Mathematics HL/SL/Studies Help

[Hinuku]

Hello,

I have a question regarding the TI-nspire calculator (with touchpad). How to solve equations using the nsolve function which have more than one solution?

Thanks,

Matt

[Slovakov]

Quick wuestion in probability (I don't get it at all):

Two players, A and B, alternately throw a fair six-sided dice, with A starting, until one of them obtains a six. Find the probability that A obtains the first six.

Edited April 26, 2011 by Slovakov

[Abu]

Hello,

I have a question regarding the TI-nspire calculator (with touchpad). How to solve equations using the nsolve function which have more than one solution?

Thanks,

Matt

Not sure if the TI-nspire is allowed or not. Definitely check the IBO calculator policy but I can't help you anyway because I don't know the answer to your question. Try googling or consulting the user manual.

Quick wuestion in probability (I don't get it at all):

Two players, A and B, alternately throw a fair six-sided dice, with A starting, until one of them obtains a six. Find the probability that A obtains the first six.

The premise of A throwing a die until A obtains a six is a negative binomial distribution. Can't help you out any further because probability isn't my strong point.

[genepeer]

Hello,

I have a question regarding the TI-nspire calculator (with touchpad). How to solve equations using the nsolve function which have more than one solution?

Thanks,

Matt

Not sure if the TI-nspire is allowed or not. Definitely check the IBO calculator policy but I can't help you anyway because I don't know the answer to your question. Try googling or consulting the user manual.

If it's NOT CAS (Computer Algebra System) then it's ok. As for using it, I don't know either but the best way on my TI-84 was graphing the function and looking at the zeroes of the function. Very quick and perfectly acceptable. All you have to do the exam is draw a quick sketch of the graph and then list down the answers!

[dessskris]

Hello,

I have a question regarding the TI-nspire calculator (with touchpad). How to solve equations using the nsolve function which have more than one solution?

Thanks,

Matt

TI-Nspire is allowed, everybody in IB1 in my school uses it.

What do you mean btw?

Solving linear systems

In calculator, click Menu, 3 Algebra, 2 Solve System of Linear Equations... Choose the no of equations and list the variables, write the equations in the empty spaces, press enter, it gives you the coordinates of the solution.

Finding roots of polynomial

In calculator, click Menu, 3 Algebra, 3 Polynomial Tools, 1 Find Roots of Polynomial... Choose the degree (if quadratic, degree=2), real roots, put in the constants and click ok. Press enter, it gives you the values of x.

[navbiha]

Would you happen to know of any good IB Math Studies revision websites or books? Sadly, the notes I had taken were in my backpack that had gotten stolen. Thank you!

[CocoPop]

Question is:

Simplify:

2a•(a×b)

The answer is 0, i'm not too sure how

You don't actually need to do the derivation to work that out. It should be clear from looking at that expression that the answer is zero.

The reason: axb is going to give you a vector perpendicular to a, let's call it vector c. When you take the dot product of vector a with this vector c, the angle between them (say, theta) is 90 degrees. So the cos(theta) term will be zero hence the dot product will also be zero.

Quick wuestion in probability (I don't get it at all):

Two players, A and B, alternately throw a fair six-sided dice, with A starting, until one of them obtains a six. Find the probability that A obtains the first six.

Haven't done probability in ages, but here's a quick stab at it (looks like an infinite geometric series).

You wan that player A gets the first six. So you could have the following combinations:

- six

- not six, not six, six

- not six, not six, not six, not six, six

- not six, not six, not six, not six, not six, not six, six

And so on.

The probability of not getting 6 is 5/6. The probability of getting six is 1/6.

So we have that the probability of A getting the first six is:

Edited April 27, 2011 by CocoPop

[Hinuku]

What I mean is that sometimes there are two answers. For example x²=4

then x can equal -2 or 2. And how can I make my calculator show both answers? In the old texas calcs you "guessed". But this is different.

[dessskris]

alright so I answered that already:

Finding roots of polynomial

In calculator, click Menu, 3 Algebra, 3 Polynomial Tools, 1 Find Roots of Polynomial... Choose the degree (if quadratic, degree=2), real roots, put in the constants and click ok. Press enter, it gives you the values of x.

In that case the constants you put in are 1, 0 and -4 (a, b and c in ax²+bx+c=0) and you'll get -2, 2 or something like that.

