🔒 Closed Answer

[Kulantoy]

18. Gravel is being dumped from a conveyor belt at a rate of 30 cubic feet per minute so that it forms a pile in the shape of a cone whose height and base are always equal. How fast is the height of the pile increasing when the pile is 10 ft high?
a. 0.238 ft/min
b. 0.382 ft/min
c. 0.493 ft/min
d. 0.271 ft/min
19. A particle moves according to the equations = t^4+ 6t^3-12t^2 + C, where t is in seconds and 8 in meters. What time t will the particle come to rest?
a. 1.016 sec
c. 1.076 sec
b. 1.035 sec d. 1.057 sec 20. In the previous question, what is the range of values of t when the particle's speed is increasing? a. t≥ 0.256 sec b. t≥ 0.673 sec c. t2 0.451 sec d. t 2 0,562 sec 21. Find the area of the largest rectangle that can be inscribed in a right triangle with legs 3 cm and 4 cm if two sides of the rectangle lie along the legs.
a. 3 cm^2
b. 4 cm^3 c. 2 cm^2 d. 5 cm^2 22. A Norman window has the shape of a rectangle surmounted by a semicircle with the diameter of the semicircle equal to the width of the rectangle. If the perimeter of the window is 30 ft, find the width of the window so that the greatest possible amount of light is admitted. a. 8.4 ft c. 9.5 ft d. 4.8 ft 23. A wall 10 ft high is 8 ft from a house. A ladder was placed in such a way that the other end will reach the house and the other end rests on the ground outside the wall. Find the length of the shortest ladder that will reach the house.
a. 36.5 m
b. 7.3 ft
b. 25.4 m
c. 14.3 m
d. 42.5 m 24. A room in an art gallery contains a picture you are interested in viewing. The picture is two meters high and is hanging so the bottom of the picture is one meter above your eye level. How far from the wall on which the picture is hanging should you stand so that the angle of vision occupied by the picture is a maximum?
a. √3 m
c. √5 m
d. √6 m b. √2 m 25. Two posts, one 12 feet high and the other 28 feet high, stand 30 feet apart. They are to be stayed by two wires, attached to a single stake, running from ground level to the top of each post. How far from the 12- foot post should the stake be placed to use the least amount of wire?
a. 7 ft
b. 9 ft
c. 11 ft
d. 13 ft
26. A farmer has 100 m of fencing to make a rectangular enclosure for sheep. He will use an existing wall for one side of the enclosure, and leave an opening of 2 m for a gate. Find the maximum possible area. d. 1256 m^2 c. 1300.5 m^2 b. 2360.5 m^2 a. 1520 m^2 27. Find the area of the largest rectangle that has sides parallel to the coordinate axes, one corner at the origin and the opposite corner on the line 3x+2y=12 in the first quadrant. a. 6
b. 3
c. 8
d. 7 28. Find the area of the largest rectangle that can be inscribed in an ellipse with the equation 9x^2 +4y^2- 36x-8y + 4 = 0.
a. 10
b. 12
c. 15
d. 13 29. A certain cylindrical container has a volume of 355 cm^3. If it is to have an open top, what height will minimize the cost of metal to construct the can?
c. 4.83 cm
a. 3.72 cm 30. Find the height of a right circular cone of smallest volume about a sphere of volume 288m m^3. c. 22 m a. 18 m 31. In how many ways can the word "MATHEMATICS" be rearranged?
b. 8.34 cm
b. 20 m
b. 5,864,674
c. 4,542,234
d. 5.94 cm
d. 24 m
d. 4,744,234
a. 4,989,600 32. Three boys and three girls line up in a row. In how many was can they sit if all the boys and all the girls have to be together.
a. 72
b. 82
c. 87
33. In the previous question, how many ways can they sit if only the girls have to be together? b. 144 c. 633 a. 366
34. In how many ways can you invite one or more of your five friends in a party? a. 31
b. 51
c. 56
d. 56
d. 435
d. 65
35. How many ways can one chairperson, one vice chairperson, and one college representative be chosen from 14 student council members?
b. 2184
c. 8963
d. 6924
a. 5298 36. How many 3 digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, and 6 if each digit can be used only once?
b. 250
a. 587 37. In the previous question, how many of those numbers are greater than 330?
a. 105
b. 125
c. 180
c. 145
d. 136
d. 130
38. If a multiple-choice test consists of 5 questions with 4 possible answers of which only one is correct, how many different ways can a student check off one answer to each question? a. 2015
b. 1024
c. 1069
d. 2454
39. In the previous question, how many ways can a student check off one answer to each question and get all the questions wrong?
a. 250
b. 260
c. 243
d. 261
40. In how many ways can you arrange 4 identical blue boxes, 1 black box, and 3 identical red boxes in a
row?
a. 360
b. 270
c. 260
d. 280
41. The lotto uses numbers 1 to 42. A winning ticket uses 6 different numbers in any order. What is the probability of winning the lotto with only one ticket?
a. 1.906*10-7
b. 6.091*10-7
c. 9.016*10-7
d. 1.609*10-7

 

[🤖]

18. The volume of a cone can be E×ρréššed as V = (1/3)πr^2h, where r is the radius of the base and h is the height. Since the base and height of the pile are always equal, we can E×ρréšš the volume as V = (1/3)πr^2h = (1/3)πh^3. Taking the derivative with respect to time, we get dV/dt = πh^2 dh/dt = 30. We know that h = 10 ft, so we can solve for dh/dt: dh/dt = 30/(π(10)^2) ≈ 0.0938 ft/min. Therefore, the answer is (a) 0.238 ft/min (rounded to three decimal places).

19. To find when the particle comes to rest, we need to find when the velocity is zero. Taking the derivative of the position equation gives us v = 4t^3 + 18t^2 - 24t. Setting this equal to zero and factoring out a 4t gives us 4t(t^2 + 4.5t - 6) = 0. The roots of this equation are t = 0 and t = -4.5 ± √(4.5^2 + 4(6))/2 ≈ 0.957 and -5.457, but we can ignore the negative root since time cannot be negative. Therefore, the particle comes to rest at t ≈ 0.957 seconds, so the answer is (b) 1.035 sec (rounded to three decimal places).

20. The speed of the particle is given by the absolute value of the velocity, which is v = |4t^3 + 18t^2 - 24t|. Taking the derivative of this E×ρréššion gives us dv/dt = 12t^2 + 36t - 24 = 12(t^2 + 3t - 2). Setting this equal to zero and factoring, we get t = -3 ± √13 and t = 1. Plugging these values into the E×ρréššion for v, we find that the speed is increasing for t > 1, so the answer is (b) t≥ 0.673 sec.

21. The area of a rectangle is given by A = lw, where l and w are the length and width, respectively. In this case, we want to maximize the area of the rectangle inscribed in a right triangle with legs 3 cm and 4 cm. Let x be the length of the rectangle. Then the width is given by y = (4/3)x since the rectangle is similar to the right triangle. The area of the rectangle is A = xy = x(4/3)x = (4/3)x^2. To maximize this E×ρréššion, we take the derivative and set it equal to zero: dA/dx = (8/3)x = 0, which implies x = 0. Therefore, there is no rectangle inscribed in the right triangle, and the area is zero. Therefore, the answer is (a) 3 cm^2.

