TrueTears question thread

[TrueTears]

any ideas?

[/0]

checked with calculator, don't think there's a correct answer.

[TrueTears]

checked with calculator, don't think there's a correct answer.

Thanks.

[TrueTears]

I think answer is wrong but can someone confirm?

I put D but apparently answer says B.

But if you think about it, looking at D, x stands for the amount of tablet dissolved, so as x gets larger then will get smaller because 15% of the amount UNDISSOLVED is continuing to get smaller.

Where as B is saying is that as x gets larger, then the rate at which the the tablet is dissolving increases. But how can this be when 15% of the amount UNDISSOLVED is continually decreasing[since the amount of tablet left undissolved is decreasing.]

[Over9000]

I would definitely say D as B isn't logical how can rate start negative, can you have a negative rate in this application. D is right, rate starts high and then slows off.
Negative probably means that the tablet undissolves (nice according to tsfx we got some LCP here)

[TrueTears]

ok thanks over9000

[TrueTears]

Why is C not right? Answer says E.

Yes I know E is right, but why is C also not one of the answers?

[/0]

The diagonals of a kite meet at 90 degrees as well. However, a kite is not a rhombus.
http://en.wikipedia.org/wiki/Kite_(geometry)

[Flaming_Arrow]

The diagonals of a kite meet at 90 degrees as well. However, a kite is not a rhombus.
http://en.wikipedia.org/wiki/Kite_(geometry)

and this is also true for a square

[/0]

The diagonals of a kite meet at 90 degrees as well. However, a kite is not a rhombus.
http://en.wikipedia.org/wiki/Kite_(geometry)

and this is also true for a square

hmm yep but a square is a subset of the rhombus so if you can prove it is a square then it is automatically a rhombus

[TrueTears]

All of a, b, c and d are vectors.

say b(t) = qt i + kt j

and d(t) = ft^2 i + ot^2 j

and a = w i + e j

where q, k, f, o, w and e are all real constants.

then how come to find t_1 ie when d and b intersect, you must do a+b (which is vector c) = d and equate i and j components to find t_1

ie, (w + qt_1)i + (kt_1 + e)j = ft_1^2 i + ot_1^2 j

so (w + qt_1) = ft_1^2 and (kt_1 + e) = ot_1^2

why cant u just do b = d and then equate the i and j components of those 2 vectors?

ie, qt_1 i + kt_1 j = ft_1^2 i + ot_1^2 j

so qt_1 = ft_1^2 and kt_1 = ot_1^2

[kamil9876]

the source of your mistake is that you probably assumed that position vector is equal to displacement vector. Really:  where is the displacement from the time to

I think this is the mistake you made since, the way you have described, b(t) it should be a striaght line through the origin. But really it has a 'psuedo-origin' (ie ) which is the position vector

[TrueTears]

ok thanks kamil

[TrueTears]

Also I noticed something in some old exams I was doing.

When a question says "...find the quadratic EQUATION that has the ROOTS b and a..."

Then do your final answer have to be in the form of  (x-b)(x-a) = 0

Because if you do not set it to 0 then it won't be an equation right?

Would marks be deducted if you just had "...the quadratic equation is (x-b)(x-a)..." ?

[/0]

I think that technically, a root of is any value of x such that f(x) = 0.
i.e. The roots of f(x) = (x-a)(x-b)  [or even just (x-a)(x-b), since f(x) can be redundant] are the solutions to the equation (x-a)(x-b) = 0.

So if they asked "...find the quadratic equation that has the solutions b and a...", I would write (x-a)(x-b)=0
But if they asked "...find the quadratic equation that has the roots b and a...", I would write h(x) = (x-a)(x-b)    (or any function)
And if they asked "...find the quadratic (expression) that has roots b and a...", I would write (x-a)(x-b)