SlideShare une entreprise Scribd logo
1  sur  36
Developing Expert Voices
     Pre Calculus 40S enriched
               2007
   Hi, my name is Sandy and I like purple(:


                   DEV PROJECT                1
Problem One
You’re a Skydiver and you’ve just jumped out from the jet plane and
   you’re heading for the ground. You pull the string to the parachute
   and start to examine a crop circle that a farmer has created.
Given that the centre of the crop circle is labeled O,
the diameter of the circle is 10km, angle BOG is
120° and that D is …

Answer the following in radians.
a) Determine the length of the arc that subtends
   an angle of 240°.
b) Determine the area of that sector using the information given above.
c) Determine the angle at the centre of the circle if the arc subtended by
    the angle is .


                                 DEV PROJECT                             2
Problem One – answers (a)
From the original question, the answers are supposed to be in radians.
The formula to change Degrees into Radians is:

Where D is Degrees and R is Radians.
Plug in the given information into the formula. It should then look like this:


Now you cross multiply:
                               240°(π) = R(180°)

Divide everything by 180° to leave R by itself and solve!
The degree signs cancel each other out, and     also reduces to     .
Therefore, 240° in RADIANS is .

                                    DEV PROJECT                                  3
Problem One – answers (a)
We have the angle in radians, so we can use that to solve for the
 arc length with this formula:



Where R is the angle in Radians, r is the radius and L is the Arc
  Length.
Now we can substitute the given information and what we found
  into the formula.                           The 5 in 2π(r) was found by
                                                 dividing the diameter by 2.
                                                 *The radius is HALF of the
                                                 diameter in a circle.


                                DEV PROJECT                                4
Problem One – answers (a)
To get rid of the fraction in a fraction you multiply the numerator
  by the reciprocal of the denominator.

Multiply the left side and then cross multiply.
                               (4π)(10π) = L(6π)
Divide everything by (6π) to leave L by itself. Therefore…

The π’s reduce, leaving you with only 1. Also 4 x 10 = 40 all
  divided by 6. Reducing that, it leaves you with



                              DEV PROJECT                             5
Problem One – answers (a)
A second way is to change an angle from degree to radians is:



Since we’ve been through this once already, it’s should be pretty
   straight forward on how to acquire the correct answer. You
   should get the same answer you got before which was:



So if you didn’t, you know you did something wrong. But don’t
  quit there, go back and try again. ☺


                             DEV PROJECT                            6
Problem One – answers (b)
The answer again must be in radians so this is the formula you use to find the
   area of a sector:



Where Θ is the angle in radians, S is the area of the sector and r is the radius of
   the circle.
Plug in the information that we already know.




Again, we have to multiply the numerator by the reciprocal of the
  denominator.


                                     DEV PROJECT                                  7
Problem One – answers (b)
Next, we simplify it and then it’s time to cross multiply.




Afterwards, we divide both sides by (6π) so that S is by itself.




We now simplify. The π’s reduce, just like before, leaving you with only one
  again, and 100 ÷ 6 reduces to




                                    DEV PROJECT                                8
Problem One – answers (b)
Another way to solve this part of the question is to use a different
  formula:

Plugging in all the information we come up with this:



And from this point, it is also pretty straight forward on how to
  acquire the answer. Also, you should arrive at the same
  answer, if not… try again!



                              DEV PROJECT                           9
Problem One – answers (c)
This is very similar to the first part of this question. However,
  instead of solving for the ARC, we’re solving for the ANGLE.



Where R is the angle in radians, L the Arc Length and r is the
  radius.
Now we can substitute the information we know into the formula.




                             DEV PROJECT                        10
Problem One – answers (c)
Knowing the routine, we multiply the numerator by the reciprocal
  of the denominator again. We simplify, then cross multiply.




                            DEV PROJECT                       11
Problem One – answers (c)
Once again, divide both sides by 9π so that R will be by itself.



Now simplify. Remembering that the π’s reduce you’re left
  with…



Therefore the angle subtended by the arc     is .




                              DEV PROJECT                          12
Problem Two
Given the graph of f(x) below, sketch the following three graphs.
a)                          b)                c)


                                                      Please note that
                                                      for tests or
                                                      examinations,
                                                      add arrows and
                                                      label your axis.




                             DEV PROJECT                          13
Problem Two – answers (a)
The first thing you should remember before starting is:


  STRETCHES BEFORE TRANSLATIONS

The basic formula for a question like this is: Af(B(x-C))+D
  Where A is a STRETCH (the y-coordinates are multiplied by
  A), B is a STRETCH (the x-coordinates are multiplied by /
  the reciprocal), C is a TRANSLATION (the x-
     coordinate is moved horizontally) and D is a
  TRANSLATION (the y-coordinate is moved vertically).

                             DEV PROJECT                      14
Problem Two – answers (a)
Looking at the graph, we can figure out all of the coordinates of
  each point.




We’ll arrive at the numbers in this order remembering that the x-
 coordinate always comes before the y-coordinate.

                              DEV PROJECT                           15
Problem Two – answers (a)
We can take those coordinates and apply what we’re given.
                Af(B(x-C))+D
Remember that A stretches the y-coordinate and B stretches the
   x-coordinate.
First point: A(-6,-3).
                        A((-6)(3)),(-3)(-2))
The 3 came from the reciprocal in the original question given
   above.
You’re then left with a new coordinate: A(-18,6).
BUT you’re not finished yet.

