Cdub100 3,903 Posted June 5, 2006 2a) Sketch the graph of the equation 2y + 3x = 1. 2b) What is the slope of the line in part (a)? 2y+3x=1 2y=-3x+1 ------------ = y= -1.5x+.5 (1.5 is your slope. Remember Y = mx+b m is your slope) 2 Set Y = 0 2(0)+3x=1 0+3x=1 3x=1 x=1/3 On your graph it should look like point at 1/2y and a point at 1/3x with a slope of 1.5 Share this post Link to post Share on other sites
Vikings4ever 550 Posted June 5, 2006 2y+3x=12y=-3x+1 ------------ = y= -1.5x+.5 (1.5 is your slope. Remember Y = mx+b m is your slope) 2 Set Y = 0 2(0)+3x=1 0+3x=1 3x=1 x=3 On your graph it should look like point at 2y and a point at 3x with a slope of 1.5 It's been a few years since I did precalc, but I'm thinking the graph part is wrong. y=.5 - 1.5x is right (easier to phrase it this way than to fock with negatives). Set x=0, y =.5 x=1, y=-1 x=2, y=-2.5 Which makes a straight line, first point at (0, .5), next point (1, -1), third point (2, -2.5). Share this post Link to post Share on other sites
Cdub100 3,903 Posted June 5, 2006 3) Your company has purchased a $12,000 machine that has a useful life of 8 years. The salvage value of the machine at the end of 8 years is $2,000. (Hint: on the y-axis indicate the value and on the x-axis the years.) Write a linear equation that describes the book value of the machine each year. What is the book value of the machine at the end of year 4? here are the numbers $12000 $2000 1 year 8 years Looks a lot like problem one right? What if I did this. (0,12000) (8, 2000) 2000-12000 -------------- = 8-0 -10000 -------- = -1250 (your slope) 8 Y=mx+b Y = Current price m = slope x= years b= Original purches price y=-1250(0)+12000 = 12000 y=-1250(4)+12000 = 7000 After 4 years the machine will be worth $7000 Share this post Link to post Share on other sites
CantTouchThis 23 Posted June 5, 2006 Just finishing up pre-calc right now, easy class, IMO, mainly just anaylzing graphs on a calculator, with quite a few formulas and stuff, most are easy. Share this post Link to post Share on other sites
nobody 2,667 Posted June 6, 2006 Just finishing up pre-calc right now, easy class, IMO, mainly just anaylzing graphs on a calculator, with quite a few formulas and stuff, most are easy. no one likes a braggart. Share this post Link to post Share on other sites
nobody 2,667 Posted June 6, 2006 here's your graph for 2a, and the answer to 2b is not 1.5. The answer is -1.5. If you leave off the negative, it's wrong. http://s62.photobucket.com/albums/h90/mjze...view¤t=line.jpg Solution 4=5(5)+-21 the answer you need to write down is actually y=5x-21 and the answer to part 1b is -1/5. That's just straight memorization. He wants you to know that if a line has a slope of m, then the perpendicular line has a slope of -1/m. Share this post Link to post Share on other sites
Birdseed 1 Posted June 6, 2006 I took Calc I, Calc II, Calc III, and Differential Equations. Reading this thread, I've come to the realization that I've wasted too many brain cells in the last 20 years. listen to nobody Share this post Link to post Share on other sites
CantTouchThis 23 Posted June 7, 2006 no one likes a braggart. Not bragging, just saying i thought it was an easy class that any one who can memorize formulas and analyze graphs can do. I had a much harder time with geometry, for some reason, i had the hardest time with that class. Share this post Link to post Share on other sites
nobody 2,667 Posted June 8, 2006 Yeah, all I'm saying is someone comes in struggling. You come in and say you think the class is easy. It makes no sense. Especially since at least half the posters here have taken much more advanced mathematics than your average joe will ever take. We got jerryskids from MIT. We got birdseed who's obviously an engineer. We got statitstics wizards all over the board. And you be knowin' Vikings4Ever gets all sexy with the math. Don't get me wrong. I like you. You're good people, but I stand by my belief that you were bragging, good buddy. Share this post Link to post Share on other sites
Vikings4ever 550 Posted June 8, 2006 And you be knowin' Vikings4Ever gets all sexy with the math. Parabolas make me horny. Share this post Link to post Share on other sites
