ib_is_a_bitch Posted December 1, 2011 Report Share Posted December 1, 2011 Hey, it would be really cool if someone could explain compound and simple interest to me. Thank you Reply Link to post Share on other sites More sharing options...
dessskris Posted December 2, 2011 Author Report Share Posted December 2, 2011 Hey, it would be really cool if someone could explain compound and simple interest to me. Thank you I honestly wasn't familiar with those terms at all (wasn't studying in English, I've only been studying in English for 3 years) and so I googled it. this page explains it well, I believe: http://www.basic-mathematics.com/simple-vs-compound-interest.html Reply Link to post Share on other sites More sharing options...
Alice M Posted December 3, 2011 Report Share Posted December 3, 2011 Hi, I'm taking HL Maths and have been given a series of questions from past papers to attempt over the weekend. Most of them are fine, but there's one question I really can't get my head around. I'm in the first year of IB, and we've covered a wide range of the course already (looking at the textbook at least...) If anyone could help, I'd be really grateful!"Find the real number k for which 1+ki, (i = root -1), is a zero of the polynomial z2 + kz + 5."I've attempted dividing the polynomial by 1+ki but I don't really know what to do with the answer... Any help would be awesome, thanks. Reply Link to post Share on other sites More sharing options...
dessskris Posted December 3, 2011 Author Report Share Posted December 3, 2011 I don't think you should attempt that question if you haven't covered complex numbers. anyway, since the coefficients of the quadratic function are real, it means the imaginary roots are conjugate of each other. meaning, the factors of the quadratic function are (1+ki) and (1-ki). now expand the brackets and compare the coefficients. 1 Reply Link to post Share on other sites More sharing options...
Alice M Posted December 3, 2011 Report Share Posted December 3, 2011 I don't think you should attempt that question if you haven't covered complex numbers. anyway, since the coefficients of the quadratic function are real, it means the imaginary roots are conjugate of each other. meaning, the factors of the quadratic function are (1+ki) and (1-ki). now expand the brackets and compare the coefficients.We have covered complex numbers, and thank you! I think I have the right answer now. Reply Link to post Share on other sites More sharing options...
Procrastination Posted December 4, 2011 Report Share Posted December 4, 2011 Sup people, I need some help with this exercises, they're quite easy but I know I'm doing something wrong. UGHFind x:(2/1-x) + 1 = (x/1-x)(a/b)x-b > a - (b/a)x, a and b cant be 0A train departs at 1:00 pm and travels at 80 km/h, An aeroplane departs from the same area at 4:00 pm and travels at 200km/h, how far will the aerplane travel before it overtakes the train? What assumptions have you made?Help would be appreciated Reply Link to post Share on other sites More sharing options...
Sabs44 Posted December 4, 2011 Report Share Posted December 4, 2011 (edited) Sup people, I need some help with this exercises, they're quite easy but I know I'm doing something wrong. UGHFind x:(2/1-x) + 1 = (x/1-x)(a/b)x-b > a - (b/a)x, a and b cant be 0A train departs at 1:00 pm and travels at 80 km/h, An aeroplane departs from the same area at 4:00 pm and travels at 200km/h, how far will the aerplane travel before it overtakes the train? What assumptions have you made?Help would be appreciated1. (2/1-x) + 1 = (x/1-x)multiplying both sides by (1-x):2 + (1-x) = x3 = 2xx = 3/2 = 1.52. (a/b)x - b > a - (b/a)xcollecting x terms on the left side and constants on the right:(a/b)x + (b/a)x > a + bMultiplying by a:a[(a/b)x + (b/a)x] > a[a+b]((a^2)/b)x + bx > (a^2) + abMultiplying by b:b[(a^2/b)x + bx] > b[(a^2) + ab](a^2)x + (b^2)x > (a^2)b + a(b^2)Or, x[(a^2)+(b^2)] > (a^2)b + a(b^2)Now we divide both sides by [(a^2)+(b^2)]:x > [(a^2)b + a(b^2)] / [(a^2)+(b^2)]3. Since the train leaves at 1 and goes 80km/h, at 4 oclock where will it be?(3 hours) x (80km/h) = 240 km from it's starting pointSo, at 4 oclock, you can set up two simultaneous equations:for the train, 240 + 80x [x is in hours from 4 oclock]and for the plane, 0 + 200x [since it hasn't gone anywhere at 4]Setting these equal, 240 + 80x = 200x240 = 120xx = 2 hours,So the plane has traveled at 200km/h for 2 hours, ie 400 km.We assume they travel in the same direction, constant speed, etc.Hope this helps! Edited December 4, 2011 by Sabs44 1 Reply Link to post Share on other sites More sharing options...
Guest hellokitty818 Posted December 7, 2011 Report Share Posted December 7, 2011 This is the whole math problem...sorry i didnt understand how to do it....A light bulb hangs form the ceiling at height h meters above the floor directly above point N. At any point A on the floor which is x meters from the light bulb the illumintation I is given by I=8^(1/2)cosangle/x^2 if NA =1 meter show that at A, I=rad8(cosangle)sin^2angle Reply Link to post Share on other sites More sharing options...