                            DEV PROJECT                          16
Problem Two – answers (a)
Now you find the Translation part.
         Af(B(x-C))+D
Knowing that C shifts the x-coordinate and D shifts the y-
   coordinate.
C is (-4) (even though it says x+4 in the question) because in the
   formula, it’s (x-C). So in order for C to be positive in the
   question, 4 must be a negative number.
                          (x+4) (x-(-4))
One negative, multiplied by another negative, always gives you a
   positive number.
Therefore: A((-18-4),(6+5))
And the new coordinate IS: A(-22,11)
                             DEV PROJECT                         17
Problem Two – answers (a)
Now there are 3 more points to do. Just for practice, try them
  yourself first and once your finished.. Go to the next slide and
  check if your answers are correct. ☺

                    The remaining points are:
                           B(-3,-3)
                            C(1,1)
                            D(4,1)




                              DEV PROJECT                        18
Problem Two – answers (a)
Since we’ve done this once before, I’m going to go through it a
   bit faster.
B(-3,-3)
Stretches first! B((-3)(3),(-3)(-2)) B(-9,6)
Now Translations! B(((-9)-4),(6+5)) (-13,11)
C(1,1)
Stretches first! C((1)(3),(1)(-2)) C(3,-2)
Now Translations! C((3-4),((-2)+5)) C(-1,3)
D(4,1)
Stretches first! D((4)(3),(1)(-2)) D(12,-2)
Now Translations! D(((12)-4),(-2+5)) D(8,3)

                             DEV PROJECT                          19
Problem Two – answers (a)
Now that we’ve calculated all the new coordinates, it’s time to plot those
  points onto a graph and connect the dots. After you’re finished, it should
  look like this:
*New Graph: Is red.




                                   DEV PROJECT                                 20
Problem Two – answers (b)
We’re looking for the INVERSE of the function.
Since we already have the coordinates, all we have to do is switch
   the y-coordinate and the x-coordinate.
          A(-6,-3) Now, SWITCH THEM. A(-3,-6)
Easy right? Okay. So lets do the others now.
B(-3,-3) B(-3,-3)
C(1,1) C(1,1)
D(4,1) D(1,4)




                             DEV PROJECT                        21
Problem Two – answers (b)
Now just like the other one, plot the points, connect the dots and
  you have your solution!




                              DEV PROJECT                            22
Problem Two – answers (c)
It’s asking for the reciprocal of the original function.
1. Look for the “invariant points.” ((-1,-1) and (1,1))
     - These points are on both the original function and the reciprocal function because
     the reciprocal of 1 is 1.
2. Examine the straight horizontal lines. One of them is y=(-3) and the other is y=(1).
     On the graph, you find the reciprocal and graph it Therefore y=(1/3) and y=(1).




                                        DEV PROJECT                                     23
Problem Two – answers (c)
Now use the “Biggering, smallering” game. If a number
  increases, its reciprocal decreases and vice versa. Once it’s
  finished, it should look like this:




                                            Don’t forget to add an
                                            asymptote where the
                                            graph has zero(s) or
                                            where x = 0.

                              DEV PROJECT                            24
Problem Three

   Prove:



     DEV PROJECT   25
Problem Three - answers
First thing to always do is to draw the “Great Wall of China.”




                                           Expand the problem.
                                           - We know that COT is the reciprocal of tan
                                           and instead of tan, you can use sin and cos.
                                           We also changed 1 to sinΘ/sinΘ because
                                           that equals one. So we don’t necessarily add
                                           anything to the original question.




                             DEV PROJECT                                           26
Problem Three - answers
                   Here we simply just multiplied everything
                   out.



                   Just like we were doing before, to get rid of the
                   double fraction we multiplied the numerator by
                   the reciprocal of the denominator.




                  The sinΘ’s reduce!




          DEV PROJECT                                           27
Problem Three - answers
                   You then end up with this after you reduce!



                   You’re probably wondering how the heck does
                   cos2Θ equal 1-sin2Θ? Well, knowing our
                   identities: sin2Θ + cos2Θ = 1
                   If we rearrange the order, we can see that
                   cos2Θ = 1-sin2Θ


                  If you notice, 1-sin2Θ is a difference of squares.
                  SO, it is easily factorable.




          DEV PROJECT                                           28
Problem Three - answers
                   The 1-sinΘ’s reduce.




                        And you’re left with 1+sinΘ!



                                          *When You’re finished
                                          solving, you must
                                          always put Q.E.D to
                                          indicate that you are
      Q.E.D                               finished.


          DEV PROJECT                                      29
Problem Four
At this very moment, there are 1135 students that attend Daniel
   McIntyre High School. 2 years ago there was only 960
   students that attended the High School.
a) At what rate is the student body population increasing at?
b) Assuming that the rate continues to increase at this rate, how
   much longer will it take for the student body population to be
   1500




                             DEV PROJECT                            30
Problem Four – answers(a)
We are looking for the rate, or the model, that the student body
  population is increasing. Here’s the formula that we use to
  solve this question:
                          P = P0(Model)t
Where P is the population at the end of the time period, P0 is the
  population at the beginning of the time period, Model is the, in
  this case, rate at which the population increases and t is the
  change in time.
So using the information we know, we can use it to plug it into
  the formula.
                       1135 = 960(Model)2

                             DEV PROJECT                         31
Problem Four – answers(a)
                      1135 = 960(Model)2
Divide both sides by 960 to leave (Model)2 by itself and reduce:

You can use “ln” to solve or you there’s an alternate way. First
  way is to use ln. Take the ln of both sides:

Multiply by ½ on both sides to remove the 2 from the right.
   Calculate the left side out.
0.0837 = ln(Model)
Therefore e0.0837 = Model. And the rate is (0.0837)(100) = 8.37%

                              DEV PROJECT                          32
Problem Four – answers(a)
                       1135 = 960(Model)2
The other way is to do it like this:



                                    These reduce.


And the result is 1.0873 = Model.
To get the actual model you minus one, because when the
  population increases, it keeps the original amount (1) and from
  there increases (+ 0.0873%).

                             DEV PROJECT                        33
Problem Four – answers(b)
We’re asked to find the time it takes for the population to
 increase from 1135 to 1500.
                      1500 = 1135(1.0873)t

Divide both sides by 1135 and reduce:




                              DEV PROJECT                     34
Problem Four – answers(b)
Now you can use ln to solve for t:

Divide both sides by ln(1.0873).




Then solve for t! and t is 3.331428783 3.3314 years.
So it will take 3.3314 years for the student body population to
  grow to 1500.