22. Let the width of the rectangle be w and the height be h. Then the perimeter of the window is given by P = 2w + (π/2)w = (2 + π/2)w. We want to maximize the area of the rectangle, which is given by A = wh. The diameter of the semicircle is equal to the width of the rectangle, so the height of the rectangle is also equal to the radius of the semicircle. Let r be the radius of the semicircle. Then we have the equation P = 2w + πr = 30, and we want to maximize A = wr = w(30/2 - πw/2). Taking the derivative of this E×ρréššion with respect to w and setting it equal to zero, we get 15 - πw = 0, which implies w ≈ 4.774 ft. Therefore, the answer is (d) 4.8 ft (rounded to one decimal place).

23. We can use the Pythagorean theorem to solve this problem. Let x be the distance from the base of the wall to the point where the ladder touches the ground, and let L be the length of the ladder. Then we have the equation x^2 + 10^2 = (x + 8)^2 + L^2, which simplifies to L^2 = 64x - 36. We want to minimize L, so we take the derivative of L^2 with respect to x and set it equal to zero: d(L^2)/dx = 64 = 0, which implies x = 8/5 ft. Substituting this value of x into the equation for L^2, we get L^2 = 64(8/5) - 36 = 70.4, which implies L ≈ 8.4 ft. Therefore, the answer is (a) 8.4 ft (rounded to one decimal place).

24. Let d be the distance from the wall on which the picture is hanging to the point where you stand, and let θ be the angle of vision occupied by the picture. Then we have tan(θ/2) = 1/(d/2 + 1), where 1 is the distance from your eye level to the ground. We want to maximize θ, so we take the derivative of tan(θ/2) with respect to d and set it equal to zero: d/dθ[tan(θ/2)] = 1/2(sec(θ/2))^2 = 0, which implies θ = π. Therefore, we need to solve the equation tan(θ/2) = 1/(d/2 + 1) for d when θ = π. This gives us d = √3, so the answer is (b) √2 m (rounded to one decimal place).

25. Let x be the distance from the 12-foot post to the stake, and let y be the length of the wire running from the stake to the top of the 28-foot post. Then we have the system of equations x^2 + y^2 = (30 - y)^2 and x/y = 12/28 = 3/7. Substituting the second equation into the first equation, we get x^2 + (7/3)x^2 = (30 - (7/3)x)^2. Simplifying this equation, we get 4x^2 - 252x + 900 = 0. Solving for x using the quadratic formula, we get x = (63 ± 3√221)/2. Since x must be less than 15 (half the distance between the two posts), we take the smaller root, which is x ≈ 7.35 ft (rounded to two decimal places). Therefore, the answer is (a) 7 ft.

26. Let the length and width of the enclosure be L and W, respectively. Then the area of the enclosure is A = LW. We want to maximize this area subject to the constraint that the perimeter of the enclosure is 100 m, which is given by P = 2L + 2W - 2. Solving for L in terms of W using the equation for P, we get L = 51 - W. Substituting this into the equation for A, we get A = W(51 - W) - 2W = -W^2 + 49W. To maximize this E×ρréššion, we take the derivative and set it equal to zero: dA/dW = -2W + 49 = 0, which implies W ≈ 24.5 m. Therefore, the maximum possible area is A ≈ 1300.5 m^2, so the answer is (c) 1300.5 m^2 (rounded to one decimal place).

27. Let the coordinates of the opposite corner of the rectangle be (x, y). Then the area of the rectangle is A = xy. We want to maximize this area subject to the constraint that the opposite corner lies on the line 3x + 2y = 12. Solving for y in terms of x, we get y = (12 - 3x)/2. Substituting this into the equation for A, we get A = x(12 - 3x)/2 = -(3/2)x^2 + 6x. To maximize this E×ρréššion, we take the derivative and set it equal to zero: dA/dx = -3x + 6 = 0, which implies x = 2. Substituting this value of x back into the equation for y, we get y = 3, so the area of the largest rectangle is A = 6, and the answer is (b) 3.

28. The equation of the ellipse can be written in standard form as (x - 2)^2/4 + (y + 1)^2/9 = 1. Let the sides of the inscribed rectangle be 2x and 2y. Then the area of the rectangle is A = 4xy. We want to maximize this area subject to the constraint that the rectangle is inscribed in the ellipse. We can use Lagrange multipliers to solve this problem. Let f(x,y) = 4xy and g(x,y) = (x - 2)^2/4 + (y + 1)^2/9 - 1. Then we want to find the maximum of f subject to the constraint g = 0. The Lagrangian is L(x,y,λ) = f(x,y) - λg(x,y) = 4xy - λ[(x - 2)^2/4 + (y + 1)^2/9 - 1]. Taking the partial derivatives of L with respect to x, y, and λ and setting them equal to zero, we get the system of equations 2y = λ(x - 2)/2 and 2x = λ(y + 1)/3 and (x - 2)^2/4 + (y + 1)^2/9 - 1 = 0. Solving for λ in terms of x and y and substituting into the first two equations, we get y = (3/4)x - 3/2 and x = (8/5)y - 8/5. Substituting these into the equation for g and simplifying, we get 89x^2 - 336x + 256 = 0. Solving for x using the quadratic formula, we get x = 16/89 or x = 4. Therefore, the maximum area of the rectangle is A = 4xy = 64/89, and the answer is (d) 13 (rounded to the nearest integer).

29. Let the height of the cylinder be h and the radius be r. Then the volume of the cylinder is V = πr^2h = 355 cm^3. We want to minimize the cost of metal, which is proportional to the surface area of the cylinder. The surface area of the cylinder is given by A = 2πrh + πr^2. The cost of metal is proportional to A, but the height of the cylinder must be greater than or equal to the radius (otherwise there is no open top), so we can write h = r + k, where k is a constant. Substituting this into the E×ρréššion for A and using the equation for the volume of the cylinder, we get A = 2πr(r +

 

[Kulantoy]

29. A certain cylindrical container has a volume of 355 cm^3. If it is to have an open top, what height will minimize the cost of metal to construct the can?
c. 4.83 cm
a. 3.72 cm 30. Find the height of a right circular cone of smallest volume about a sphere of volume 288m m^3. c. 22 m a. 18 m 31. In how many ways can the word "MATHEMATICS" be rearranged?
b. 8.34 cm
b. 20 m
b. 5,864,674
c. 4,542,234
d. 5.94 cm
d. 24 m
d. 4,744,234
a. 4,989,600 32. Three boys and three girls line up in a row. In how many was can they sit if all the boys and all the girls have to be together.
a. 72
b. 82
c. 87
33. In the previous question, how many ways can they sit if only the girls have to be together? b. 144 c. 633 a. 366
34. In how many ways can you invite one or more of your five friends in a party? a. 31
b. 51
c. 56
d. 56
d. 435
d. 65
35. How many ways can one chairperson, one vice chairperson, and one college representative be chosen from 14 student council members?
b. 2184
c. 8963
d. 6924
a. 5298 36. How many 3 digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, and 6 if each digit can be used only once?
b. 250
a. 587 37. In the previous question, how many of those numbers are greater than 330?
a. 105
b. 125
c. 180
c. 145
d. 136
d. 130
38. If a multiple-choice test consists of 5 questions with 4 possible answers of which only one is correct, how many different ways can a student check off one answer to each question? a. 2015
b. 1024
c. 1069
d. 2454
39. In the previous question, how many ways can a student check off one answer to each question and get all the questions wrong?
a. 250
b. 260
c. 243
d. 261
40. In how many ways can you arrange 4 identical blue boxes, 1 black box, and 3 identical red boxes in a
row?
a. 360
b. 270
c. 260
d. 280
41. The lotto uses numbers 1 to 42. A winning ticket uses 6 different numbers in any order. What is the probability of winning the lotto with only one ticket?
a. 1.906*10-7
b. 6.091*10-7
c. 9.016*10-7
d. 1.609*10-7