Cdub100 3,903 Posted June 8, 2006 It's been a few years since I did precalc, but I'm thinking the graph part is wrong. y=.5 - 1.5x is right (easier to phrase it this way than to fock with negatives). Set x=0, y =.5 x=1, y=-1 x=2, y=-2.5 Which makes a straight line, first point at (0, .5), next point (1, -1), third point (2, -2.5). You're right I fixed my math I should of got y=1/2 x=1/3 Of course that's not a very good line so finding a third point is a good idea Share this post Link to post Share on other sites
bigbadbuff 0 Posted June 8, 2006 I have one more I need explained... how do you approach this problem Two people set out simultaneously from two locations 12 miles apart and walk toward each other. One person walks 5 miles an hour faster than the other. Find the rate of speed of each person if they meet in one hour and ten minutes. Share this post Link to post Share on other sites
Birdseed 1 Posted June 8, 2006 I have one more I need explained... how do you approach this problem Two people set out simultaneously from two locations 12 miles apart and walk toward each other. One person walks 5 miles an hour faster than the other. Find the rate of speed of each person if they meet in one hour and ten minutes. Let x = miles per hour of the slower person so, x + 5 = miles per hour of the faster person Known equation: 1.166667 hours [ (x) + (x + 5) ] = 12 miles so, 1.166667 (2x + 5) = 12 miles so, 2.333333x + 5.833333 = 12 miles so, 2.333333x = 6.166667 miles so, x = 2.64286 miles per hour = speed of slower person so, x + 5 = 7.64286 miles per hour = speed of faster person HTH Share this post Link to post Share on other sites
bigbadbuff 0 Posted June 12, 2006 Did I do this one right? Solve the following equation for x. 8(x – 1) -2(7x + 10) – 16 = 0 8x – 8 -14x – 4 = 0 -6x – 12 = 0 -12 = 6x -2 Share this post Link to post Share on other sites
bigbadbuff 0 Posted June 12, 2006 How about this one... the ^ indicates a raised power Expand the following 1) (2x^2-3)^2 2) (2e^x + 3)(2e^x - 3) (2e^x + 3)2e^x + (2e^x + 3)(-3) I don't know what to do from here... Share this post Link to post Share on other sites
Birdseed 1 Posted June 12, 2006 How about this one... the ^ indicates a raised powerExpand the following 1) (2x^2-3)^2 2) (2e^x + 3)(2e^x - 3) (2e^x + 3)2e^x + (2e^x + 3)(-3) I don't know what to do from here... F-O-I-L First, Outside, Inside, Last 1) (2x^2-3)^2 = (2x^2-3)(2x^2-3) So, 4x^4 - 6x^2 -6x^2 + 9 So, 4x^4 - 12x^2 +9 Share this post Link to post Share on other sites
Bill E. 703 Posted June 12, 2006 I got the first two and guessed zero on the 3rd which zeros out all of the functions. Did not get 4. Is #5 correct? A^2 +B^2 = c^2 If the hypotenuse is 18 inches long, then c^2 = 324 and A and B would each be like 12.7". Good advice in here. I agree that the calculus is not that rough, however the algebra involved can be a kick in the shorts. if you struggle with algebra basics you may be toast. Test yourself on these basics... 1. Solve for x: .4x + 8 = 0 2. Solve for x: x^2 - 25 = 0 3. Solve for x: 2x^3 + 11x^2 + 15x=0 4. Graphically, do you understand what each of the following functions look like? Find x and y intercepts? vertex points? y = .4x + 8 y = x^2 - 25 y = 2x^3 + 11x^2 + 15x=0 5. Knowing pythagorean theorem forewords and backwards is key for success in the trigonometry you'll encounter. How long are the legs of a right isosceles triangle if the hypotenuse is 18 inches long? There's obvously more than these few questions, but it should be a good start for you to gauge where you're at. Answers below in white. 1. x = -20 2. x = 5 or -5 3. x = 0, -3, or -2.5 4. The solutions to 1 through 3 are the x intercepts of the functions. 5. Each leg = 3 inches. Share this post Link to post Share on other sites
bigbadbuff 0 Posted June 12, 2006 thanks dude... how about the one above, the solve the equation for x I showed... is that right? Share this post Link to post Share on other sites
Birdseed 1 Posted June 12, 2006 thanks dude... how about the one above, the solve the equation for x I showed... is that right? Yeah, looks right to me. To ckeck yourself, just stick the (-2) back in the original equation for x and see if it works out to zero. Another thing to remember, when you multiply powers, you add the exponents. Share this post Link to post Share on other sites
G-Men (formerly Big Blue) 0 Posted June 18, 2006 Did you learn partial fractions yet? I remember learning that in Pre-calc and using it again in calc. so if you learn it, make sure you remember it becuase you will use it again. Share this post Link to post Share on other sites
