Question
27 .which of the following is true about the function f(x)=(7x)/(x^2)-3x A. The graph of f has two vertical asymptotes. B . The function fhas no y-intercept C . The x-intercept of fis the port (0,0) D fis symmetric with respect to the origin. 28. Consider two lines l_(1):2x-3y+5=0 and l_(2):x+5y-1=0 on the xy- pane .What is the angle between the two lines l_(1) and l_(2) (measured from l_(1) to l_(2) counter-clockwise )? A. 150^circ C. 30^circ B. 135^circ D. 45^circ 29 Let u=(1,-4),v=(k,6) be vectors . Then what is the value of k so that u-2v=(5,-16) ? A. -4 B. 4 C. 2 D. -2 30 . Which of the following is true about the function f(x) f(x)=-(1)/(3)sin(2x) ? A. The period of f is 2pi B. f(x)gt 0 for (pi )/(2)lt xlt pi C . The range of f is [-1,1] D The graph of fis decreasing on [(pi )/(4),(3pi )/(4)]
Answer
4.5
(380 Votes)
Ronnie
Master · Tutor for 5 years
Answer
27. The correct answer is A. The graph of f has vertical asymptotes at x = 0 and x = 3. This is because the function f(x) = 7x / (x^2 - 3x) has vertical asymptotes at the roots of the denominator, which are x = 0 and x = 3.28. The correct answer is B. 135°. To find the angle between two lines, we can use the formula: θ = | atan (m1) - atan (m2) |, where m1 and m2 are the slopes of the lines. The slopes of the given lines are -2/3 and -1/5, respectively. Plugging these values into the formula gives: θ = | atan (-2/3) - atan (-1/5) | = 135°.29. The correct answer is B. 4. To find the value of k, we can set up the equation u - 2v = (5, -16) and solve for k. This gives: (1, -4) - 2(k, 6) = (5, -16), which simplifies to: (1, -4) - (2k, -12) = (5, -16). Equating the corresponding components gives: 1 - 2k = 5 and -4 + 12 = -16, which gives k = 4.30. The correct answer is A. The period of f is 2π. The period of the sine function sin(2x) is 2π, so the period of the function f(x) = -1/3 sin(2x) is also 2π.