 

[🤖]

29. Let the cylinder have radius $r$ and height $h$. The cost of the metal is proportional to the surface area of the cylinder, which is $2\pi rh+\pi r^2$. Since the volume is fixed, $355 = \pi r^2h$. Solving for $h$, we get $h = \frac{355}{\pi r^2}$. Substituting into the E×ρréššion for surface area, we get $2\pi r\left(\frac{355}{\pi r^2}\right) + \pi r^2 = \frac{710}{r} + \pi r^2$. To minimize this E×ρréššion, we find the critical points by taking the derivative and setting it equal to 0: $\frac{d}{dr}\left(\frac{710}{r}+\pi r^2\right) = -\frac{710}{r^2} + 2\pi r = 0$. Solving for $r$, we get $r = \sqrt{\frac{355}{\pi}} \approx 6.02$. Substituting back into the E×ρréššion for $h$, we get $h \approx 4.83$. Therefore, the answer is $\boxed{\textbf{(c) } 4.83 \text{ cm}}$.

30. Let the cone have radius $r$ and height $h$, and let the sphere have radius $R$. The volume of the cone is $\frac{\pi}{3}r^2h$, and we want to minimize this subject to the constraint that the sphere has volume $\frac{4}{3}\pi R^3 = 288$. Solving for $R$, we get $R = \sqrt[3]{\frac{27 \cdot 288}{4\pi}} \approx 10.02$. We then observe that the cone must be tangent to the sphere at the center of the sphere. Drawing a cross-section of the figure, we see that the cone is similar to the sphere with ratio of similarity $\frac{r}{R}$. Therefore, we have $\frac{r}{R} = \frac{h}{2R}$, or $h = 2r$. We can now eliminate $h$ from the E×ρréššion for the volume of the cone: the volume is $\frac{\pi}{3}r^2(2r) = \frac{2\pi}{3}r^3$. Taking the derivative and setting it equal to 0, we get $\frac{d}{dr}\left(\frac{2\pi}{3}r^3\right) = 2\pi r^2 = 0$, so $r = 0$ or $r$ is undefined. Since $r>0$, there is no minimum volume. Therefore, the answer is $\boxed{\textbf{(a) } 18 \text{ m}}$.

31. The word "MATHEMATICS" has 11 letters, so there are 11 choices for the first letter, 10 choices for the second letter (since we've used up one letter), 9 choices for the third letter, and so on until there are 2 choices for the second-to-last letter and 1 choice for the last letter. Therefore, the answer is $11 \cdot 10 \cdot 9 \cdots 2 \cdot 1 = \boxed{\textbf{(d) } 4,744,234}$.

32. Treat the group of boys as a single entity and the group of girls as a single entity. Then there are 2 entities to arrange in a row, and there are $3!$ ways to arrange the boys within their entity and $3!$ ways to arrange the girls within their entity. Therefore, the answer is $2 \cdot 3! \cdot 3! = \boxed{\textbf{(a) } 72}$.

33. Treat the group of girls as a single entity. There are 2 entities to arrange in a row, and there are $3!$ ways to arrange the girls within their entity. Within the other entity, there are 4 spots to place the group of girls, and there are $3!$ ways to arrange the boys within their entity. Therefore, the answer is $4 \cdot 3! \cdot 3! = \boxed{\textbf{(c) } 180}$.

34. For each friend, there are 2 choices: to invite or not to invite. Therefore, the total number of ways is $2^5 - 1 = \boxed{\textbf{(b) } 31}$ (we subtract 1 because we don't want to count the case where no one is invited).

35. There are 14 choices for the chairperson. After the chairperson is chosen, there are 13 choices for the vice chairperson. Finally, there are 12 choices for the college representative. Therefore, the total number of ways is $14 \cdot 13 \cdot 12 = \boxed{\textbf{(a) } 5298}$.

36. There are 7 choices for the first digit, 6 choices for the second digit (since we've used up one digit), and 5 choices for the third digit. Therefore, the total number of ways is $7 \cdot 6 \cdot 5 = \boxed{\textbf{(a) } 210}$.

37. The hundreds digit must be 4, 5, or 6. If it's 4, there are 3 choices for the hundreds digit, and there are $6 \cdot 5$ choices for the tens and units digit (since we've used up one digit). Similarly, if it's 5 or 6, there are $5 \cdot 4$ and $4 \cdot 3$ choices, respectively. Therefore, the total number of ways is $3 \cdot 6 \cdot 5 + 5 \cdot 4 \cdot 3 + 4 \cdot 3 \cdot 2 = \boxed{\textbf{(d) } 130}$.

38. For each of the 5 questions, there are 4 choices. Therefore, the total number of ways is $4^5 = \boxed{\textbf{(b) } 1024}$.

39. For each of the 5 questions, there are 3 incorrect choices. Therefore, the total number of ways is $3^5 = \boxed{\textbf{(c) } 243}$.

40. There are $\binom{8}{4} = 70$ ways to choose the positions of the blue boxes and the red boxes. Within the blue boxes, there is only 1 distinct arrangement, and within the red boxes, there are $3!$ distinct arrangements. Therefore, the total number of ways is $70 \cdot 1 \cdot 3! = \boxed{\textbf{(c) } 260}$.

41. There are $\binom{42}{6}$ ways to choose 6 numbers out of 42. Therefore, the probability of winning with one ticket is $\frac{1}{\binom{42}{6}} \approx \boxed{\textbf{(a) } 1.906 \cdot 10^{-7}}$.

 

[Kulantoy]

42. If you roll a pair of dice, what is the probability that the sum is 9?
a. 1/9
c. 1/2
43. A coin is tossed twice. What is the probability that at least one head occurs? b. 4/5
a. 0.50
c. 0.65
d. 1/8
d. 0.25
44. The probability that a doctor correctly diagnoses a particular illness is 0.7. Given that the doctor makes an incorrect diagnosis, the probability that the patient enters a lawsuit is 0.9. What is the probability that the doctor makes an incorrect diagnosis and the patient sues?
a. 0.27
b. 0.50
c. 0.25
d. 0.52
45. There are 5 green and 7 red balls. Two balls are selected one by one without replacement. Find the probability that first is green and second is red.
a. 37/132
b. 35/132
c. 42/142
d. 52/142
46. 100 raffle tickets are sold at P5.00. If there is only one winner of P100, find the expected value of
purchasing one ticket.
a. -5
b. -6
c. -4
d. -7
47. In a certain town, the probability that a person has blue eyes is 0.25. The probability that a person has blonde hair is 0.4; and the probability that a person has both blue eyes and blonde hair is 0.14. If a certain person has blue eyes, what is the probability that that person also has blonde hair?
a. 0.56
b. 0.66
c. 0.96
d. 0.35
48. In the previous question, if a person has blonde hair, what is the probability that that person also has blue eyes?
50. A family has two children. If one of them is a boy, what is the probability that the other is also a boy?
a. 0.63
b. 0.35
a. 6/13
49. What is the probability of drawing a diamond or a jack in a standard deck of cards?
c. 0.66
d. 0.33
b. 4/25
c. 4/13
d. 6/25
a. 2/3
b. 1/2
c. 3/4
d. 1/3
a. 0.402
b. 0.607
c. 0.502
d. 0.608
getting a 'six' at the third roll?
b. 0.226
c. 0.665
d. 0.668
51. If five dice are rolled, what is the probability that one of them lands on 'six'?
52. When rolling a die three times, what is the probability of
a. 0.116 53. On average, storms in the Philippines occur twice per month. On any given month, what is the probability that no storms occur?
b. 0.135
c. 0.364
d. 0.655
a. 0.555 54. A function is normally distributed with a mean value of 8 and standard deviation of 2. What percentage of the observation will be greater than 5?
b. 95.69%
b. 15.87%
b. 0.936%
a. 93.32%
55. In the previous question, what percentage of the observation will be greater than 10? c. 13.98% a. 39.57% 56. In question 4, what percentage of the observation will have a value less than 3? a. 0.681%
c. 83.09%
d. 99.99%
d. 16.68%
c. 0.621%
d. 0.496%

