FaultyClockwork Report post Posted April 29, 2008 So my Math exam is Thursday and Friday and I'm having issues with some of the practice problems. Here's some I need done and explained: 1) Find the value of x: e^x + 1 = 4. 2) What is the period of the function y = 5 sin 3x? 52(pi)/532(pi)/3 3) If Jamar can run at 3/5 of a mile in 2 minutes 30 seconds, what is his rate in miles per minute? 4/56/253 and 1/104 and 1/6 4) A box contains one 2-inch rod, one 3-inch rod, one 4-inch rod, and one 5-inch rod. What is the maximum number of different triangles that can be made using these rods as sides? 1234 5) An archer shoots an arrow into the air such that its height at any time, t, is given by the function h(t) = -16t^2 + kt + 3. If the maximum height of the arrow occurs at t = 4, what is the value of k? 1286484 Share this post Link to post

Siendra Report post Posted May 1, 2008 It's been a long time... I'll give it a whirl, though. 1) Find the value of x: e^x + 1 = 4. log(e)^x = log(4) X log(e)= log (4) X = log (4)/log (x) make both sides a logarithm. Then bring down the X. Collect both logs to one side. When you have a log subtract a log it's treated as division. 3) If Jamar can run at 3/5 of a mile in 2 minutes 30 seconds, what is his rate in miles per minute? 4/56/253 and 1/104 and 1/6 6/25 He can run .6 miles in 2 minutes 30 seconds. .6/1 = 2.5/x and cross multiply it out. 4) A box contains one 2-inch rod, one 3-inch rod, one 4-inch rod, and one 5-inch rod. What is the maximum number of different triangles that can be made using these rods as sides? 1234 4. The possible combinations with fundamental counting are: 1x1x2 = 2 1x1x1 = 1 1x1x1 = 1 543 542 532 432 Your hypotenuse must ALWAYS be the 5 or 4 inch bar. After that you need one smaller bar followed again by another smaller bar. So if your hypotenuse is the 5 inch bar, the next bar must be four or three as. The last bar must always be three or two. Wish I could remember how the rest work. Share this post Link to post

David Report post Posted May 7, 2008 2) What is the period of the function y = 5 sin 3x? 52(pi)/532(pi)/3 The period of a function is, well, the length traveled until the function repeats itself. The period of y = sinx is 360 degrees, or 2pi. However, since the equation here is 5sin3x, the length of x necessary for one full period is cut in third. The period of y = 5sin3x is 120 degrees, or 2pi/3, which would be choice D. Simply remember the period of the basic functions, then divide that by any multipliers you have in the function. Or, you can just graph the function out and look at it. Periods of basic functions: sinx: 360 degrees secx: 360 degrees cosx: 360 degrees cscx: 360 degrees tanx: 180 degrees cotx: 180 degrees I can't remember the non-calculus way to do #5, and I'm sure you don't know calculus yet. Share this post Link to post

Siendra Report post Posted May 7, 2008 Oh, yeah. Memorize the unit circle. That thing is annoying as all hell, but really damn helpful. Share this post Link to post

Svenska Aeroplan Report post Posted May 7, 2008 I have no idea. You totally don't need this shit in real life. Share this post Link to post

David Report post Posted May 7, 2008 5) An archer shoots an arrow into the air such that its height at any time, t, is given by the function h(t) = -16t^2 + kt + 3. If the maximum height of the arrow occurs at t = 4, what is the value of k? 1286484 Since the equation given is a parabolic formula and it is a concave down parabola, the maximum height of the arrow occurs at the vertex of the parabola (if it is concave up, the minimum height occurs at the vertex instead). Use y = ax^2 + bx + c and what's known about the vertex to find k... y = -16x^2 + kx + 3 a = -16 b = k c = 3 The x coordinate of a vertex is -b/2a. It's given to us that it is also 4, so set -b/2a = 4 and use that to solve for b (which is also k). (-k)/(2)(-16) = 4 (-k)/(-32) = 4 k = 4(32) = 128 Share this post Link to post

Anime Gee Report post Posted May 7, 2008 Excellent work my mini geniuses! ( ^o^) <3 Share this post Link to post