Problem no.1 Trig Identities
Can u help me to prove this trig identity? In solving these problems, you have to work on the complicated side first. But in this case both of them are complicated so we will solve them together.
Firstly, you have to put a boundary  between them or what we called our  “ Red Great Wall of China”. Left side : We recognized that sin2x is a double  angle identity and so as the cos2x Right side : cot²x+1=csc²x
Left side: I will write the long term for double  angle identities to make things clear  for you. Right side: We’ll leave cotx the way it is (there’s a reason for that) Then, sec²x=1/cos²x csc²x=1/sin²x
Left side: (Numerator)  We can add sinxcosx+cosxsinx since they are same terms.. (Denominator) cosx · cosx= cos²x  sinx · sinx= sin²x So it will equal to 1+cos²x-sin²x We could recognize that  1-sin²x=cos²x Right side: We can already change cotx=cosx/sinx The equation is divided by 1/sin²x, we can take the reciprocal of it and multiply it to the equation.
Left side: 2 is reduced and cos as well. We’ll  get sinx/cosx Right side: so the equation is reduced to  sinx/cosx Eventhough they are already the  same we can still reduce it…
Sinx/cosx = tanx Thanks for your help..hehe..
Problem no. 2 Permutations
I’m J.r. Manleigh and I am the manager of the new  band called “Private”. I have here the list of the possible provinces that we can have the concert. Pangasinan Bataan Bukidnon Leyte Batangas Sulu Romblon La Union Surigao Del Norte Bohol Cavite Surigao Del Sur Cebu Laguna Mindanao(green) Visayas(yellow) Luzon(red)
 
If the band plays 6 provinces in total, how many different concert schedules can be created if there is no restrictions in which they play? Order is important thing in a touring schedule, so use permutations.  Out of 12 possible provinces, the band wants to create a schedule with  only 6.  Pick Formula : n P r  =  n ! ( n  -  r )! nPr is read as “n Pick r” where  n  is the number of objects to pick from r  is the no. of objects to be arranged
n P r   =  n ! ( n - r )! 14 P 6  =  14 ! ( 14 - 6 )! 14 ! 8! =  2,162,160 or using your calculator… 14 P 6   =  2,162,160  different concert schedules
If the band the band must play all provinces in Luzon, 3 in Visayas and 2 in Mindanao how many different schedules can be created? The 6 provinces in  Luzon  can be ordered in  6 P 6  ways.  The 3 provinces in  Visayas  can be ordered in  3 P 4  ways. The 2 provinces in  Mindanao  can be ordered in  2 P 4  ways. 6P 6  · 4P 3  · 4P 2  = 207,360 different schedules
If the band decides not to go to  Bataan  and  Sulu  (because many rumors about those provinces that they are anti-Private people). How many different schedules can be created if the band plays all provinces in  Luzon  and  Mindanao , and only 3 for  Visayas ? We can already ignore  Bataan  and  Sulu  from the selection pool. The first 5 cities in  Luzon  can now be ordered in 5P5 ways. The 3 cities in  Visayas  can be ordered in 4P3 ways. The 3 cities in  Mindanao  can be ordered in 3P3 ways. So… 5P5 · 4P3 · 3P3 = 17280 ways that the concert schedule can be created
Problem no. 3 Logarithms
Rio really likes the model of Honda Civic 2007 which will cost you a brand new one $24, 950 but cannot afford to buy it now. She’s planning to save up some money first then buy it. After 3 years, she will have enough money to buy a used car Honda Civic 2007. But she has 3 dealers to choose from if she wants to get a low priced car.  Andrei who offers a Civic that depreciates 0.02% daily for 3 years. Danny who offers a Civic that depreciates 0.5% monthly for 3 years. Ronnie who offers a Civic that depreciates for 0.15% weekly for 3 years. Who will be the best to go to if she wants the cheapest one?
Andrei who offers a Civic that depreciates 0.02% daily for 3 years. A o =  24, 950 model = 0.02%  ->  0.0002 (percent to decimal) t = 365 x 3(because 365 days in a year and it is for 3 years) A = ? p = 1 A = 20, 042.4379 or $20, 042.44
Danny who offers a Civic that depreciates 0.5% monthly for 3 years. A o =  24, 950 model = 0.5%  ->  0.005 (percent to decimal) t = 12 x 3 (because 12 months in one year and it is for 3 years) A = ? p = 1 A = 20, 830.5845 or $20, 830.58
Ronnie who offers a Civic that depreciates for 0.15% weekly for 3  years. A o =  24, 950 model = 0.15%  ->  0.0015 (percent to decimal) t = 52 x 3 (because 52 weeks in one year and it is for 3 years) A = ? p = 1 A = 19, 741.0090 or $19, 741.01
We all know that Ronnie is the cheapest one  So Riolyn will choose him.