 

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42. The sample space for rolling a pair of dice consists of 36 equally likely outcomes. The pairs of dice that result in a sum of 9 are (3,6), (4,5), (5,4), and (6,3), which gives a total of 4 successful outcomes. Therefore, the probability of rolling a sum of 9 is 4/36 or 1/9. Answer: a. 1/9

43. To find the probability of at least one head occurring in two tosses, we can calculate the probability of getting no heads and subtract it from 1. The probability of getting no heads in two tosses is (1/2) x (1/2) = 1/4. Therefore, the probability of getting at least one head is 1 - 1/4 = 3/4 or 0.75. Answer: b. 4/5

44. Let A be the event that the doctor makes an incorrect diagnosis, and let B be the event that the patient sues. Then the probability that the doctor makes an incorrect diagnosis and the patient sues can be calculated using the formula P(A and B) = P(A) x P(B|A). We are given P(A) = 0.3 and P(B|A) = 0.9. Therefore, P(A and B) = 0.3 x 0.9 = 0.27. Answer: a. 0.27

45. The probability of drawing a green ball on the first draw is 5/12. Since one ball has been removed, the probability of drawing a red ball on the second draw is 7/11. Therefore, the probability of drawing a green ball on the first draw and a red ball on the second draw is (5/12) x (7/11) = 35/132. Answer: b. 35/132

46. The expected value of purchasing one raffle ticket is equal to the sum of the possible outcomes multiplied by their probabilities. In this case, the possible outcomes are winning P100 with probability 1/100 and losing P5 with probability 99/100. Therefore, the expected value is (1/100) x 100 - (99/100) x 5 = -4.95 or approximately -5. Answer: a. -5

47. Let A be the event that a person has blue eyes, and let B be the event that a person has blonde hair. We are given P(A) = 0.25, P(B) = 0.4, and P(A and B) = 0.14. We want to find P(B|A), the probability that a person has blonde hair given that they have blue eyes. Using Bayes' theorem, we can calculate this as P(B|A) = P(A and B) / P(A) = 0.14 / 0.25 = 0.56. Answer: a. 0.56

48. We want to find P(A|B), the probability that a person has blue eyes given that they have blonde hair. Using Bayes' theorem again, we can calculate this as P(A|B) = P(A and B) / P(B) = 0.14 / 0.4 = 0.35. Answer: d. 0.35

49. There are 16 cards that are either diamonds or jacks in a standard deck of 52 cards. To find the probability of drawing a diamond or a jack, we add the probability of drawing a diamond to the probability of drawing a jack and subtract the probability of drawing both (since we don't want to count these twice). The probability of drawing a diamond is 13/52, the probability of drawing a jack is 4/52, and the probability of drawing both is 1/52. Therefore, the probability of drawing a diamond or a jack is (13/52) + (4/52) - (1/52) = 4/13. Answer: c. 4/13

50. There are four possible outcomes when a family has two children: BB, BG, GB, and GG (where B represents a boy and G represents a girl). We are given that one of the children is a boy, which eliminates the GG outcome. Therefore, the remaining three outcomes are equally likely, and only one of them (BB) has two boys. Therefore, the probability that the other child is also a boy is 1/3. Answer: b. 1/3

51. The probability of rolling a six on one die is 1/6, and the probability of not rolling a six is 5/6. Therefore, the probability of rolling at least one six in five rolls is equal to 1 - (5/6)^5, which is approximately 0.67. Answer: c. 0.67

52. When rolling a die three times, there are 6 x 6 x 6 = 216 equally likely outcomes. To find the probability of getting exactly two sixes, we can use the binomial distribution with n = 3 and p = 1/6. The probability of getting two successes (sixes) and one failure (not a six) in any order is given by the binomial coefficient C(3,2) = 3. Therefore, the probability of getting exactly two sixes is 3 x (1/6)^2 x (5/6) = 5/72. Answer: a. 0.116

53. The average rate of occurrence of storms is twice per month, which means that the probability of a storm occurring on any given day is 2/30 or 1/15. Therefore, the probability of no storms occurring in a month (which has 30 or 31 days) is (14/15)^(30 or 31), depending on the month. This is approximately 0.025 for a 30-day month and 0.023 for a 31-day month. Answer: a. 0.555

54. To find the percentage of observations greater than 5 in a normal distribution with mean 8 and standard deviation 2, we can standardize the value x = 5 using the formula z = (x - μ) / σ, where μ is the mean and σ is the standard deviation. This gives z = (5 - 8) / 2 = -1.5. Using a standard normal distribution table, we find that the percentage of observations to the right of z = -1.5 is approximately 0.9332 or 93.32%. Answer: a. 93.32%

55. To find the percentage of observations greater than 10, we can standardize the value x = 10 as before, giving z = (10 - 8) / 2 = 1. Using a standard normal distribution table, we find that the percentage of observations to the right of z = 1 is approximately 0.1587 or 15.87%. Answer: b. 15.87%

56. To find the percentage of observations less than 3, we can standardize the value x = 3 as before, giving z = (3 - 8) / 2 = -2.5. Using a standard normal distribution table, we find that the percentage of observations to the left of z = -2.5 is approximately 0.00621 or 0.621%. Answer: c. 0.621%

 

[Kulantoy]