[hel]

Hey guys I need help how to solve this equation... Express your answer in the form alnb

9e^4x-e^2x=0

This is as far as I got

9e^4x=e^2x

ln(9e^4x)=ln(e^2x)

[HelloooG]

e^2x (9e^2x -1 ) = 0

So e^2x = 0 ==> x=0

And 9e^2x -1 = 0

e^2x = 1/9

2x e^ln = ln (1/9)

2x = ln 9^-1

2x = -ln 9

x = (-ln 9) / 2

S= ( x = 0; x= (-ln9)/2 or -1,10)

I'm not sure it is the right answer..

[hel]

e^2x (9e^2x -1 ) = 0

So e^2x = 0 ==> x=0

And 9e^2x -1 = 0

e^2x = 1/9

2x e^ln = ln (1/9)

2x = ln 9^-1

2x = -ln 9

x = (-ln 9) / 2

S= ( x = 0; x= (-ln9)/2 or -1,10)

I'm not sure it is the right answer..

no I don't think it is...

thanks anyway!

if it helps the mark scheme gives the answers as...

x=1/2ln1/9, x=-1/2ln9, x=ln1/3, a=-1/2 and b=9, x=-ln3 (accept a=-1 and b=3)

I don't understand why there are so many options? and which ones are right?

Edited April 28, 2011 by hel

[Abu]

Here's how I see it. I've worked through it step by step so it should be self explanatory. Let me know if you don't understand it:

[hel]

The lines (AC) and (BD) intersect at the point P(3,K)

Show K=1.

AC=(4,2) or (x,y)=(5,2)+s(4,2)

BD=(3,-6) or (x,y)=(1,5)+t(3,-6)

A(1,0), B(1,5), C(5,2), D(4,-1)

[Hexa]

Set them equal to each other and solve.

(5,2)+s(4,2)=(1,5)+t(3,-6)

Becomes the equations:

5+4s=1+3t

2+2s=5-6t

I solved from here by multiplying the top one by 2 and then adding the two equations to solve for s.

2(5+4s=1+3t)+ (2+2s=5-6t)=

12+10s=7

10s=-5

s=-.5

Then plug the -.5 back into the original s equation:

(5, 2)+(-.5)(4, 2)=

(3, 1)

Edited April 29, 2011 by Hexa

[Summer Glau]

You can create parametric equations just by multiplying s and t into your equations.

x1 = 5+4s

y1 = 2+2s

x2 = 1+3t

y2 = 5-6t

Now, since you want them to intersect at the same point, equate the two x equations and the two y equations (since when the two lines intersect, they will have the same x and y coordinates).

x1 = x2

5+4s = 1+3t

4s-3t = -4

y1 = y2

2+2s = 5-6t

2s+6t = 3

Now you have a 2 equations where you can solve for s or t. I am going to solve for t, but you can solve for s if you like. Either way you should get the same answer at the end

4s-3t = -4

2(2s+6t) = 3

4s+12t = 6

So then we have

4s-3t = -4

4s+12t = 6

Subtracting the two equations, the 4s cancels out, which leaves us with...

-15t = -10

t = 2/3

You can solve for s if you want to, but since you're looking for a point of intersection, you only really need to solve for one variable (even if you solve for s and subsitute it into the other set of equations, you should get the same answer because the lines intersect at the same point).

So now substitute t into the original y2 equation:

y2 = 5-6(2/3)

y2 = 1

This y2 value is the y corrdinate of where the point of intersection occurs. So therefore it's the K corrdinate you were looking for, and K = 1

[crodio]

Could you please help me with this? :

Use your calculator to determine the coordinates of the point on the unit circle corresponding to an angle of:

a) 320° b) 163°

I don't know how to use the calculator for this... thanks!!!

Edited April 29, 2011 by Summer Glau
no text speak =)

[genepeer]

x-coordinate = cos(θ)

y-coordinate = sin(θ)

[hel]

The equation of the tangent at P is y=3 x - 2 . Let T be the region enclosed by

the graph of f , the tangent [PR] and the line x= k , between x = −2 and x= k

where − 2< k < 1.

Given that are of T is 2k+4, show that k satisfies the equation k^4-6k^2+8=0.

So I know you have to integrate x^3-3x-2 for the x=k and x=-2 but I struggled to integrate because I got k^4/4-3k^2/2-4k-10=0

I tried equating the integrated equation to 2k+4 but I didn't get the right answer, k^4-6k^2+8=0.

(if you want the whole question is Q9.b, Nov 10/TZ0 P1)

[genepeer]

Is there a missing picture?