                              DEV PROJECT                         35
Toodle-
Toodle-Loo Kangaroo!
         =)


        DEV PROJECT    36

Contenu connexe

Tendances

Assignments for class XII
Assignments for class XIIAssignments for class XII
Assignments for class XIIindu thakur
 
Circle & solid geometry f3
Circle & solid geometry f3Circle & solid geometry f3
Circle & solid geometry f3normalamahadi
 
Função quadrática resumo teórico e exercícios - celso brasil
Função quadrática   resumo teórico e exercícios - celso brasilFunção quadrática   resumo teórico e exercícios - celso brasil
Função quadrática resumo teórico e exercícios - celso brasilCelso do Rozário Brasil Gonçalves
 
Chapter 12 vectors and the geometry of space merged
Chapter 12 vectors and the geometry of space mergedChapter 12 vectors and the geometry of space merged
Chapter 12 vectors and the geometry of space mergedEasyStudy3
 
F4 10 Angles Of Elevation Dep
F4 10 Angles Of Elevation   DepF4 10 Angles Of Elevation   Dep
F4 10 Angles Of Elevation Depguestcc333c
 
F4 05 The Straight Line
F4 05 The Straight LineF4 05 The Straight Line
F4 05 The Straight Lineguestcc333c
 
Pptpersamaankuadrat 150205080445-conversion-gate02
Pptpersamaankuadrat 150205080445-conversion-gate02Pptpersamaankuadrat 150205080445-conversion-gate02
Pptpersamaankuadrat 150205080445-conversion-gate02MasfuahFuah
 
267 1 3 d coordinate system-n
267 1 3 d coordinate system-n267 1 3 d coordinate system-n
267 1 3 d coordinate system-nmath260
 
Lesson 25: Evaluating Definite Integrals (slides)
Lesson 25: Evaluating Definite Integrals (slides)Lesson 25: Evaluating Definite Integrals (slides)
Lesson 25: Evaluating Definite Integrals (slides)Matthew Leingang
 
Inverse trigonometric functions xii[1]
Inverse trigonometric functions xii[1]Inverse trigonometric functions xii[1]
Inverse trigonometric functions xii[1]indu thakur
 

Tendances (17)

Assignments for class XII
Assignments for class XIIAssignments for class XII
Assignments for class XII
 
Week 6
Week 6Week 6
Week 6
 
Circle & solid geometry f3
Circle & solid geometry f3Circle & solid geometry f3
Circle & solid geometry f3
 
Função quadrática resumo teórico e exercícios - celso brasil
Função quadrática   resumo teórico e exercícios - celso brasilFunção quadrática   resumo teórico e exercícios - celso brasil
Função quadrática resumo teórico e exercícios - celso brasil
 
Chapter 12 vectors and the geometry of space merged
Chapter 12 vectors and the geometry of space mergedChapter 12 vectors and the geometry of space merged
Chapter 12 vectors and the geometry of space merged
 
F4 10 Angles Of Elevation Dep
F4 10 Angles Of Elevation   DepF4 10 Angles Of Elevation   Dep
F4 10 Angles Of Elevation Dep
 
F4 05 The Straight Line
F4 05 The Straight LineF4 05 The Straight Line
F4 05 The Straight Line
 
Notes up to_ch7_sec3
Notes up to_ch7_sec3Notes up to_ch7_sec3
Notes up to_ch7_sec3
 
F4 03 Sets
F4 03 SetsF4 03 Sets
F4 03 Sets
 
C4 January 2012 QP
C4 January 2012 QPC4 January 2012 QP
C4 January 2012 QP
 
FUNÇÃO EXPONENCIAL E LOGARÍTMICA
FUNÇÃO EXPONENCIAL E LOGARÍTMICAFUNÇÃO EXPONENCIAL E LOGARÍTMICA
FUNÇÃO EXPONENCIAL E LOGARÍTMICA
 
Pptpersamaankuadrat 150205080445-conversion-gate02
Pptpersamaankuadrat 150205080445-conversion-gate02Pptpersamaankuadrat 150205080445-conversion-gate02
Pptpersamaankuadrat 150205080445-conversion-gate02
 
267 1 3 d coordinate system-n
267 1 3 d coordinate system-n267 1 3 d coordinate system-n
267 1 3 d coordinate system-n
 
Lesson 25: Evaluating Definite Integrals (slides)
Lesson 25: Evaluating Definite Integrals (slides)Lesson 25: Evaluating Definite Integrals (slides)
Lesson 25: Evaluating Definite Integrals (slides)
 
Função afim resumo teórico e exercícios - celso brasil
Função afim   resumo teórico e exercícios - celso brasilFunção afim   resumo teórico e exercícios - celso brasil
Função afim resumo teórico e exercícios - celso brasil
 
Inverse trigonometric functions xii[1]
Inverse trigonometric functions xii[1]Inverse trigonometric functions xii[1]
Inverse trigonometric functions xii[1]
 
Probability theory
Probability theoryProbability theory
Probability theory
 

En vedette

قرارداد صلح امام حسن(علیه السلام) با معاویه و شرطهاى اساسى آن
قرارداد صلح امام حسن(علیه السلام) با معاویه و شرطهاى اساسى آنقرارداد صلح امام حسن(علیه السلام) با معاویه و شرطهاى اساسى آن
قرارداد صلح امام حسن(علیه السلام) با معاویه و شرطهاى اساسى آنDr Fereidoun Dejahang
 
اشاره قرآن به جنسیت ملکه زنبور عسل هزار و صد سال قبل از کشف این حقیفت
اشاره قرآن به جنسیت ملکه زنبور عسل هزار و صد سال قبل از کشف این حقیفتاشاره قرآن به جنسیت ملکه زنبور عسل هزار و صد سال قبل از کشف این حقیفت
اشاره قرآن به جنسیت ملکه زنبور عسل هزار و صد سال قبل از کشف این حقیفتDr Fereidoun Dejahang
 