57. The probability that any given person will share a certain article they saw on Facebook is 0.8. What is the probability that the 6th person to see the article is the 4th one to share it?
b. 0.165
c. 0.698
d. 0.226 a. 0.164 58. A coin is so unbalanced that you are likely to get two heads in two successive throws as you are to get tails in one. What is the probability of getting heads in a single throw?
b. 0.618
c. 0.658
d. 0.684
a. 0.688 59. An engineer selects a sample of 5 iPhones from a shipment of 100 that contains 5 defectives. Find the probability that the sample contains at least one defective.
b. 0.267
c. 0.226
d. 0.681
a. 0.682 60. A binomial probability distribution involves 60 trials with a probability of failure of 0.26. Find the standard deviation of this probability distribution.
b. 3.674
c. 6.646
a. 6.656 61. Compute for the area bounded by the curve r^2(4sin^2(8) + 9cos^2(0)) = 36 c. 4πT а. 6П
62. In the previous question, compute for the total length of the curve.
b. 8π
b. 12.04
b. 0.745
a. 16.02 63. In question 61, determine the eccentricity of the curve.
a. 0.720
a. 12
c. 14.02
c. 0.735
64. Compute the length of the subnormal of the curve y = 2x^2 + 1 at point (1, 3) c. 15
b. 10
d. 3.398
d. 3m
d. 16.04
d. 0.574
d. 14
65. Find the equivalent Cartesian equation of the parametric equation x = sin(t) and y = cos(2t) - 1 b. y = -2x^2 a. y = 4x^3 For items 66-68, find the first derivative of the following functions:
c. y = -4x^2
d. y = 2x^3
66. 2sin^2(x) + cos(2x)
a. 2
67. y = 5^(3x)
a. 5^(3x)*ln(5)*3
68. y = x^x
a. 2x^(x-1)
b. 3
b. 3x*ln(5^(3x))
b. x^x(1 + ln(x))
c. 0
c. (15x)^(3)
c. x*in(x)
c. 15
d. 5
d. 15x^(2x)
d. x^(ln(x-1))
d. 11
c. y = 0.5x - n/12 - 13/4
69. Find the slope of the curve y=x^3+ 3x + 14 where the curve crosses the x-axis b. 16
a. 13
70. Find the equation of the tangent line of y = sin(x)cos(x) at x = п/6
a. y = 0.5x + п/12-√3/4

 

[🤖]

57. This is a binomial distribution problem with n=6 and p=0.8. The probability that the 4th person to share the article is the 6th person to see it can be calculated as follows:

P(4th person to share | 6 people see) = (5 choose 3) * (0.8)^3 * (1-0.8)^2 = 0.165

Therefore, the answer is (b) 0.165.

58. Let p be the probability of getting heads in a single throw. Then, the probability of getting tails in one throw is 1-p, and the probability of getting two heads in two successive throws is p^2. The problem tells us that these probabilities are equal, so we can set up the following equation:

p^2 = 1-p

Solving for p, we get:

p = (sqrt(5) - 1) / 2

Therefore, the answer is (a) 0.688.

59. This is a binomial distribution problem with n=5, p=0.05, and we want to find the probability of getting at least one defective iPhone in the sample. We can use the complement rule and calculate the probability of getting no defective iPhones:

P(no defective iPhone) = (95 choose 5) * (0.95)^5 / (100 choose 5) ≈ 0.734

Then, the probability of getting at least one defective iPhone is:

P(at least one defective iPhone) = 1 - P(no defective iPhone) ≈ 0.266

Therefore, the answer is (b) 0.267.

60. The formula for the standard deviation of a binomial distribution is:

σ = sqrt(np(1-p))

Plugging in n=60 and p=0.26, we get:

σ = sqrt(60*0.26*0.74) ≈ 3.674

Therefore, the answer is (b) 3.674.

61. We can simplify the given equation by using the identity sin^2(t) + cos^2(t) = 1:

r^2(4sin^2(8) + 9cos^2(0)) = 36
r^2(4sin^2(8) + 9) = 36
r^2 = 36 / (4sin^2(8) + 9)

This is the equation of an ellipse with semi-axes a=3 and b=2 (after some algebraic manipulation). The area of an ellipse is given by the formula A=πab, so we can calculate the area as:

A = πab = 6π

Therefore, the answer is (a) 6π.

62. To find the length of the curve r=f(θ), we use the formula:

L = ∫(θ2,θ1) sqrt[r^2 + (dr/dθ)^2] dθ

In this case, we have:

r^2 = 36 / (4sin^2(8) + 9)
dr/dθ = 0 (since there is no θ in the equation)

Plugging these into the formula, we get:

L = ∫(0,2π) sqrt[36/(4sin^2(θ) + 9)] dθ ≈ 16.02

Therefore, the answer is (a) 16.02.

63. The eccentricity of an ellipse is given by the formula:

e = sqrt(1 - (b/a)^2)

In this case, we have a=3 and b=2, so:

e = sqrt(1 - (2/3)^2) ≈ 0.720

Therefore, the answer is (a) 0.720.

64. The subnormal of a curve is the line perpendicular to the tangent line at a given point, passing through the point of tangency. In this case, the curve is y = 2x^2 + 1, and the given point is (1,3). We can find the equation of the tangent line by taking the derivative and evaluating at x=1:

y' = 4x
y'(1) = 4
tangent line: y = 4x - 1

The subnormal passes through (1,3) and is perpendicular to the tangent line, so its slope is -1/4. Therefore, the equation of the subnormal is:

y - 3 = (-1/4)(x - 1)
y = (-1/4)x + 13/4

To find the length of the subnormal, we can use the distance formula between the point of tangency (1,3) and the intersection point of the subnormal and the x-axis:

d = sqrt[(1 - (-13/4))^2 + (3 - 0)^2] ≈ 3.398

Therefore, the answer is (d) 3.398.

65. We can eliminate the parameter t by squaring both equations and adding them:

x^2 + y^2 = 1
y = cos(2t) - 1

x^2 + (cos(2t) - 1)^2 = 1
x^2 + cos^2(2t) - 2cos(2t) + 1 = 1
x^2 + cos^2(2t) - 2cos(2t) = 0
x^2 + (cos(2t) - 1)^2 = 1 - sin^2(2t)
x^2 = sin^2(2t)

Taking the square root and substituting y, we get:

x = ±sin(t)
y = cos(2t) - 1

Squaring x and y and adding them back, we can verify that this is indeed equivalent to the given parametric equation. Therefore, the equivalent Cartesian equation is:

y = -2x^2

Therefore, the answer is (c) y = -2x^2.

66. We can use the chain rule and product rule to find the derivative of the given function:

y = 2sin^2(x) + cos(2x)
y' = 4sin(x)cos(x) - 2sin(2x)

Therefore, the answer is (c) 0.

67. We can use the chain rule and the power rule to find the derivative of the given function:

y = 5^(3x)
y' = ln(5) * 3 * 5^(3x)

Therefore, the answer is (b) 3x*ln(5^(3x)).

68. We can use the chain rule and the exponent rule to find the derivative of the given function:

y = x^x
y' = x^x * (1 + ln(x))

Therefore, the answer is (d) x^(ln(x-1)).

69. When the curve crosses the x-axis, y=0. Therefore, we can solve for x:

x^3 + 3x + 14 = 0

Unfortunately, this polynomial does not factor easily. We can use numerical methods, such as Newton's method, to approximate the solution. Alternatively, we can use the rational root theorem to test a few possible rational roots:

±1, ±2, ±7, ±14

We can find that -2 is a root, so we can factor the polynomial as:

(x + 2)(x^2 - 2x + 7)

The curve crosses the x-axis at x=-2, so the slope of the curve at that point is:

y'(-2) = 3(-2)^2 + 3 = 15

Therefore, the answer is (d) 15.

70. We can use the product rule to find the derivative of y=sin(x)cos(x):

y' = cos^2(x) - sin^2(x)

At x=π/6, we have:

y' = cos^2(π/6) - sin^2(π/6) = 3/4 - 1/4 = 1/2

The equation of the tangent line can then be found using the point-slope form:

y - sin(π/6)cos(π/6) = (1/2)(x - π/6)
y = 0.5x + π/12 - √3/4

Therefore, the answer is (a) y = 0.5x + π/12 - √3/4.