Problem no. 4 Conics
What you see above is a tunnel in Winnipeg that has a length of 162 m and a height of 56   m.
If the arch is modeled as an ellipse with the standard form:   (x - h)² + (y - k)²  = 1 a²  b² Determine the equation: First, is to use the measures that was given to you. Major axis: 2a = 162  a=  81 Semi-minor axis: b = 54  Centre = (0, 0) Therefore:  (x - h)² + (y - k)² = 1  (x - 0)²+ (y – 0)²  = 1 a²  b²  (81)²  (54)² x²  +  y²  = 1 6561  2916
If the arch is modeled as a parabola with the standard form:  y – k = a(x - h)² Determine the equation: First state what you know: Vertex = (0 , 54) point in the graph = (81, 0) Try to plug these values in the equation to find the “a” value. y - k  = a( x - h )² 0 - 54 = a(81 - 0) ² -54 = a(81) ² -54  =  6561 a 6561  6561 a = -54  6561 y – 54 = -54 x² 6561
Determine the height of the tunnel at a point 40 m away from the center using  the equations given above. If the actual height at this point is 40.2 m,  determine which model makes the closest approximation. Ellipse:  Parabola: x²  +  y²  = 1  y-54 = -54x ² 6561  2916  6561 1600  +  y²  = 1  y-54 = -54(40) ² 6561  2916  6561 y²  = 1 –  0.244  y-54 = -13.169 2916  y=40.83 y²  = 0.756 2916 √  y² = √ 2204.496 y =46.952
Ellipse = 46.952 Parabola = 40.83 Which makes parabola the closest one to the actual height.

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Project Ko!

  • 1. Problem no.1 Trig Identities
  • 2. Can u help me to prove this trig identity? In solving these problems, you have to work on the complicated side first. But in this case both of them are complicated so we will solve them together.
  • 3. Firstly, you have to put a boundary between them or what we called our “ Red Great Wall of China”. Left side : We recognized that sin2x is a double angle identity and so as the cos2x Right side : cot²x+1=csc²x
  • 4. Left side: I will write the long term for double angle identities to make things clear for you. Right side: We’ll leave cotx the way it is (there’s a reason for that) Then, sec²x=1/cos²x csc²x=1/sin²x
  • 5. Left side: (Numerator) We can add sinxcosx+cosxsinx since they are same terms.. (Denominator) cosx · cosx= cos²x sinx · sinx= sin²x So it will equal to 1+cos²x-sin²x We could recognize that 1-sin²x=cos²x Right side: We can already change cotx=cosx/sinx The equation is divided by 1/sin²x, we can take the reciprocal of it and multiply it to the equation.
  • 6. Left side: 2 is reduced and cos as well. We’ll get sinx/cosx Right side: so the equation is reduced to sinx/cosx Eventhough they are already the same we can still reduce it…
  • 7. Sinx/cosx = tanx Thanks for your help..hehe..
  • 8. Problem no. 2 Permutations
  • 9. I’m J.r. Manleigh and I am the manager of the new band called “Private”. I have here the list of the possible provinces that we can have the concert. Pangasinan Bataan Bukidnon Leyte Batangas Sulu Romblon La Union Surigao Del Norte Bohol Cavite Surigao Del Sur Cebu Laguna Mindanao(green) Visayas(yellow) Luzon(red)
  • 10.  