003 balanced scorecord & value chain
003 balanced scorecord & value chain003 balanced scorecord & value chain
003 balanced scorecord & value chainDr Fereidoun Dejahang
 
039 the mathematical composition part 2
039 the mathematical composition part 2039 the mathematical composition part 2
039 the mathematical composition part 2Dr Fereidoun Dejahang
 
عجز معاویه از پاسخ به پادشاه روم
عجز معاویه از پاسخ به پادشاه رومعجز معاویه از پاسخ به پادشاه روم
عجز معاویه از پاسخ به پادشاه رومDr Fereidoun Dejahang
 
Proses konflik
Proses konflikProses konflik
Proses konflikdhipan
 
Change of Address Form PAMB
Change of Address Form PAMBChange of Address Form PAMB
Change of Address Form PAMBDiyana Arus
 
English Essay Comparison Action Comedy movies
English Essay Comparison Action Comedy moviesEnglish Essay Comparison Action Comedy movies
English Essay Comparison Action Comedy moviesSolomonTangerine
 
Unidad 4 Mercadotecnia Electronica
Unidad 4 Mercadotecnia Electronica Unidad 4 Mercadotecnia Electronica
Unidad 4 Mercadotecnia Electronica Janahí Villanueva
 

En vedette (17)

055 the relativity of time
055 the relativity of time055 the relativity of time
055 the relativity of time
 
قرارداد صلح امام حسن(علیه السلام) با معاویه و شرطهاى اساسى آن
قرارداد صلح امام حسن(علیه السلام) با معاویه و شرطهاى اساسى آنقرارداد صلح امام حسن(علیه السلام) با معاویه و شرطهاى اساسى آن
قرارداد صلح امام حسن(علیه السلام) با معاویه و شرطهاى اساسى آن
 
اشاره قرآن به جنسیت ملکه زنبور عسل هزار و صد سال قبل از کشف این حقیفت
اشاره قرآن به جنسیت ملکه زنبور عسل هزار و صد سال قبل از کشف این حقیفتاشاره قرآن به جنسیت ملکه زنبور عسل هزار و صد سال قبل از کشف این حقیفت
اشاره قرآن به جنسیت ملکه زنبور عسل هزار و صد سال قبل از کشف این حقیفت
 
003 balanced scorecord & value chain
003 balanced scorecord & value chain003 balanced scorecord & value chain
003 balanced scorecord & value chain
 
039 the mathematical composition part 2
039 the mathematical composition part 2039 the mathematical composition part 2
039 the mathematical composition part 2
 
DEV
DEVDEV
DEV
 
018 the earth creation in six days
018 the earth creation in six days018 the earth creation in six days
018 the earth creation in six days
 
سنگ حضرت موسي(ع)
سنگ حضرت موسي(ع)سنگ حضرت موسي(ع)
سنگ حضرت موسي(ع)
 
041 ozone layer , venus and mars,..
041 ozone layer , venus and mars,..041 ozone layer , venus and mars,..
041 ozone layer , venus and mars,..
 
047 the movement of mountains
047 the movement of mountains047 the movement of mountains
047 the movement of mountains
 
قرارگاه خورشيد
قرارگاه خورشيدقرارگاه خورشيد
قرارگاه خورشيد
 
عجز معاویه از پاسخ به پادشاه روم
عجز معاویه از پاسخ به پادشاه رومعجز معاویه از پاسخ به پادشاه روم
عجز معاویه از پاسخ به پادشاه روم
 
Proses konflik
Proses konflikProses konflik
Proses konflik
 
007 benefits management-use of it
007 benefits management-use of it007 benefits management-use of it
007 benefits management-use of it
 
Change of Address Form PAMB
Change of Address Form PAMBChange of Address Form PAMB
Change of Address Form PAMB
 
English Essay Comparison Action Comedy movies
English Essay Comparison Action Comedy moviesEnglish Essay Comparison Action Comedy movies
English Essay Comparison Action Comedy movies
 
Unidad 4 Mercadotecnia Electronica
Unidad 4 Mercadotecnia Electronica Unidad 4 Mercadotecnia Electronica
Unidad 4 Mercadotecnia Electronica
 

Similaire à Developing Expert Voices

Notes Day 6
Notes   Day 6Notes   Day 6
Notes Day 6Megan
 
2012 Mathacre JV Written Test
2012 Mathacre JV Written Test2012 Mathacre JV Written Test
2012 Mathacre JV Written Testkayleigh_lane
 
Ppt kbsm t4 matematik k1
Ppt kbsm t4 matematik k1Ppt kbsm t4 matematik k1
Ppt kbsm t4 matematik k1Saya Unigfx
 
UPSEE - Mathematics -2003 Unsolved Paper
UPSEE - Mathematics -2003 Unsolved PaperUPSEE - Mathematics -2003 Unsolved Paper
UPSEE - Mathematics -2003 Unsolved PaperVasista Vinuthan
 
April 13, 2015
April 13, 2015April 13, 2015
April 13, 2015khyps13
 
LAC2013 UNIT preTESTs!
LAC2013 UNIT preTESTs!LAC2013 UNIT preTESTs!
LAC2013 UNIT preTESTs!A Jorge Garcia
 
10 Mathematics Standard.pdf
10 Mathematics Standard.pdf10 Mathematics Standard.pdf
10 Mathematics Standard.pdfRohitSindhu10
 
Maths Problems - Qwizdom ppt
Maths Problems - Qwizdom pptMaths Problems - Qwizdom ppt
Maths Problems - Qwizdom pptQwizdom UK
 
Pre-Test: Applications of Integrals
Pre-Test: Applications of IntegralsPre-Test: Applications of Integrals
Pre-Test: Applications of Integralsk_ina
 
Sample test paper JEE mains 2013
Sample test paper JEE mains 2013Sample test paper JEE mains 2013
Sample test paper JEE mains 2013APEX INSTITUTE
 
Helping your child with gcse maths
Helping your child with gcse mathsHelping your child with gcse maths
Helping your child with gcse mathscountesthorpecc
 