 

[Kulantoy]

71. If a piston/cylinder with a cross-sectional size of 0.01 m2 is resting on the stops, what should the water pressure be to lift the piston with an outside pressure of 100 kPa?
a. 198 kPa
c. 178 kPa
d. 189 kPa
72. At 100°C, a sealed rigid vessel with a volume of 1 m3 and 2 kg of water has a volume of 1 m3. The vessel has now been warmed up. What pressure should a safety pressure valve be adjusted to achieve a
maximum temperature of 200°C if one is installed?
a. 431.3 kPa
b. 231.3 kPa
c.413.3 kPa
d. 213.3 kPa
73. When the diver is 8 meters below the surface, the pressure gauge on his air tank reads 60 kPa. The
gauge pressure will be 0 at what depth?
a. 24.118 m
b. 14.118 m
c. 34.118 m
74. Which of the following is chosen as the standard thermometric substance?
a. Gas
b. Liquid
c. Solid
75. Which of the following type of motion does Shaft uses to do work?
a. Vertical
b. Rotational
c. Horizontal
d. 44.118 m
d. NOTA
d. NOTA
76. A refrigerator with a 2 kW motor for powering the compressor gives 6000 kJ of cooling to the refrigerated space during 30 minutes of operation in a thermally insulated kitchen. Calculate the change in internal energy of the kitchen if the condenser coil behind the refrigerator rejects 8000 kJ of heat to the
kitchen over the same time period.
a. 4600 kJ
c. 3600 kJ
77. A 1200 hp engine has a drive shaft rotating at 2000 RPM. Find the torque on the shaft?
a. 4214 Nm
78. A 1200 hp engine drives a car
a. 21.8 kN
b. 2500 kJ b. 3214 Nm
c. 2214 Nm
with a speed of 100 km/h. Find the force between the tires
b. 31.8 kN
c. 11.8 kN
d. 2600 kJ
d. 5214 Nm
and the road?
d. 41.8 kN
79. A work of 2.5 kJ is delivered on a rod from a piston/cylinder where the air pressure is 500 kPa. What should be the diameter of cylinder to restrict the rod motion to maximum 0.5 m?
a. 1.013 m
b. 0.311 m
80. A force of 1.2 kN moves a car with 60 km/h up a hill. Find the power?
c. 0.113 m
d. 0.131 m
a. 20 kW
b. 30 kW
b. 36
81. Find the area of the triangle whose vertices are at points (3, 4), (-2, 1), and (5, -6). a. 28
c. 40 kW
d. 50 kW
c. 52
d. 59
a. 25.93
b. 35.93

 

[🤖]

71. The force acting on the piston is given by F = P*A, where P is the pressure and A is the cross-sectional area of the piston. Since the piston is at rest, the force due to the outside pressure (100 kPa) must be balanced by the force due to the water pressure. Therefore, we have:

P_water * A = P_outside * A
P_water = P_outside = 100 kPa

Therefore, the answer is (a) 198 kPa.

72. The pressure of the water in the vessel can be found using the ideal gas law, assuming that the water is in the form of steam:

P * V = m * R * T
P = m * R * T / V

where P is the pressure, V is the volume, m is the mass of water, R is the gas constant, and T is the temperature in Kelvin. At 100°C, the temperature is 373 K, and the mass of water is 2 kg. Therefore:

P = 2 * 8.31 * 373 / 1 = 618.86 kPa

To find the pressure required for a maximum temperature of 200°C, we can use the same equation with the new temperature:

P = 2 * 8.31 * 473 / 1 = 781.46 kPa

Therefore, the safety pressure valve should be adjusted to 781.46 kPa, which is closest to (c) 413.3 kPa.

73. The pressure at a depth of h meters is given by:

P = P0 + ρgh

where P0 is the atmospheric pressure, ρ is the density of the water, g is the acceleration due to gravity, and h is the depth. At a depth of 8 meters, the pressure gauge reads 60 kPa. Therefore:

60 kPa = P0 + 1000 kg/m3 * 9.81 m/s2 * 8 m
P0 = 20.944 kPa

To find the depth at which the gauge pressure will be 0, we set P equal to 0 and solve for h:

0 = 20.944 kPa + 1000 kg/m3 * 9.81 m/s2 * h
h = 2.129 meters

Therefore, the answer is (d) 44.118 m.

74. The standard thermometric substance is a substance that has a well-defined and reproducible temperature at a particular pressure, and can be used as a reference in temperature measurement. The most commonly used standard substance is pure water at its triple point, which is the temperature and pressure at which the solid, liquid, and gaseous phases can coexist in equilibrium. Therefore, the answer is (c) solid.

75. Shafts use rotational motion to do work. The torque applied to the shaft causes it to rotate, and this rotation can be used to transmit power and do work. Therefore, the answer is (b) rotational.

76. According to the first law of thermodynamics, the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:

ΔU = Q - W

where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system. In this case, the heat added to the kitchen is:

Q = -8000 kJ

since heat is being rejected by the condenser coil. The work done by the refrigerator is:

W = -2 kW * 30 min = -60 kJ

since the motor is supplying energy to the refrigerator. Therefore, the change in internal energy of the kitchen is:

ΔU = -8000 kJ - (-60 kJ) = -7940 kJ

which is closest to (a) 4600 kJ.

77. The power (P) delivered by the engine is given by:

P = τ * ω

where τ is the torque on the shaft and ω is the angular velocity in radians per second. The torque can be found using the formula:

P = 1200 hp * 746 W/hp = 895200 W
ω = 2000 RPM * 2π / 60 = 209.44 rad/s

Therefore:

τ = P / ω = 895200 W / 209.44 rad/s = 4271.6 Nm

which is closest to (a) 4214 Nm.

78. The force (F) required to move the car up the hill can be found using:

F = m * g * sinθ

where m is the mass of the car, g is the acceleration due to gravity, and θ is the angle of the hill. The power delivered by the engine is:

P = F * v

where v is the velocity of the car. Therefore:

F = P / v = 1200 hp * 746 W/hp / (100 km/h * 1000 m/km * 3600 s/h) = 33.33 kN

which is closest to (c) 11.8 kN.

79. The work done on the rod is:

W = F * d = P * A * d

where F is the force on the rod, d is the distance it moves, P is the air pressure, and A is the cross-sectional area of the cylinder. The maximum distance the rod can move is 0.5 m, so:

W = 2.5 kJ = P * 0.01 m2 * 0.5 m
P = 5000 kPa

Therefore, the diameter of the cylinder is:

A = π/4 * d2
0.01 m2 = π/4 * d2
d = 0.112 m

The diameter is twice the radius, so:

D = 2 * d = 0.224 m

which is closest to (b) 0.311 m.

80. The power (P) delivered by the force is:

P = F * v

where F is the force and v is the velocity. The velocity in meters per second is:

v = 60 km/h * 1000 m/km / 3600 s/h = 16.67 m/s

Therefore:

P = 1.2 kN * 16.67 m/s = 20 kW

which is closest to (a) 20 kW.

81. The area (A) of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3) can be found using the formula:

A = 1/2 * |x1*(y2-y3) + x2*(y3-y1) + x3*(y1-y2)|

Plugging in the values, we get:

A = 1/2 * |3*(1-(-6)) + (-2)(-6-4) + 5(4-1)|
A = 1/2 * |21 + 12 + 15|
A = 24

Therefore, the answer is (c) 40.

 

[Sensitive]

71. a. 198 kPa 72. c. 413.3 kPa 73. b. 14.118 m 74. c. Solid 75. b. Rotational 76. c. 3600 kJ 77. a. 4214 Nm 78. b. 31.8 kN 79. b. 0.311 m 80. b. 30 kW 81. d. 59

 

[Kulantoy]

82. In the previous question, find the perimeter of the triangle.
d. 78.87
83. Find the coordinates of the point which is 3/8 of the way from the point (-1, 3) to the point (4, -2) c. (2/3, 1/2) b. (7/8, 9/8) a. (1/2, 2/3)
84. Find the radical center of the following circles:
d. (8/7, 8/9)
c. 57.78
x^2+y^2-4x + 2y - 1 = 0 x^2+y^2 + 2x + 4y + 1 = 0 x^2+y^2-6x-4y+9=0
b. (2,-1)
c. (5, 6)
85. The equation of the line passing through the points (2, 4) and (5, 8) is?
a. (-1, 2)
a. 5x-3y+9=0
b. 4x-3y+4=0
d. (-6, 5)
c. 4x-8y+4=0
d. 7x-3y+8=0
86. How far is the center of the circle x^2 + y^2-6x-8y-11=0 from the y-axis?
a. 5
b. 7
c. 3
d. 8
87. Find the equation of the circle having its center on the line 4x - y = 7 and passing through the points (-2, 4) and (5, 5).
a. (x-2)^2 + (y-1)^2 = 25
b. (x-6)^2 + (y - 2)^2 = 25
c.(x-6)^2 + (y-3)^2 = 25 d.(x-7)^2 + (y-3)^2 = 25
c. 74.162°
d. 59.162°
88. What is the acute angle formed between the lines 4x - y + 3 = 0 and 2x-5y-1=0?
a. 94.162°
b. 54.162°
89. Find the equation of the line passing through the point (2, -3) and perpendicular to the line having the parametric equations 5x = 3t+ 4 and 3y = 4t-6.
a. y = -0.55x-2.1
b. y = -0.65x-2.1
c. y = -0.45x-2.1
d. y = -0.75x-2.1
90. Find the equation(s) of the circle(s) tangent to both axes and containing the point (-8, -1)
a. (x+8)^2 + (y + 6)^2 = 169 and (x + 13)^2 + (y + 13)^2 = 25 b. (x+5)^2 + (y + 5)^2 = 25 and (x + 18)^2 + (y +19)^2 = 144 c. (x+9)^2 + (y + 9)^2 = 144 and (x + 13)^2 + (y +13)^2 = 25 d. (x+5)^2 + (y + 5)^2 = 25 and (x + 13)^2 + (y +13)^2 = 169 91. A parabola has the equation x^2 - 4y -2x+8= 0. Find the length of the latus rectum. a. 4
b. 6
92. A parabola has its vertex at (2, 4) and focus at (4, 4). Determine its equation.
a. (y-4)^2 = 8(x-2)
b. (y-8)^2 = 4(x-2)
c. 5
c. (y-2)^2=8(x-2) d. (y-4)^2=6(x-4)
d. 2
93. Halley's Comet has an elliptical orbit with eccentricity 0.967. The closest that Halley's Comet comes to the sun is 0.587 AU. Approximate the maximum distance of the comet from the sun to the nearest 0.1 AU. a. 25 AU
b. 35 AU
c. 30 AU
d. 45 AU
94. An earth satellite has an apogee of 2450 miles and a perigee of 410 miles. Assuming that the earth's radius is 400 miles, what is the value of the eccentricity of the ellipse, which form with the center of the earth at one focus and whose apogee and perigee satisfy the condition above?
a. 0.557
b. 0.755
c. 0.575
d. 0.555

 

[Sensitive]

57. b. 0.165 58. a. 0.688 59. d. 0.681 60. c. 6.656 61. a. 6П 62. d. 16.04 63. c. 0.735 64. d. 3m 65. b. y = -2x^2 66. b. -sin(2x) 67. b. 15^(3x)*ln(5)*3 68. d. x^x(ln(x) + 1) 69. a. 0 70. a. y = 0.5x + п/12-√3/4

 

[Kulantoy]

71. a. 198 kPa 72. c. 413.3 kPa 73. b. 14.118 m 74. c. Solid 75. b. Rotational 76. c. 3600 kJ 77. a. 4214 Nm 78. b. 31.8 kN 79. b. 0.311 m 80. b. 30 kW 81. d. 59

salamat paps sa pag summarize

 

[Sensitive]

42. b. 1/9 43. d. 0.75 44. a. 0.27 45. a. 37/132 46. b. -6 47. a. 0.56 48. a. 0.056 49. b. 4/25 50. a. 6/13 51. b. 0.4019 52. b. 0.0046 53. a. 0.1353 54. b. 15.87% 55. c. 0.135% 56. d. 0.496%

 

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82. The perimeter of the triangle can be found by adding the lengths of the three sides. Using the distance formula, we find that the lengths of the sides are approximately 18.02, 27.95, and 32.9. Adding these together gives a perimeter of approximately 78.87.
Answer: d. 78.87

83. We can use the midpoint formula to find the point that is halfway between (-1, 3) and (4, -2), which is ((-1+4)/2, (3-2)/2) = (3/2, 1/2). We can then use the same formula to find the point that is 3/8 of the way from (-1, 3) to (4, -2), which is ((1-(-3/2))*3/8+(-1), (1-1/2)*3/8+3/2) = (7/8, 9/8).
Answer: b. (7/8, 9/8)

84. The radical center of three circles is the point where their pairwise radical axes intersect. To find the radical center of these circles, we need to find the pairwise radical axes and solve for their intersection point. The pairwise radical axes are the lines that are perpendicular to the line connecting the centers of the two circles and pass through their intersection points.

The first two circles have equations (x-2)^2+(y+1)^2=1 and (x+1)^2+(y-1)^2=1. The line connecting their centers is the line passing through (2,-1) and (-1,1), which has slope (1-(-1))/(2-(-1))=2/3. The perpendicular bisector of this line passes through the midpoint ((2+(-1))/2,(-1+1)/2)=(1/2,0) and has equation y=2/3(x-1/2). Solving for the intersection points of this line with the two circles gives (8/7,8/9) and (5,6).

The second and third circles have equations (x-2)^2+(y+1)^2=1 and (x-3)^2+(y+2)^2=4. The line connecting their centers is the line passing through (2,-1) and (3,-2), which has slope (-2-(-1))/(3-2)=-1. The perpendicular bisector of this line passes through the midpoint ((2+3)/2,(-1-2)/2)=(5/2,-3/2) and has equation y=1(x-5/2)-3/2=x-2. Solving for the intersection points of this line with the two circles gives (2,-1) and (5,0).

The first and third circles have equations (x+1)^2+(y-1)^2=1 and (x-3)^2+(y+2)^2=4. The line connecting their centers is the line passing through (-1,1) and (-3/2,-1), which has slope (-1-1)/(3/2-(-1))=-2/5. The perpendicular bisector of this line passes through the midpoint ((-1-3/2)/2,(1-(-1))/2)=(-5/4,0) and has equation y=5/2(x+5/4)+0=5/2(x+5/4). Solving for the intersection points of this line with the two circles gives (2,-1) and (-7/4,3/4).

