Specialist 1/2 Question Thread!

[Scribe]

I don't know where to start with this question.

If the roots of are complex numbers and z is one of them, find:

i) Re(z) in terms of a
ii) The values of a such that Re(z) < 0.5

Thanks 

[RuiAce]

I don't know where to start with this question.

If the roots of are complex numbers and z is one of them, find:

i) Re(z) in terms of a
ii) The values of a such that Re(z) < 0.5

Thanks 

The next part is now easy - you're just solving \( -\frac{a}{6} < 0.5\)

[unicornvce99.95]

HELP: 1. The number of ‘type A’ apple bugs present in an orchard is estimated to be 40 960, and the number is reducing by 50% each week. At the same time it is estimated that there are 40 ‘type B’ apple bugs, whose number is doubling each week. After how many weeks will there be the same number of each type of bug?

2.Consider the geometric sequence 1,a,a2,a3,.... Suppose that the sum of two consecutive terms in the sequence gives the next term in the sequence. Find a.

[S_R_K]

HELP: 1. The number of ‘type A’ apple bugs present in an orchard is estimated to be 40 960, and the number is reducing by 50% each week. At the same time it is estimated that there are 40 ‘type B’ apple bugs, whose number is doubling each week. After how many weeks will there be the same number of each type of bug?

The numbers of Type A and Type B bugs each follows a geometric sequence. For Type A bugs, the initial value is 40960 and the ratio between consecutive terms is 0.5. Hence, the number of Type A bugs at week n is 40960*(0.5)^(n-1). Similarly, for Type B bugs, the initial value is 40 and the ratio between consecutive terms is 2. Hence the number of Type B bugs at week n is 40*(2)^(n-1). The number of each type of bug is the same when 40960*(0.5)^(n-1) = 40*(2)^(n-1).

2.Consider the geometric sequence 1,a,a2,a3,.... Suppose that the sum of two consecutive terms in the sequence gives the next term in the sequence. Find a.

If the sum of two consecutive terms gives the next term, then the sum of the first and second terms is equal to the third term. So 1 + a = a^2. (We could also use a + a^2 = a^3, and so on, but these all reduce to the first case).

[unicornvce99.95]

The numbers of Type A and Type B bugs each follows a geometric sequence. For Type A bugs, the initial value is 40960 and the ratio between consecutive terms is 0.5. Hence, the number of Type A bugs at week n is 40960*(0.5)^(n-1). Similarly, for Type B bugs, the initial value is 40 and the ratio between consecutive terms is 2. Hence the number of Type B bugs at week n is 40*(2)^(n-1). The number of each type of bug is the same when 40960*(0.5)^(n-1) = 40*(2)^(n-1).

If the sum of two consecutive terms gives the next term, then the sum of the first and second terms is equal to the third term. So 1 + a = a^2. (We could also use a + a^2 = a^3, and so on, but these all reduce to the first case).

I got it...thankyou so much 

[Gogurt]

having trouble with this question. Any thoughts?
“If tanƟ=4/3 and 180˚<Ɵ<270˚, evaluate (sinƟ-2cos(-Ɵ))/(cotƟ-sinƟ)”
*no calculator

-thanks

[mzhao]

having trouble with this question. Any thoughts?
“If tanƟ=4/3 and 180˚<Ɵ<270˚, evaluate (sinƟ-2cos(-Ɵ))/(cotƟ-sinƟ)”
*no calculator

-thanks

Once you have the value of any one of sin x, cos x or tan x, as well as which quadrant x is in, you can work out the values of all of the others. For instance, you can draw a right angled triangle, plug in two of the edge lengths and find the length of the missing edge with Pythagoras. Alternatively, you can use any of the following formulae:

\begin{align}
\sin^2(x) + \cos^2(x) = 1\\
1+\tan^2(x) = \sec^2(x)\\
1+\cot^2(x) = \csc^2(x)\\
\end{align}

Spoiler: Values of sin θ and cos θ for this question

\begin{align}
\sin(\theta) = -\frac{4}{5}\\
\cos(\theta) = -\frac{3}{5}\\
\end{align}

With the value of sin θ and cos θ, you should be able to substitute them into the expression

[Evolio]

Hello guys.
I was wondering if anyone knew how to do  question b?

[Jaijai]

For number systems and sets, 2D question 11. Image attached.
How do you do this question? Thanks

[DBA-144]

For number systems and sets, 2D question 11. Image attached.
How do you do this question? Obviously provide working out, thanks.

Just expand out the brackets. You should be able to recognise that this is the difference of two cubes formula. I don't know latex so sorry, can't provide a full solution for you.

[Jaijai]

Hello guys.
I was wondering if anyone knew how to do  question b?

To share the kindness DBA shared through helping me, in part a of this question you determined BD.
As it suggests to do, use the sine rule to work out CD.
Determine the angle opposite CD by determining the other unknown angle since this is possible using the known information. Once two angles are known, obviously the remaining angle can be found.
Refer to the attachment for the actual worked solution, I was merely explaining the process / what was done and why.

[DBA-144]

To share the kindness DBA shared through helping me, in part a of this question you determined BD.
As it suggests to do, use the sine rule to work out CD.
Determine the angle opposite CD by determining the other unknown angle since this is possible using the known information. Once two angles are known, obviously the remaining angle can be found.
Refer to the attachment for the actual worked solution, I was merely explaining the process / what was done and why.

Ahahaha it would have been more kind if I gave you a fully worked solution but im too lazy to learn latex. Next time I might write it and attach a photo if time allows.

However, for this question, why disnt you consider the ambiguous case? Just something to ponder over. Not trying to sound smart or whatever, but this might help you with your understanding.

[Jaijai]

Ahahaha it would have been more kind if I gave you a fully worked solution but im too lazy to learn latex. Next time I might write it and attach a photo if time allows.

However, for this question, why disnt you consider the ambiguous case? Just something to ponder over. Not trying to sound smart or whatever, but this might help you with your understanding.

The ambigiuous case? Hasn't done trigonometry for a while but wouldn't that break the triangle - sum of internal angles = 180° and if we considered the ambigiuous case, this angle plus the other known angle would already exceed 180°. Hence the ambiguous case isn't in this scenario as the angle has to be below 180 less the known angle (rough inequality)
Also, workings out for my question weren't actually necessary - I was attempting to turn a - b into the right hand side but I should have just done what you'd mentioned. Thanks for your support to the students (including myself)

[Ionic Doc]

So anyone learning sequences and series right now???

What chapter are you guys  ( and girls ) on
What topic are you studying
And What book are you using
My school uses Cambridge

[Ionic Doc]

When all eight factors of 30 are multiplied together, the product is 30k. What is the value of k?


someone help me .........................