Similaire à Developing Expert Voices (20)

IIT JEE Mathematics 2005
IIT JEE Mathematics   2005IIT JEE Mathematics   2005
IIT JEE Mathematics 2005
 
Notes Day 6
Notes   Day 6Notes   Day 6
Notes Day 6
 
CEER 2012 Math Lecture
CEER 2012 Math LectureCEER 2012 Math Lecture
CEER 2012 Math Lecture
 
2012 Mathacre JV Written Test
2012 Mathacre JV Written Test2012 Mathacre JV Written Test
2012 Mathacre JV Written Test
 
Complex Variables Assignment Help
Complex Variables Assignment HelpComplex Variables Assignment Help
Complex Variables Assignment Help
 
Complex Variables Assignment Help
Complex Variables Assignment HelpComplex Variables Assignment Help
Complex Variables Assignment Help
 
Calculus Assignment Help
Calculus Assignment HelpCalculus Assignment Help
Calculus Assignment Help
 
Calculus Assignment Help
Calculus Assignment HelpCalculus Assignment Help
Calculus Assignment Help
 
CAT -2010 Unsolved Paper
CAT -2010 Unsolved PaperCAT -2010 Unsolved Paper
CAT -2010 Unsolved Paper
 
Ppt kbsm t4 matematik k1
Ppt kbsm t4 matematik k1Ppt kbsm t4 matematik k1
Ppt kbsm t4 matematik k1
 
UPSEE - Mathematics -2003 Unsolved Paper
UPSEE - Mathematics -2003 Unsolved PaperUPSEE - Mathematics -2003 Unsolved Paper
UPSEE - Mathematics -2003 Unsolved Paper
 
April 13, 2015
April 13, 2015April 13, 2015
April 13, 2015
 
April 1
April 1April 1
April 1
 
LAC2013 UNIT preTESTs!
LAC2013 UNIT preTESTs!LAC2013 UNIT preTESTs!
LAC2013 UNIT preTESTs!
 
10 Mathematics Standard.pdf
10 Mathematics Standard.pdf10 Mathematics Standard.pdf
10 Mathematics Standard.pdf
 
Maths Problems - Qwizdom ppt
Maths Problems - Qwizdom pptMaths Problems - Qwizdom ppt
Maths Problems - Qwizdom ppt
 
Pre-Test: Applications of Integrals
Pre-Test: Applications of IntegralsPre-Test: Applications of Integrals
Pre-Test: Applications of Integrals
 
Sample test paper JEE mains 2013
Sample test paper JEE mains 2013Sample test paper JEE mains 2013
Sample test paper JEE mains 2013
 
Helping your child with gcse maths
Helping your child with gcse mathsHelping your child with gcse maths
Helping your child with gcse maths
 
1509 circle- coordinate geometry
1509 circle- coordinate geometry1509 circle- coordinate geometry
1509 circle- coordinate geometry
 

Dernier

Cybersecurity Workshop #1.pptx
Cybersecurity Workshop #1.pptxCybersecurity Workshop #1.pptx
Cybersecurity Workshop #1.pptxGDSC PJATK
 
activity_diagram_combine_v4_20190827.pdfactivity_diagram_combine_v4_20190827.pdf
activity_diagram_combine_v4_20190827.pdfactivity_diagram_combine_v4_20190827.pdfactivity_diagram_combine_v4_20190827.pdfactivity_diagram_combine_v4_20190827.pdf
activity_diagram_combine_v4_20190827.pdfactivity_diagram_combine_v4_20190827.pdfJamie (Taka) Wang
 
AI You Can Trust - Ensuring Success with Data Integrity Webinar
AI You Can Trust - Ensuring Success with Data Integrity WebinarAI You Can Trust - Ensuring Success with Data Integrity Webinar
AI You Can Trust - Ensuring Success with Data Integrity WebinarPrecisely
 
Videogame localization & technology_ how to enhance the power of translation.pdf
Videogame localization & technology_ how to enhance the power of translation.pdfVideogame localization & technology_ how to enhance the power of translation.pdf
Videogame localization & technology_ how to enhance the power of translation.pdfinfogdgmi
 
VoIP Service and Marketing using Odoo and Asterisk PBX
VoIP Service and Marketing using Odoo and Asterisk PBXVoIP Service and Marketing using Odoo and Asterisk PBX
VoIP Service and Marketing using Odoo and Asterisk PBXTarek Kalaji
 
Machine Learning Model Validation (Aijun Zhang 2024).pdf
Machine Learning Model Validation (Aijun Zhang 2024).pdfMachine Learning Model Validation (Aijun Zhang 2024).pdf
Machine Learning Model Validation (Aijun Zhang 2024).pdfAijun Zhang
 
Comparing Sidecar-less Service Mesh from Cilium and Istio
Comparing Sidecar-less Service Mesh from Cilium and IstioComparing Sidecar-less Service Mesh from Cilium and Istio
Comparing Sidecar-less Service Mesh from Cilium and IstioChristian Posta
 
UiPath Solutions Management Preview - Northern CA Chapter - March 22.pdf
UiPath Solutions Management Preview - Northern CA Chapter - March 22.pdfUiPath Solutions Management Preview - Northern CA Chapter - March 22.pdf
UiPath Solutions Management Preview - Northern CA Chapter - March 22.pdfDianaGray10
 
UiPath Studio Web workshop series - Day 6
UiPath Studio Web workshop series - Day 6UiPath Studio Web workshop series - Day 6
UiPath Studio Web workshop series - Day 6DianaGray10
 
Building AI-Driven Apps Using Semantic Kernel.pptx
Building AI-Driven Apps Using Semantic Kernel.pptxBuilding AI-Driven Apps Using Semantic Kernel.pptx
Building AI-Driven Apps Using Semantic Kernel.pptxUdaiappa Ramachandran
 
