algebra home work

timer Asked: Nov 15th, 2016

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need help please

1. Simplify the trigonometric expression. Show your work. Solution Combining the two expressions under a common denominator gives: 1 1+𝑠𝑖𝑛 πœƒ 1 + 1βˆ’π‘ π‘–π‘›πœƒ = 1βˆ’π‘ π‘–π‘›πœƒ+1+π‘ π‘–π‘›πœƒ (1+π‘ π‘–π‘›πœƒ)(1βˆ’π‘ π‘–π‘›πœƒ) = 2π‘ π‘–π‘›πœƒ 1βˆ’π‘ π‘–π‘›2 πœƒ But from the trigonometric identity: sin2 πœƒ + cos2 πœƒ = 1 Then, cos2 πœƒ = 1 - sin2 πœƒ 1 1 Hence, 1+𝑠𝑖𝑛 πœƒ + 1βˆ’π‘ π‘–π‘›πœƒ = 2π‘ π‘–π‘›πœƒ π‘π‘œπ‘ 2 πœƒ = 2π‘‘π‘Žπ‘›πœƒ π‘π‘œπ‘ πœƒ = 2π‘ π‘’π‘πœƒπ‘‘π‘Žπ‘›πœƒ , 2. In is a right angle. Find the remaining sides and angles. Round your answers to the nearest tenth. Show your work. a = 3, c = 19 Solution From Pythagoras theorem: AC2 + CB2 = AB2 Then, Side b =βˆšπ’„πŸ βˆ’ π’‚πŸ = βˆšπŸπŸ—πŸ βˆ’ πŸ‘πŸ = βˆšπŸ‘πŸ“πŸ= 18.8 𝑏 < 𝐴 = π‘π‘œπ‘  βˆ’1 𝑐 = π‘π‘œπ‘  βˆ’1 18.8 19 = 9.1Β° Also, < 𝐡 = 180βˆ’< π΄βˆ’< 𝐢 = 180 βˆ’ 90 βˆ’ 9.1 = 80.9Β° 3. Please show your work to find the mean and standard deviation of the data. Round to the nearest tenth. 20, 16, 18, 14, 9, 20, 16 Solution Mean is calculated as follows, π‘š = Mean= 20+16+18+14+9+20+16 7 = π‘₯1+π‘₯2+π‘₯3+β‹― 𝑛 113 7 =16.1 βˆ‘(π‘₯βˆ’π‘š)2 Standard deviation is calculated as follows, s.d=√ 𝑛 (20βˆ’16.1)2 +(16βˆ’16.1)2 +(18βˆ’16.1)2 +(14βˆ’16.1)2 +(9βˆ’16.1)2+(20βˆ’16.1)2 +(16βˆ’16.1)2 7 Thus, s.d=√ 15.21+0.01+3.61+4.41+50.41+15.21+0.01 = 7 S.d=√ 3.54 4. What are the points of discontinuity? Are they all removable? Please show your work. Solution A point of discontinuity is a point where the function increases to infinity and/or decreases to negative infinity i.e it does not exist. For the equation, consider the denominator is zero. Then, the solution for x2 – 6x + 5 = 0 is as follows, (x - 5)(x - 1) = 0 Hence, x = 5 or x =1 makes the equation discontinuous at this points. They are not removable, since they are included in the solution 5. A sound wave is modeled with the equation . a. Find the period. Explain your method. b. Find the amplitude. Explain your method. c. What is the equation of the midline? What does it represent? Solution A periodic equation is given as, y = A cos𝝎𝜽 a) The period, T=2Ο€/𝝎 =6Ο€/2Ο€ = 3 sec b) Amplitude = A =0.25 c) Midline = Maximum value – Amplitude = 0 The equation of the midline of a sinusoidal function represents the line that passes exactly in the middle of its extreme values. 