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[m1nx]

ahh, thank you!!
just one question, tho~ it's the right answer, but i'm not sure how you got the 29d??

 

[Calculon]

The next ten terms are from 11 to 20

Term 11 = a + 10d

Term 20 = a + 19d

Term 11 + Term 20 = (a+10d) + (a+19d)
_______________= 2a + 29d

 

[CM_Tutor]

Originally posted by KeypadSDM
I would have to disagree.

The maximum area is nearly, but not equal to 2.25 m^2 as the question states that some wire must be used to form both the square and the rectangle.

I.e. 0 < x < 6, not 0 <= x <= 6

You have raised an interesting point, one that I have given some thought, and here is my response:

The problem here lies with the interpretation of the word "cut". The question simply says that the wire is cut (ie. partitioned) into two parts - it does not explicitly state that each piece is non-zero in length.

You are interpreting "cut" as requiring non-zero parts. This leads (as you suggest) to 0 < x < 6, and hence to the maximum area being as close as you like to 2.25 m^2, which could really only be expressed mathematically as a limit, ie. lim (x ---> 0) A(max) = 2.25 m^2.

I would contend that "cut" does not require non-zero parts. Mathematically, partitioning an object into two parts, one of size zero, is perfectly valid. In this case, the part of length 0 m is used to form a rectangle of area 0 m^2, having a width of 0 m and a length of 3 * 0 = 0 m, and the maximum area is 2.25 m^2.

I have checked on this with a mate of mine, who has a first class honours degree in pure mathematics, and he confirmed that partitioning or cutting the wire as I have described is valid.

I should point out that the point we are discussing is hair splitting - I think any reasonable marker would accept an answer based on either interpretation, provided the explanation accompanying it was sufficiently clear.

 

[RyddeckerSMP]

easy trig question- i have no idea tho:
if x= 5cos@ and y=5sin@ find an equation connecting x and y by eliminating @:

 

[Calculon]

since sin²@ + cos²@ = 1

then (y/5)² + (x/5)² = 1

(y² + x²)/25 = 1

y² + x² = 25

 

[AGB]

its not really a trig question....more of a parametric question

 

[sikeveo]

i need help with deriving the addition theorms(trig).....i did it b4 but now i dont understand....

 

[AGB]

is that the compound angle formulae you are referring to?? e.g. sin ( a + b ) = sina.cosb + sinb.cosa etc etc

if it is, then you just have to learn them....(hope im talking bout the right thing...)

 

[Heinz]

Originally posted by sikeveo
i need help with deriving the addition theorms(trig).....i did it b4 but now i dont understand....

or are you talking about the fundamental identity ones.
(sin[x])^2 + (cos[x])^2 = 1

divide both sides by (sin[x])^2 to get
1 + (cot[x])^2 = (cosec[x])^2

or dividing by (cos[x])^2 to get
(tan[x])^2 + 1 = (sec[x])^2

 

[sikeveo]

no the ones like i said above... they are in the 3unit book(sorry i should have posted it in the ext 1 topic)

 

[jellybean_girl]

Hey i was just wondering if u could help with this point for 2unit i have to explain/teach it to the class as an assessment and it has to include 5 questions
"Review the graphs of trig. functions in terms of domain, range, periodicity and frequency in terms of radians. Include solving equations graphically, for domain in radians.

 

[Winston]

Originally posted by jellybean_girl
Hey i was just wondering if u could help with this point for 2unit i have to explain/teach it to the class as an assessment and it has to include 5 questions
"Review the graphs of trig. functions in terms of domain, range, periodicity and frequency in terms of radians. Include solving equations graphically, for domain in radians.

well you can talk about how the core items of the graphs, like the amplitude, how to work that out, the period of the trig function.

 

[blam_babe]

Agh I hate integration

The area enclosed by the graph y=(1/2x+3), the x axis and the lines x=0 and x=3 is revolved around the x axis. Prove that the volume of the solid formed is (pi/9) cubic units

I got my answer to be 3pi which obviously isn't right and there's no answer for it on the back of the sheet

 

[Heinz]

Originally posted by blam_babe
Agh I hate integration

The area enclosed by the graph y=(1/2x+3), the x axis and the lines x=0 and x=3 is revolved around the x axis. Prove that the volume of the solid formed is (pi/9) cubic units

I got my answer to be 3pi which obviously isn't right and there's no answer for it on the back of the sheet

Im not sure why you got 3pi but heres the solution

 

[blam_babe]

Thanks Heinz!!! Ur a champion!! I forgot to square the original value which is how i got 3pi

 

[rckl]

Hey a i got a question with a solution on the text book..
but i don't get it.. can anyone pls help..


Differentiate

2x^5(5x+3)^3

Solution:

dy/dx=uv'+vu'

=2x^5 *3(5x+3)^2 *5 + (5x+3)^3 *10x^4 1st Line

=30x^5 *(5x+3)^2 + 10x^4 *(5x+3)^3 2nd Line

=10x^4 *(5x+3)^2[3x+(5x+3)] 3rd Line
------------------------------------------------------------------
can anyone tell me how did they do it?
i can't understand how did they got the third line.....

If they expanded the brackets, then the solution must be very long
but the third line doesn't seen they have expanded the brackets.

help pls. thank you

 

[Xayma]

For the third line they took out the common factor of 10x⁴(5x+3)² since both of them contain it.

 

[rckl]

ahh.. ok thx dude..
so that's mean i need to factories....

 

[rckl]

hello.. i got another question

how to intergrate this

x^2+x+3/ 3x^5

 

[Heinz]

Originally posted by rckl
hello.. i got another question

how to intergrate this

x^2+x+3/ 3x^5

im guessing thats (x^2+x+3)/ 3x^5 If so, just split the fraction up into x^2/3x^5 + x/3x^5 + 3/3x^5 which becomes 1/3x^3 + 1/3x^4 + 1/x^5. change to index and then integrate that.