Empowering Africa's Next Generation: The AI Leadership Blueprint
Empowering Africa's Next Generation: The AI Leadership BlueprintEmpowering Africa's Next Generation: The AI Leadership Blueprint
Empowering Africa's Next Generation: The AI Leadership BlueprintMahmoud Rabie
 
Linked Data in Production: Moving Beyond Ontologies
Linked Data in Production: Moving Beyond OntologiesLinked Data in Production: Moving Beyond Ontologies
Linked Data in Production: Moving Beyond OntologiesDavid Newbury
 
9 Steps For Building Winning Founding Team
9 Steps For Building Winning Founding Team9 Steps For Building Winning Founding Team
9 Steps For Building Winning Founding TeamAdam Moalla
 
Artificial Intelligence & SEO Trends for 2024
Artificial Intelligence & SEO Trends for 2024Artificial Intelligence & SEO Trends for 2024
Artificial Intelligence & SEO Trends for 2024D Cloud Solutions
 
UiPath Community: AI for UiPath Automation Developers
UiPath Community: AI for UiPath Automation DevelopersUiPath Community: AI for UiPath Automation Developers
UiPath Community: AI for UiPath Automation DevelopersUiPathCommunity
 
UiPath Platform: The Backend Engine Powering Your Automation - Session 1
UiPath Platform: The Backend Engine Powering Your Automation - Session 1UiPath Platform: The Backend Engine Powering Your Automation - Session 1
UiPath Platform: The Backend Engine Powering Your Automation - Session 1DianaGray10
 
COMPUTER 10 Lesson 8 - Building a Website
COMPUTER 10 Lesson 8 - Building a WebsiteCOMPUTER 10 Lesson 8 - Building a Website
COMPUTER 10 Lesson 8 - Building a Websitedgelyza
 
Secure your environment with UiPath and CyberArk technologies - Session 1
Secure your environment with UiPath and CyberArk technologies - Session 1Secure your environment with UiPath and CyberArk technologies - Session 1
Secure your environment with UiPath and CyberArk technologies - Session 1DianaGray10
 

Dernier (20)

Cybersecurity Workshop #1.pptx
Cybersecurity Workshop #1.pptxCybersecurity Workshop #1.pptx
Cybersecurity Workshop #1.pptx
 
activity_diagram_combine_v4_20190827.pdfactivity_diagram_combine_v4_20190827.pdf
activity_diagram_combine_v4_20190827.pdfactivity_diagram_combine_v4_20190827.pdfactivity_diagram_combine_v4_20190827.pdfactivity_diagram_combine_v4_20190827.pdf
activity_diagram_combine_v4_20190827.pdfactivity_diagram_combine_v4_20190827.pdf
 
AI You Can Trust - Ensuring Success with Data Integrity Webinar
AI You Can Trust - Ensuring Success with Data Integrity WebinarAI You Can Trust - Ensuring Success with Data Integrity Webinar
AI You Can Trust - Ensuring Success with Data Integrity Webinar
 
Videogame localization & technology_ how to enhance the power of translation.pdf
Videogame localization & technology_ how to enhance the power of translation.pdfVideogame localization & technology_ how to enhance the power of translation.pdf
Videogame localization & technology_ how to enhance the power of translation.pdf
 
VoIP Service and Marketing using Odoo and Asterisk PBX
VoIP Service and Marketing using Odoo and Asterisk PBXVoIP Service and Marketing using Odoo and Asterisk PBX
VoIP Service and Marketing using Odoo and Asterisk PBX
 
Machine Learning Model Validation (Aijun Zhang 2024).pdf
Machine Learning Model Validation (Aijun Zhang 2024).pdfMachine Learning Model Validation (Aijun Zhang 2024).pdf
Machine Learning Model Validation (Aijun Zhang 2024).pdf
 
20150722 - AGV
20150722 - AGV20150722 - AGV
20150722 - AGV
 
Comparing Sidecar-less Service Mesh from Cilium and Istio
Comparing Sidecar-less Service Mesh from Cilium and IstioComparing Sidecar-less Service Mesh from Cilium and Istio
Comparing Sidecar-less Service Mesh from Cilium and Istio
 
UiPath Solutions Management Preview - Northern CA Chapter - March 22.pdf
UiPath Solutions Management Preview - Northern CA Chapter - March 22.pdfUiPath Solutions Management Preview - Northern CA Chapter - March 22.pdf
UiPath Solutions Management Preview - Northern CA Chapter - March 22.pdf
 
20230104 - machine vision
20230104 - machine vision20230104 - machine vision
20230104 - machine vision
 
UiPath Studio Web workshop series - Day 6
UiPath Studio Web workshop series - Day 6UiPath Studio Web workshop series - Day 6
UiPath Studio Web workshop series - Day 6
 
Building AI-Driven Apps Using Semantic Kernel.pptx
Building AI-Driven Apps Using Semantic Kernel.pptxBuilding AI-Driven Apps Using Semantic Kernel.pptx
Building AI-Driven Apps Using Semantic Kernel.pptx
 
Empowering Africa's Next Generation: The AI Leadership Blueprint
Empowering Africa's Next Generation: The AI Leadership BlueprintEmpowering Africa's Next Generation: The AI Leadership Blueprint
Empowering Africa's Next Generation: The AI Leadership Blueprint
 
Linked Data in Production: Moving Beyond Ontologies
Linked Data in Production: Moving Beyond OntologiesLinked Data in Production: Moving Beyond Ontologies
Linked Data in Production: Moving Beyond Ontologies
 
9 Steps For Building Winning Founding Team
9 Steps For Building Winning Founding Team9 Steps For Building Winning Founding Team
9 Steps For Building Winning Founding Team
 
Artificial Intelligence & SEO Trends for 2024
Artificial Intelligence & SEO Trends for 2024Artificial Intelligence & SEO Trends for 2024
Artificial Intelligence & SEO Trends for 2024
 
UiPath Community: AI for UiPath Automation Developers
UiPath Community: AI for UiPath Automation DevelopersUiPath Community: AI for UiPath Automation Developers
UiPath Community: AI for UiPath Automation Developers
 