Sabs44 Posted December 7, 2011 Report Share Posted December 7, 2011 (edited) This is the whole math problem...sorry i didnt understand how to do it....A light bulb hangs form the ceiling at height h meters above the floor directly above point N. At any point A on the floor which is x meters from the light bulb the illumintation I is given by I=8^(1/2)cosangle/x^2if NA =1 meter show that at A, I=rad8(cosangle)sin^2angleCan you put some brackets around the "I=8^(1/2)cosangle/x^2" and "I=rad8(cosangle)sin^2angle" so we can help you? Edited December 7, 2011 by Sabs44 Reply Link to post Share on other sites More sharing options...
Guest hellokitty818 Posted December 7, 2011 Report Share Posted December 7, 2011 so sorry...A light bulb hangs form the ceiling at height h meters above the floor directly above point N. At any point A on the floor which is x meters from the light bulb the illumintation I is given by I=[√8(cosӨ)]/(x^2) if NA =1 meter show that at A, I=√8(cosӨ)sin^2Ө Reply Link to post Share on other sites More sharing options...
foggy_mornings Posted December 9, 2011 Report Share Posted December 9, 2011 (edited) Dear, clever IB-students! Please help out with this question. Cheers"It took a train 45 seconds to pass through a 1320 ft tunnel.At the same speed, it took the train 15 seconds to pass a a watchman.How long was the train?"well if it gets past a single point in 15 seconds then in three times the amount of time it would travel three times it's length so if is goes 1320 feet in 45 seconds it would be 440 feet long Edited December 9, 2011 by jenene Reply Link to post Share on other sites More sharing options...
rFumachi Posted December 9, 2011 Report Share Posted December 9, 2011 (edited) V= s/t V= 1320/45 Speed of the train = 29.33 s=V*t s= 29.33*15 Distance traveled by the train in 15 seconds = 440ft Thus the train is 440ft long, supposing the watchman is not fat Edit: You already answered your question lol Edited December 9, 2011 by rFumachi Reply Link to post Share on other sites More sharing options...
MariusIBDP Posted December 11, 2011 Report Share Posted December 11, 2011 (edited) How is range found in functions like these? If possible, could you tell me how to do it with and without calculator(TI-84 plus)? P.S. I know that x-(-4) is x+4, it's just that everytime I wrote x+4, it came out as x(4), so I wrote it this way. Edited December 11, 2011 by MariusIBDP Reply Link to post Share on other sites More sharing options...
dessskris Posted December 11, 2011 Author Report Share Posted December 11, 2011 nice message title Marius okay so: with calculator: just plot the graph and look at the min and max values of y without calculator: find maximum/minimum points, and try the least and the most possible values of x. for example if 0<=x<=5, try finding y when x=0 and when x=5. as simple as that. Reply Link to post Share on other sites More sharing options...
Guest hellokitty818 Posted December 12, 2011 Report Share Posted December 12, 2011 I have no idea how to do this....please help The curve y=(x^3)/3 - x^2 -3x+4 has a local maximum point at P and a local minimum point at Q. Determine the equation of the straight line passing through P and Q in the form ax+by+c=0 where a,b,c are all real numbers Reply Link to post Share on other sites More sharing options...
Drake Glau Posted December 13, 2011 Report Share Posted December 13, 2011 Find your derivative. Use the derivative to find the x values for P and Q (when the derivative equals 0) and then use those values to find their corresponding Y values using the original function. Then you have 2 points and you can use point slope form and then do some algebra to get it into the form you need.I'd try it but your function doesn't have enough parenthesis for me to know exactly that that function is, it could be a really easy chain rule or a stupid quotient rule =/ Reply Link to post Share on other sites More sharing options...
Guest hellokitty818 Posted December 14, 2011 Report Share Posted December 14, 2011 okay peeps last question.....a particle is moving along a straight line so that after t seconds after passing through a fixed point O on the line, its velocity is given by v(t)=t* [sin( (3/pi)*t) ]. find the values of t for which v(t)=0 given that 0 is less than or equal to t is less than or equal to 6 Reply Link to post Share on other sites More sharing options...
Drake Glau Posted December 14, 2011 Report Share Posted December 14, 2011 set the function they gave you equal to 0 and solve for t...then pick the answers that stay within your interval. Reply Link to post Share on other sites More sharing options...
hraouf1 Posted December 14, 2011 Report Share Posted December 14, 2011 People i need help with my infinite summation IA , the part which is talking about Tn (a,x) , the validity , general statement , alot !!Help please Reply Link to post Share on other sites More sharing options...
ILM Posted December 14, 2011 Report Share Posted December 14, 2011 People i need help with my infinite summation IA , the part which is talking about Tn (a,x) , the validity , general statement , alot !!Help pleaseThere is a complete thread discussing this portfolio. Reply Link to post Share on other sites More sharing options...
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