6. Paula spots a glider located at an angle of elevation of 42Β°. The distance between the glider and Paula is 3280 feet. To the nearest foot, what is the height of the glider work. h from the ground? Show your 3280 ft h 42 o Paula Solution using the definition of sine then, sin 42 = h/3280 Hence, height of glider, h =3280 sin 42 = 2195 ft 7. What is the product in simplest form? State any restrictions on the variable. Please show your work. y2 y y2 3 y2 y 6 y Solution Factoring the following quadratic expressions gives the following, y2 – y – 6 = (y-3) (y+2) y2 + y = y(y+1) then the expression simplifies to: 𝑦2 (π‘¦βˆ’3)(𝑦+2) π‘₯ π‘¦βˆ’3 𝑦(𝑦+1) = 𝑦(𝑦+2) 𝑦+1 The restriction is for y not to be equal to -1. 8. Verify the identity. Justify each step. Solution Since tan(ΞΈ) = sin(ΞΈ)/cos(ΞΈ) and cot(ΞΈ) = cos(ΞΈ)/sin(ΞΈ) Then, tanΞΈ + cotΞΈ = sinΞΈ/cosΞΈ +cosΞΈ/sinΞΈ Expressing these under a common denominator becomes, tanΞΈ + cotΞΈ = 𝑠𝑖𝑛2 πœƒ+π‘π‘œπ‘ 2 πœƒ π‘ π‘–π‘›πœƒπ‘π‘œπ‘ πœƒ But, 𝑠𝑖𝑛2 πœƒ + π‘π‘œπ‘  2 πœƒ = 1 Hence, 9. Is there any bias in the survey question? Explain. What do think would help students pay more attention in class? Solution Yes, the survey can contain bias. If a student is answering the question or someone related or friends with a student. They could then answer the survey to better fit them, rather than all of the students as a whole 10. Find the values of the 30th and 90th percentiles of the data. Please show your work. 129, 113, 200, 100, 105, 132, 100, 176, 146, 152 11. What is the quotient in simplified form? State any restrictions on the variable. Show Work. a 2 a 1 a5 a2 8a 15 Solution First, factorize the quadratic equation as follows: a2 – 8a + 15 = a2 -3a -5a +15 = (a-5) (a-3) Then the expression is evaluated as follows: π‘Ž + 2 (π‘Ž βˆ’ 5)(π‘Ž βˆ’ 3) (π‘Ž + 2)(π‘Ž βˆ’ 3) π‘₯ = π‘Žβˆ’5 π‘Ž+1 π‘Ž+1 Restrictions on variable is that a not be equal to -1. 12. Vance is designing a garden in the shape of an isosceles triangle. The base of the garden is 36 feet y long. The function 36 feet models the height of the triangular garden. not drawn to scale a. What is the height of the triangle when b. What is the height of the triangle when c. Vance is considering using either for his garden. Compare the areas of the two possible gardens. or Explain how you found the areas. Soln a) When πœƒ = 45, 𝑦 = 18π‘‘π‘Žπ‘›45 = 18 ft b) When πœƒ = 55, 𝑦 = 18π‘‘π‘Žπ‘›55 = 25.7 𝑓𝑑 c) Area of triangle A=1/2 b*y Area whenπœƒ = 45, A=0.5 x 36 x 18=324 ft2 Area whenπœƒ = 55, A=0.5 x 36 x 25.7=462.6 ft2 Hence the area is bigger when πœƒ = 55 π‘‘β„Žπ‘Žπ‘› π‘€β„Žπ‘’π‘› πœƒ = 45 13. Verify the identity. Justify each step. Solution Expressing left hand side under one denominator gives π‘ π‘’π‘πœƒ(π‘π‘ π‘πœƒ+π‘π‘œπ‘‘πœƒ)βˆ’π‘ π‘’π‘πœƒ(π‘π‘ π‘πœƒβˆ’π‘π‘œπ‘‘πœƒ) (π‘π‘ π‘πœƒβˆ’π‘π‘œπ‘‘πœƒ)(π‘π‘ π‘πœƒ+π‘π‘œπ‘‘πœƒ) Since, cot2 ΞΈ = csc2 ΞΈ βˆ’ 1 = 2π‘ π‘’π‘πœƒπ‘π‘œπ‘‘πœƒ 𝑐𝑠𝑐 2 πœƒβˆ’π‘π‘œπ‘‘ 2 πœƒ 2π‘ π‘’π‘πœƒπ‘π‘œπ‘‘πœƒ Thus it simplifies to: 𝑐𝑠𝑐2 πœƒβˆ’π‘π‘ π‘2 πœƒ+1 = 2π‘ π‘’π‘πœƒπ‘π‘œπ‘‘πœƒ Also, cot(ΞΈ) = cos(ΞΈ)/sin(ΞΈ) which gives: 2 cscπœƒ 14. Compare the graphs of the inverse variations. Please provide at least 3 comparisons and Solution Consider the following points of x a) When x=0, gives y1=y2 being at infinity b) When x= 2, gives y1= -0.1 and y2=-0.15 c) When x=-2, gives y1= 0.1 and y2 = 0.15 15. Use a graphing calculator to solve the equation Round to the nearest hundredth. in the interval from Solution t=70.52, 250.52 16. The equation models the height h in centimeters after t seconds of a weight attached to the end of a spring that has been stretched and then released. a. Solve the equation for t. . b. Find the times at which the weight is first at a height of 1 cm, of 3 cm, and of 5 cm above the rest position. Round your answers to the nearest hundredth. c. Find the times at which the weight is at a height of 1 cm, of 3 cm, and of 5 cm below the rest position for the second time. Round your answers to the nearest hundredth. Solution 3 a) Making t the subject of the formula:𝑑 = πœ‹ π‘π‘œπ‘  βˆ’1 (β„Ž/7) 3 1 3 3 3 πœ‹ 5 7 b) When h=1, t=πœ‹ π‘π‘œπ‘  βˆ’1 (7) = 78.10 sec When h=3, t=πœ‹ π‘π‘œπ‘  βˆ’1 (7) = 61.71 sec When h=5, t= π‘π‘œπ‘  βˆ’1 ( ) = 42.41 sec 3 1 3 3 3 πœ‹ 5 7 c) When h=-1, t=πœ‹ π‘π‘œπ‘  βˆ’1 (βˆ’ 7) = 78.10 sec When h=-3, t=πœ‹ π‘π‘œπ‘  βˆ’1 (βˆ’ 7) = 61.71 sec When h=-5, t= π‘π‘œπ‘  βˆ’1 (βˆ’ ) = 42.41 sec 17. Consider the graph of the cosine function shown below. y 4 2 O 2 –2 –4 a.Find the period and amplitude of the cosine function. for b.At what values of do the maximum value(s), minimum values(s), and zeros occur? Solution a) Period = πœ‹ Amplitude= 4 b) Maximum values at 0, πœ‹ π‘Žπ‘›π‘‘ 2πœ‹ πœ‹ Minimum values at 2 π‘Žπ‘›π‘‘ 3πœ‹/2 πœ‹ 3πœ‹ 5πœ‹ , 4 4 Zeros at 4 , π‘Žπ‘›π‘‘ 7πœ‹ 4 18. Use the graph of the sine function shown below. a. How many cycles occur in the graph? b. Find the period of the graph. c. Find the amplitude of the graph. Solution a) 1 complete cycle b) Period = 2πœ‹ c) Amplitude = 2 19. Verify the Pythagorean Identity. Solution From the identity,sin2 πœƒ + cos2 πœƒ = 1 π‘π‘œπ‘ 2 πœƒ 1 Divide both sides by sin2 πœƒ, we get 1 + 𝑠𝑖𝑛2 ΞΈ = 𝑠𝑖𝑛2 πœƒ Hence, 20. Howard is flying a kite and wants to find its angle of elevation. The string on the kite is 32 meters long and the kite is level with the top of a building that he knows is 28 meters high. To the nearest tenth of a degree, find the angle of elevation. Show your work. Solution Using the definition of sine, Sin Ξ± = height of building/length of string =28/32=0.875 Thus angle of elevation, Ξ±= sin-1 0.875 = 61

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