UiPath Platform: The Backend Engine Powering Your Automation - Session 1
UiPath Platform: The Backend Engine Powering Your Automation - Session 1UiPath Platform: The Backend Engine Powering Your Automation - Session 1
UiPath Platform: The Backend Engine Powering Your Automation - Session 1
 
COMPUTER 10 Lesson 8 - Building a Website
COMPUTER 10 Lesson 8 - Building a WebsiteCOMPUTER 10 Lesson 8 - Building a Website
COMPUTER 10 Lesson 8 - Building a Website
 
Secure your environment with UiPath and CyberArk technologies - Session 1
Secure your environment with UiPath and CyberArk technologies - Session 1Secure your environment with UiPath and CyberArk technologies - Session 1
Secure your environment with UiPath and CyberArk technologies - Session 1
 

Developing Expert Voices

  • 1. Developing Expert Voices Pre Calculus 40S enriched 2007 Hi, my name is Sandy and I like purple(: DEV PROJECT 1
  • 2. Problem One You’re a Skydiver and you’ve just jumped out from the jet plane and you’re heading for the ground. You pull the string to the parachute and start to examine a crop circle that a farmer has created. Given that the centre of the crop circle is labeled O, the diameter of the circle is 10km, angle BOG is 120° and that D is … Answer the following in radians. a) Determine the length of the arc that subtends an angle of 240°. b) Determine the area of that sector using the information given above. c) Determine the angle at the centre of the circle if the arc subtended by the angle is . DEV PROJECT 2
  • 3. Problem One – answers (a) From the original question, the answers are supposed to be in radians. The formula to change Degrees into Radians is: Where D is Degrees and R is Radians. Plug in the given information into the formula. It should then look like this: Now you cross multiply: 240°(π) = R(180°) Divide everything by 180° to leave R by itself and solve! The degree signs cancel each other out, and also reduces to . Therefore, 240° in RADIANS is . DEV PROJECT 3
  • 4. Problem One – answers (a) We have the angle in radians, so we can use that to solve for the arc length with this formula: Where R is the angle in Radians, r is the radius and L is the Arc Length. Now we can substitute the given information and what we found into the formula. The 5 in 2π(r) was found by dividing the diameter by 2. *The radius is HALF of the diameter in a circle. DEV PROJECT 4
  • 5. Problem One – answers (a) To get rid of the fraction in a fraction you multiply the numerator by the reciprocal of the denominator. Multiply the left side and then cross multiply. (4π)(10π) = L(6π) Divide everything by (6π) to leave L by itself. Therefore… The π’s reduce, leaving you with only 1. Also 4 x 10 = 40 all divided by 6. Reducing that, it leaves you with DEV PROJECT 5
  • 6. Problem One – answers (a) A second way is to change an angle from degree to radians is: Since we’ve been through this once already, it’s should be pretty straight forward on how to acquire the correct answer. You should get the same answer you got before which was: So if you didn’t, you know you did something wrong. But don’t quit there, go back and try again. ☺ DEV PROJECT 6
  • 7. Problem One – answers (b) The answer again must be in radians so this is the formula you use to find the area of a sector: Where Θ is the angle in radians, S is the area of the sector and r is the radius of the circle. Plug in the information that we already know. Again, we have to multiply the numerator by the reciprocal of the denominator. DEV PROJECT 7
  • 8. Problem One – answers (b) Next, we simplify it and then it’s time to cross multiply. Afterwards, we divide both sides by (6π) so that S is by itself. We now simplify. The π’s reduce, just like before, leaving you with only one again, and 100 ÷ 6 reduces to DEV PROJECT 8
  • 9. Problem One – answers (b) Another way to solve this part of the question is to use a different formula: Plugging in all the information we come up with this: And from this point, it is also pretty straight forward on how to acquire the answer. Also, you should arrive at the same answer, if not… try again! DEV PROJECT 9
  • 10. Problem One – answers (c) This is very similar to the first part of this question. However, instead of solving for the ARC, we’re solving for the ANGLE. Where R is the angle in radians, L the Arc Length and r is the radius. Now we can substitute the information we know into the formula. DEV PROJECT 10
  • 11. Problem One – answers (c) Knowing the routine, we multiply the numerator by the reciprocal of the denominator again. We simplify, then cross multiply. DEV PROJECT 11
  • 12. Problem One – answers (c) Once again, divide both sides by 9π so that R will be by itself. Now simplify. Remembering that the π’s reduce you’re left with… Therefore the angle subtended by the arc is . DEV PROJECT 12
  • 13. Problem Two Given the graph of f(x) below, sketch the following three graphs. a) b) c) Please note that for tests or examinations, add arrows and label your axis. DEV PROJECT 13
  • 14. Problem Two – answers (a) The first thing you should remember before starting is: STRETCHES BEFORE TRANSLATIONS The basic formula for a question like this is: Af(B(x-C))+D Where A is a STRETCH (the y-coordinates are multiplied by A), B is a STRETCH (the x-coordinates are multiplied by / the reciprocal), C is a TRANSLATION (the x- coordinate is moved horizontally) and D is a TRANSLATION (the y-coordinate is moved vertically). DEV PROJECT 14
  • 15. Problem Two – answers (a) Looking at the graph, we can figure out all of the coordinates of each point. We’ll arrive at the numbers in this order remembering that the x- coordinate always comes before the y-coordinate. DEV PROJECT 15
  • 16. Problem Two – answers (a) We can take those coordinates and apply what we’re given. Af(B(x-C))+D Remember that A stretches the y-coordinate and B stretches the x-coordinate. First point: A(-6,-3). A((-6)(3)),(-3)(-2)) The 3 came from the reciprocal in the original question given above. You’re then left with a new coordinate: A(-18,6). BUT you’re not finished yet. DEV PROJECT 16