  • 11. If the band plays 6 provinces in total, how many different concert schedules can be created if there is no restrictions in which they play? Order is important thing in a touring schedule, so use permutations. Out of 12 possible provinces, the band wants to create a schedule with only 6. Pick Formula : n P r = n ! ( n - r )! nPr is read as “n Pick r” where n is the number of objects to pick from r is the no. of objects to be arranged
  • 12. n P r = n ! ( n - r )! 14 P 6 = 14 ! ( 14 - 6 )! 14 ! 8! = 2,162,160 or using your calculator… 14 P 6 = 2,162,160 different concert schedules
  • 13. If the band the band must play all provinces in Luzon, 3 in Visayas and 2 in Mindanao how many different schedules can be created? The 6 provinces in Luzon can be ordered in 6 P 6 ways. The 3 provinces in Visayas can be ordered in 3 P 4 ways. The 2 provinces in Mindanao can be ordered in 2 P 4 ways. 6P 6 · 4P 3 · 4P 2 = 207,360 different schedules
  • 14. If the band decides not to go to Bataan and Sulu (because many rumors about those provinces that they are anti-Private people). How many different schedules can be created if the band plays all provinces in Luzon and Mindanao , and only 3 for Visayas ? We can already ignore Bataan and Sulu from the selection pool. The first 5 cities in Luzon can now be ordered in 5P5 ways. The 3 cities in Visayas can be ordered in 4P3 ways. The 3 cities in Mindanao can be ordered in 3P3 ways. So… 5P5 · 4P3 · 3P3 = 17280 ways that the concert schedule can be created
  • 15. Problem no. 3 Logarithms
  • 16. Rio really likes the model of Honda Civic 2007 which will cost you a brand new one $24, 950 but cannot afford to buy it now. She’s planning to save up some money first then buy it. After 3 years, she will have enough money to buy a used car Honda Civic 2007. But she has 3 dealers to choose from if she wants to get a low priced car. Andrei who offers a Civic that depreciates 0.02% daily for 3 years. Danny who offers a Civic that depreciates 0.5% monthly for 3 years. Ronnie who offers a Civic that depreciates for 0.15% weekly for 3 years. Who will be the best to go to if she wants the cheapest one?
  • 17. Andrei who offers a Civic that depreciates 0.02% daily for 3 years. A o = 24, 950 model = 0.02% -> 0.0002 (percent to decimal) t = 365 x 3(because 365 days in a year and it is for 3 years) A = ? p = 1 A = 20, 042.4379 or $20, 042.44
  • 18. Danny who offers a Civic that depreciates 0.5% monthly for 3 years. A o = 24, 950 model = 0.5% -> 0.005 (percent to decimal) t = 12 x 3 (because 12 months in one year and it is for 3 years) A = ? p = 1 A = 20, 830.5845 or $20, 830.58
  • 19. Ronnie who offers a Civic that depreciates for 0.15% weekly for 3 years. A o = 24, 950 model = 0.15% -> 0.0015 (percent to decimal) t = 52 x 3 (because 52 weeks in one year and it is for 3 years) A = ? p = 1 A = 19, 741.0090 or $19, 741.01
  • 20. We all know that Ronnie is the cheapest one So Riolyn will choose him.
  • 21. Problem no. 4 Conics
  • 22. What you see above is a tunnel in Winnipeg that has a length of 162 m and a height of 56 m.
  • 23. If the arch is modeled as an ellipse with the standard form: (x - h)² + (y - k)² = 1 a² b² Determine the equation: First, is to use the measures that was given to you. Major axis: 2a = 162 a= 81 Semi-minor axis: b = 54 Centre = (0, 0) Therefore: (x - h)² + (y - k)² = 1 (x - 0)²+ (y – 0)² = 1 a² b² (81)² (54)² x² + y² = 1 6561 2916
  • 24. If the arch is modeled as a parabola with the standard form: y – k = a(x - h)² Determine the equation: First state what you know: Vertex = (0 , 54) point in the graph = (81, 0) Try to plug these values in the equation to find the “a” value. y - k = a( x - h )² 0 - 54 = a(81 - 0) ² -54 = a(81) ² -54 = 6561 a 6561 6561 a = -54 6561 y – 54 = -54 x² 6561
  • 25. Determine the height of the tunnel at a point 40 m away from the center using the equations given above. If the actual height at this point is 40.2 m, determine which model makes the closest approximation. Ellipse: Parabola: x² + y² = 1 y-54 = -54x ² 6561 2916 6561 1600 + y² = 1 y-54 = -54(40) ² 6561 2916 6561 y² = 1 – 0.244 y-54 = -13.169 2916 y=40.83 y² = 0.756 2916 √ y² = √ 2204.496 y =46.952
  • 26. Ellipse = 46.952 Parabola = 40.83 Which makes parabola the closest one to the actual height.