB = s and BC = hProve:AABCD = sh D6AABCD= 6shCStatementsReasons1. ABCD is a rectangle AB=s and BC=hGiven2.AABCD=lwThe area of a rectangle is the product of its length and width3. AABCD = shSubstitution4. 6AABCD = 6shMPECONJECTURE 5The surface area of the regular hexagonal prism with side s units and height h units is equal to the sum of the areas of the two bases and the 6 faces. In symbols, SA =3√3s² + 6sh.ProofGiven: The figure at the rightProve: SA=3 √3s²+6shs hStatementsReasons1. AHEXAGON= ½ aPThe area of a regular polygon is one -half the product of its apothem and its perimeter2. a = √3/2sThe side opposite to 60˚ is a 30˚-60˚-90˚ triangle is one-half the hypotenuse times √3.3. P = 6sThe perimeter of a regular polygon is the sum of all sides.4. AHEXAGON = ½ (√3s)(6s)2Substitution5. A HEXAGON = 3 √3s²2Mathematical fact6. 2A HEXAGON= 3 √3s²MPE7. A RECTANGULAR FACES = shThe area of a rectangle is equal to length (h) times the width (s).8. 6ARECTANGULAR FACES = 6shMPE9. SA = 2A HEXAGON + 6A RECTANGULAR FACESDefinition of surface area10. SA = 3 √3s² + 6shSubstitutionCHAPTER VSUMMARY/CONCLUSIONSAfter the investigation, the question of the third year student on "What is the surface area of the regular hexagonal prism whose side and height were given" was cleared and answered. Indeed, God is so good because of the benefits that the investigators gained like the discovery of various formulas and conjectures based on the patterns observed in the data gathered and most of all, the friendship that rooted between the hearts of the investigators and the third year students could not be bought by any gold.The main problem of this investigation was: "What is the formula in finding the surface area of a regular hexagonal prism with side s units and height h units?"shSpecifically, the researchers would like to answer the following questions:1. What are the formulas in finding the areas of the regular hexagon and of the rectangle as lateral faces?2. What is the formula in finding the surface area of the regular hexagonal prism?Based on the results, the investigators found out the following conjectures:CONJECTURE 1:The perimeter of a regular hexagon with side s is equal to 6 times the side s. In symbols P = 6s.CONJECURE 2The measure of the apothem of the base of the regular hexagonal prism is ½ the side s times √3.In symbols: ½ √3 s or √3 s.2CONJECTURE 3The total areas of the two bases of the regular hexagonal prism is 3√3 times the square of its side s. In symbols, A=3√3 s².CONJECTRE 4The total areas of the six rectangular faces of a regular hexagonal prism with side s units and height h units is equal to 6 times the product of its side s and height h. In symbols, A = 6sh.CONJECTURE 5The surface area of the regular hexagonal prism with side s units and height h units is equal to the sum of the areas of the two bases and the 6 faces. In symbols, SA =3√3s²+6sh.These conjectures were proven based on the gathered data on different sources like books, practical applications, and internet. The formulas also followed the rules in finding the surface area of a prism.CHAPTER VIPOSSIBLE EXTENSIONSThe investigators would like to elicit answers of the readers by applying the conjectures discovered and formulated through this study.A. Find the surface area of the following regular hexagonal prism.1. 8 cm 2. 7 cm 3.9.8 cm12 cm10 cm 50 cm4. .a = 8 √322 cmB. Derive a formula in finding the surface area of:1. regular hexagonal prism whose side equals x cm and height equals y cm.2. regular hexagonal prism whose side equals (x-1) cm and height equals (x2+4x+4) cm.C. Derive the formula for the surface area of a regular octagonal prism. (Hint: Use Trigonometric Functions and Pythagorean Theorem)
No. Conjectures are "good" guesses.
In geometry, deductive rules can be used to prove conjectures.
prove conjectures
14
Some words that rhyme with "lectures" are textures, conjectures, and ruptures.
surmises
inductive
hypotheses or more generally conjectures should be capable of being refuted see: Karl Popper - Conjectures and Refutations
Twenty Conjectures in Geometry:Vertical Angle Conjecture: Non-adjacent angles formed by two intersecting lines.Linear Pair Conjecture: Adjacent angles formed by two intersecting lines.Triangle Sum Conjecture: Sum of the measures of the three angles in a triangle.Quadrilateral Sum Conjecture: Sum of the four angles in a convex four-sided figure.Polygon Sum Conjecture: Sum of the angles for any convex polygon.Exterior Angles Conjecture: Sum of exterior angles for any convex polygon.Isosceles Triangle Conjectures: Isosceles triangles have equal base angles.Isosceles Trapezoid Conjecture: Isosceles trapezoids have equal base angles.Midsegment Conjectures: Lengths of midsegments for triangles and trapezoids.Parallel Lines Conjectures: Corresponding, alternate interior, and alternate exterior angles.Parallelogram Conjectures: Side, angle, and diagonal relationships.Rhombus Conjectures: Side, angle, and diagonal relationships.Rectangle Conjectures: Side, angle, and diagonal relationships.Congruent Chord Conjectures: Congruent chords intercept congruent arcs.Chord Bisector Conjecture: The bisector of a chord passes through the center of the circle.Tangents to Circles Conjectures: A tangent to a circle is perpendicular to the radius.Inscribed Angle Conjectures: An inscribed angles has half the measure of intercepted arc.Inscribed Quadrilateral Conjecture: Opposite angles are supplements.The Number "Pi" Conjectures: Circumference and diameter relationship for a circle.Arc Length Conjecture: Formula to calculate the length of an arc on a circle.
false
By creating a strong inference, you can then put your ideas to the test. After close observation, you can then rule-out any incorrect guesses.
Usually not. If you do use conjectures, you should make it quite clear that the proof stands and falls with the truth of the conjecture. That is, if the conjecture happens to be false, then the proof of your statement turns out to be invalid.