  • 17. Problem Two – answers (a) Now you find the Translation part. Af(B(x-C))+D Knowing that C shifts the x-coordinate and D shifts the y- coordinate. C is (-4) (even though it says x+4 in the question) because in the formula, it’s (x-C). So in order for C to be positive in the question, 4 must be a negative number. (x+4) (x-(-4)) One negative, multiplied by another negative, always gives you a positive number. Therefore: A((-18-4),(6+5)) And the new coordinate IS: A(-22,11) DEV PROJECT 17
  • 18. Problem Two – answers (a) Now there are 3 more points to do. Just for practice, try them yourself first and once your finished.. Go to the next slide and check if your answers are correct. ☺ The remaining points are: B(-3,-3) C(1,1) D(4,1) DEV PROJECT 18
  • 19. Problem Two – answers (a) Since we’ve done this once before, I’m going to go through it a bit faster. B(-3,-3) Stretches first! B((-3)(3),(-3)(-2)) B(-9,6) Now Translations! B(((-9)-4),(6+5)) (-13,11) C(1,1) Stretches first! C((1)(3),(1)(-2)) C(3,-2) Now Translations! C((3-4),((-2)+5)) C(-1,3) D(4,1) Stretches first! D((4)(3),(1)(-2)) D(12,-2) Now Translations! D(((12)-4),(-2+5)) D(8,3) DEV PROJECT 19
  • 20. Problem Two – answers (a) Now that we’ve calculated all the new coordinates, it’s time to plot those points onto a graph and connect the dots. After you’re finished, it should look like this: *New Graph: Is red. DEV PROJECT 20
  • 21. Problem Two – answers (b) We’re looking for the INVERSE of the function. Since we already have the coordinates, all we have to do is switch the y-coordinate and the x-coordinate. A(-6,-3) Now, SWITCH THEM. A(-3,-6) Easy right? Okay. So lets do the others now. B(-3,-3) B(-3,-3) C(1,1) C(1,1) D(4,1) D(1,4) DEV PROJECT 21
  • 22. Problem Two – answers (b) Now just like the other one, plot the points, connect the dots and you have your solution! DEV PROJECT 22
  • 23. Problem Two – answers (c) It’s asking for the reciprocal of the original function. 1. Look for the “invariant points.” ((-1,-1) and (1,1)) - These points are on both the original function and the reciprocal function because the reciprocal of 1 is 1. 2. Examine the straight horizontal lines. One of them is y=(-3) and the other is y=(1). On the graph, you find the reciprocal and graph it Therefore y=(1/3) and y=(1). DEV PROJECT 23
  • 24. Problem Two – answers (c) Now use the “Biggering, smallering” game. If a number increases, its reciprocal decreases and vice versa. Once it’s finished, it should look like this: Don’t forget to add an asymptote where the graph has zero(s) or where x = 0. DEV PROJECT 24
  • 25. Problem Three Prove: DEV PROJECT 25
  • 26. Problem Three - answers First thing to always do is to draw the “Great Wall of China.” Expand the problem. - We know that COT is the reciprocal of tan and instead of tan, you can use sin and cos. We also changed 1 to sinΘ/sinΘ because that equals one. So we don’t necessarily add anything to the original question. DEV PROJECT 26
  • 27. Problem Three - answers Here we simply just multiplied everything out. Just like we were doing before, to get rid of the double fraction we multiplied the numerator by the reciprocal of the denominator. The sinΘ’s reduce! DEV PROJECT 27
  • 28. Problem Three - answers You then end up with this after you reduce! You’re probably wondering how the heck does cos2Θ equal 1-sin2Θ? Well, knowing our identities: sin2Θ + cos2Θ = 1 If we rearrange the order, we can see that cos2Θ = 1-sin2Θ If you notice, 1-sin2Θ is a difference of squares. SO, it is easily factorable. DEV PROJECT 28
  • 29. Problem Three - answers The 1-sinΘ’s reduce. And you’re left with 1+sinΘ! *When You’re finished solving, you must always put Q.E.D to indicate that you are Q.E.D finished. DEV PROJECT 29
  • 30. Problem Four At this very moment, there are 1135 students that attend Daniel McIntyre High School. 2 years ago there was only 960 students that attended the High School. a) At what rate is the student body population increasing at? b) Assuming that the rate continues to increase at this rate, how much longer will it take for the student body population to be 1500 DEV PROJECT 30
  • 31. Problem Four – answers(a) We are looking for the rate, or the model, that the student body population is increasing. Here’s the formula that we use to solve this question: P = P0(Model)t Where P is the population at the end of the time period, P0 is the population at the beginning of the time period, Model is the, in this case, rate at which the population increases and t is the change in time. So using the information we know, we can use it to plug it into the formula. 1135 = 960(Model)2 DEV PROJECT 31
  • 32. Problem Four – answers(a) 1135 = 960(Model)2 Divide both sides by 960 to leave (Model)2 by itself and reduce: You can use “ln” to solve or you there’s an alternate way. First way is to use ln. Take the ln of both sides: Multiply by ½ on both sides to remove the 2 from the right. Calculate the left side out. 0.0837 = ln(Model) Therefore e0.0837 = Model. And the rate is (0.0837)(100) = 8.37% DEV PROJECT 32
  • 33. Problem Four – answers(a) 1135 = 960(Model)2 The other way is to do it like this: These reduce. And the result is 1.0873 = Model. To get the actual model you minus one, because when the population increases, it keeps the original amount (1) and from there increases (+ 0.0873%). DEV PROJECT 33
  • 34. Problem Four – answers(b) We’re asked to find the time it takes for the population to increase from 1135 to 1500. 1500 = 1135(1.0873)t Divide both sides by 1135 and reduce: DEV PROJECT 34
  • 35. Problem Four – answers(b) Now you can use ln to solve for t: Divide both sides by ln(1.0873). Then solve for t! and t is 3.331428783 3.3314 years. So it will take 3.3314 years for the student body population to grow to 1500. DEV PROJECT 35
  • 36. Toodle- Toodle-Loo Kangaroo! =) DEV PROJECT 36