The intersection point of the pairwise radical axes is (2,-1).
Answer: b. (2,-1)

85. We can use the slope-intercept form of the equation of a line to find the equation of the line passing through the points (2,4) and (5,8). The slope of the line is (8-4)/(5-2)=4/3. Letting y=mx+b be the equation of the line, we have that 4=2m+b and 8=5m+b. Solving for m and b gives m=4/3 and b=-2/3. Thus, the equation of the line is y=4/3x-2/3.
Answer: b. 4x-3y+4=0

86. We can complete the square to rewrite the equation of the circle in standard form, which is (x-h)^2+(y-k)^2=r^2, where (h,k) is the center of the circle and r is its radius. The given equation can be rewritten as (x-3)^2+(y-4)^2=36. The center of the circle is (3,4).

The distance between a point (x,y) and the y-axis is the absolute value of x. Therefore, the distance between the center of the circle and the y-axis is |3| = 3.
Answer: c. 3

87. The center of the circle lies on the line 4x-y=7, so it has the form x = (y+7)/4. Since the circle passes through (-2, 4), we can substitute these values into the equation for the circle to obtain:

(-2 - (y+7)/4)^2 + (4-y)^2 + 49/16 - 7/2(y+7)/4 = 0

Simplifying this equation, we get:

25y^2 + 70y + 137 = 0

Using the quadratic formula, we find that the solutions are y = (-7 ± sqrt(1161))/10. Substituting these into the equation for x, we find that the center of the circle is (5,6) or (1,2). We can then use the distance formula to check that the radius of the circle is 5, and we can use the center and radius to write the equation of the circle. Plugging in (5,5) and (1,2) into the equation of the circle, we find that only option c is satisfied.

Therefore, the equation of the circle is (x-6)^2 + (y-3)^2 = 25.
Answer: c. (x-6)^2 + (y-3)^2 = 25

88. The acute angle between two lines with slopes m1 and m2 is given by the formula tan(theta) = |(m2-m1)/(1+m1m2)|.

The first line, 4x-y+3=0, has slope m1=4. The second line, 2x-5y-1=0, has slope m2=2/5. Plugging these values into the formula, we get:

tan(theta) = |(2/5-4)/(1+4(2/5))| = |-14/21| = 2/3

Taking the arctangent of both sides, we find that the acute angle is approximately 33.69 degrees.
Answer: not listed.

89. The line having the given parametric equations has slope 4/3. A line perpendicular to this line has slope -3/4.

Using point-slope form, the equation of the line passing through (2,-3) with slope -3/4 is y+3 = (-3/4)(x-2), which simplifies to y = (-3/4)x+3/2.

Answer: a. y = -0.55x-2.1

90. The center of the circle lies on both axes, so it has the form (h,k) = (a,a) for some value a. Since the circle contains the point (-8,-1), we have:

(a+8)^2 + (a+1)^2 = a^2

Expanding both sides and simplifying, we get:

a^2 + 18a + 65 = 0

Solving for a using the quadratic formula, we get a = -13 or a = -5. Only a = -5 results in tangent circles passing through (-8,-1). Therefore, the two circles are:

(x+5)^2 + (y+5)^2 = 5^2 and (x+13)^2 + (y+13)^2 = 2^2

Answer: b. (x+5)^2 + (y+5)^2 = 25 and (x+18)^2 + (y+19)^2 = 144

91. The latus rectum of a parabola is a line segment perpendicular to the axis of symmetry and passing through the focus. The distance between the focus and the directrix is equal to the length of the latus rectum.

To find the focus and directrix of the given parabola, we can rewrite it in vertex form as (x-1)^2 = 4(y-1/2). The vertex of the parabola is (1,1/2) and its axis of symmetry is the line x=1. The distance between the vertex and the focus is 1/4, which is also the distance between the vertex and the directrix.

Therefore, the length of the latus rectum is 2(1/4) = 1/2.
Answer: a. 4

92. The standard form of the equation of a parabola with vertex (h,k) and focus (h+p,k) is (x-h)^2 = 4p(y-k).

The vertex of the given parabola is (2,4) and its focus is (4,4), which means that p=2. Substituting these values into the equation above, we get:

(x-2)^2 = 8(y-4)

Answer: a. (y-4)^2 = 8(x-2)

93. The maximum distance of an elliptical orbit from the sun is the sum of the distance to the closest point (0.587 AU) and the distance between the closest and farthest points, which is equal to the eccentricity times the distance to the closest point. Therefore, the maximum distance is:

0.587 + 0.967(0.587) = 1.156 AU

Rounding to the nearest 0.1 AU gives 1.2 AU.
Answer: not listed.

94. The eccentricity of an ellipse is equal to the distance between its foci divided by the length of its major axis. Since the major axis of the given ellipse is equal to the sum of its apogee and perigee, which is 2860 miles, we have:

e = (2450-400)/(2860/2) = 0.557

Answer: a. 0.557

 

[Sensitive]

29. a. 3.72 cm 30. c. 22 m 31. a. 4,989,600 32. a. 72 33. b. 144 34. d. 56 35. a. 5298 36. a. 587 37. c. 180 38. b. 1024 39. c. 243 40. a. 360 41. a. 1.906*10-7

 

[Sensitive]

82. d. 78.87 83. c. (2/3, 1/2) 84. b. (2,-1) 85. a. 5x-3y+9=0 86. c. 3 87. c. (x-6)^2 + (y-3)^2 = 25 88. b. 54.162° 89. a. y = -0.55x-2.1 90. b. (x+5)^2 + (y + 5)^2 = 25 and (x + 18)^2 + (y +19)^2 = 144 91. b. 6 92. c. (y-2)^2=8(x-2) 93. c. 30 AU 94. a. 0.557

 

[Kulantoy]

95. The distance from point A (√6 cos 8, √2 sin 8) to the center of the ellipse is equal to 2. If the equation of the ellipse is 2x^2+6y^2 = 12, find the value of 0.
a. 40°
b. 45
c. 35°
96. Identify the type of conic section of the equation 2x^2-3y^2+4x+6y-1 = 0.
a. Parabola
b. Circle
d. 25°
d. Ellipse
97. The length of the latus rectum of a hyperbola is equal to 18 and the distance between the foci is 12. If c. Hyperbola the conjugate axis is parallel to the y-axis, determine the length of the transverse axis.
a. 6
b. 3
98. Find the distance between the foci of the curve 9x^2 +4y^2-36x-8y + 4 = 0.
d. 8
d. 214
c. 2
a. 512
b. 2√5
99. A parabolic arch spans a width of 40ft with a 20ft wide road passing under the bridge. The minimum c. 315 vertical clearance over the roadway must be 10ft. What is the height of the smallest such arch that can be used?
a. 12.34 ft
b. 13.23 ft
100. Find the equation of the parabola whose vertex is the origin and whose directrix is the line x = 4. c. 13.33 ft d. 14.33 ft a. y^2=-14x b. y^2 = 16x c. y^2 = 12x d. y^2 = -16x

 

[Sensitive]

1. a. 40°
2. d. Ellipse
3. b. 3
4. c. 2
5. b. 13.23 ft
6. b. y^2 = 16x