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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

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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.

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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.

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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.

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Sup people, I need some help with this exercises, they're quite easy but I know I'm doing something wrong. UGH

Find x:

  • (2/1-x) + 1 = (x/1-x)
  • (a/b)x-b > a - (b/a)x, a and b cant be 0
  • A 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

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Sup people, I need some help with this exercises, they're quite easy but I know I'm doing something wrong. UGH

Find x:

  • (2/1-x) + 1 = (x/1-x)
  • (a/b)x-b > a - (b/a)x, a and b cant be 0
  • A 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

1. (2/1-x) + 1 = (x/1-x)

multiplying both sides by (1-x):

2 + (1-x) = x

3 = 2x

x = 3/2 = 1.5

2. (a/b)x - b > a - (b/a)x

collecting x terms on the left side and constants on the right:

(a/b)x + (b/a)x > a + b

Multiplying by a:

a[(a/b)x + (b/a)x] > a[a+b]

((a^2)/b)x + bx > (a^2) + ab

Multiplying 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 point

So, 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 = 200x

240 = 120x

x = 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 by Sabs44
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Guest hellokitty818

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

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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

Can 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 by Sabs44
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Guest hellokitty818

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

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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 by jenene
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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 XD

Edit: You already answered your question lol

Edited by rFumachi
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nice message title Marius :P 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.

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Guest hellokitty818

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

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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 =/

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Guest hellokitty818

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

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