O-Level Additional Math

[g131271]

Hi parents,


Does anyone of you can help me in this question? A million thanks to those who help to solve.

[b]A regular polygon has n sides. The size of each interior angle is 4 times the size of each exterior angle.
a. calculate the value of each exterior angle.
b. calculate the value of n.
c. hence find the sum of interior angles of polygon

[tianzhu]

Herbie:

hi uncle tianzhu. Thanks very much.

Hi

You're welcome.

Best wishes

[Herbie]

hi 131271,


A regular polygon has n sides. The size of each interior angle is 4 times the size of each exterior angle.
a. calculate the value of each exterior angle.
Exterior angle = 180/5 = 36 deg

b. calculate the value of n.
Each interior angle for a regular polygon=
144= {(n-2)x 180}/n
144n = (n-2) x180
180n-360= 144n
180n-144n=360
36n =360
n=10

c. hence find the sum of interior angles of polygon= 9x144 = 1296 deg

Hope it helps.

[Herbie]

hi i hv one qn which need help.


A lorry travels at 50km per hour. Given that diameter of itd wheel is 88Cm, find how many revolutions pwe minut the wheel is turning. Give yr answer to the nearest whole no.

Tq

[FrekiWang]

Herbie:

hi i hv one qn which need help.


A lorry travels at 50km per hour. Given that diameter of itd wheel is 88Cm, find how many revolutions pwe minut the wheel is turning. Give yr answer to the nearest whole no.

Tq

one revolution = 2 pi r = 5.5292m

In one hour, the wheel turns 50km / 5.5292m = 9042.9

In one minute, the wheel turns 9042.9 / 60 = 150.7 = 151

151rev/min

[listener]

Q1) A particle P travels in a straight line so that its distance, s m, from a fixed point O is given by s = 2t + 18/(t+1), where t is the time in seconds measured from the start of the motion. Calculate

i) the initial acceleration of P.
ii) the velocity of P when it is next at starting point.
iii) the value of t when the particle is instantaneously at rest.
iv) the total distance travelled in the first 5 seconds.

Q2) The diagram shows parts of the curve y = 2e^(x/2) (not drawn to scale) and the lines x=1, x=k and x=4.
i) Find the total area of regions A, B and C giving your answer correct to 1 decimal place.
ii) Given that the area of Region A is equal to Region B, calculate the value of k giving your answer correct to 4 significant figures.

http://i56.tinypic.com/nya59z.jpg\">

[FrekiWang]

listener:

Q1) A particle P travels in a straight line so that its distance, s m, from a fixed point O is given by s = 2t + 18/(t+1), where t is the time in seconds measured from the start of the motion. Calculate

i) the initial acceleration of P.
ii) the velocity of P when it is next at starting point.
iii) the value of t when the particle is instantaneously at rest.
iv) the total distance travelled in the first 5 seconds.

Q2) The diagram shows parts of the curve y = 2e^(x/2) (not drawn to scale) and the lines x=1, x=k and x=4.
i) Find the total area of regions A, B and C giving your answer correct to 1 decimal place.
ii) Given that the area of Region A is equal to Region B, calculate the value of k giving your answer correct to 4 significant figures.

http://i56.tinypic.com/nya59z.jpg\">

Q1 (I assume there is no () for 2t+18)
i)
v= ds/dt = 2 - 18(t+1)^(-2), a = dv/dt = 36(t+1)^(-3)
when t=0, a = 36(1)^(-3)=36m/s2
ii)
for the position of the starting point, t=0, s = 2(0) + 18/(0+1)= 18
when it is at the starting point again, s=18
2t+18/(t+1)=18
2t(t+1)+18=18(t+1)
2t^2 -16t = 0
2t(t - = 0
t = 0(when it starts) or t = 8(next at starting point)
v = 2 - 18(8+1)^(-2) = 16/9 m/s
iii)
v=0 (at rest)
2-18(t+1)^(-2)=0
(t+1)^(-2)=1/9
(t+1)^2=9
t+1 = -3 or t+1 = 3
t = -4(reject) or t = 2
so t = 2
iv) it makes a turn at t = 2, so consider two segments 0 to 2, and 2 to 5.
0 to 2:
when t=0, s = 18
when t=2, s = 2(2)+18/(2+1)=10
distance travelled=|10-18|=8
0 to 5\"
when t=5, s = 2(5) + 18/(5+1)=13
distance travelled=|13-10|=3
So total distance travelled is 8+3=11m

[FrekiWang]

listener:

Q2) The diagram shows parts of the curve y = 2e^(x/2) (not drawn to scale) and the lines x=1, x=k and x=4.
i) Find the total area of regions A, B and C giving your answer correct to 1 decimal place.
ii) Given that the area of Region A is equal to Region B, calculate the value of k giving your answer correct to 4 significant figures.

http://i56.tinypic.com/nya59z.jpg\">

Q2.
i) when x=4, y=2e^2, total area of A,B,C and small unknown area = 4 x 2e^2=8e^2.
The small unknown area
=Integrate[2e^(x/2)]dx (from 0 to 1)
=4e^(x/2) (from 0 to 1)
=4e^0.5 - 4e^0
=4e^0.5 - 4
So total area of A+B+C
= 8e^2 - (4e^0.5 - 4)
= 56.52
= 56.5(1dp)
ii) Area of A
= Integrate[2e^(x/2)]dx (from 1 to k)
= 4e^(x/2) (from 1 to k)
= 4e^(k/2) - 4e^0.5
Area of B
= Integrate[2e^(x/2)]dx (from k to 4)
= 4e^(x/2) (from k to 4)
= 4e^(2) - 4e^(k/2)
Area of A = Area of B, we have
4e^(k/2) - 4e^0.5 = 4e^(2) - 4e^(k/2)
e^(k/2) - e^0.5 = e^(2) - e^(k/2)
2e^(k/2) = e^0.5 + e^2
e^(k/2) = 4.518889
k/2 = ln(4.518889)
k = 3.017(4sf)

[Xiao Hu]

Hi Freki,

Appreciate it if you could help to solve this Add Maths question. It's from the \"Double-Angle Formulae\" trigo topic.

If 270<x<360, simplify sqrt(2+sqrt(2+2cosx)).
Ans:2sin(x/4).

My answer is 2cos(x/4), different from txtbook's.
http://i56.tinypic.com/jrdxxs.jpg\">
Thanks in advance,
Xiao Hu

[FrekiWang]

Xiao Hu:

Hi Freki,

Appreciate it if you could help to solve this Add Maths question. It's from the \"Double-Angle Formulae\" trigo topic.

If 270<x<360, simplify sqrt(2+sqrt(2+2cosx)).
Ans:2sin(x/4).

My answer is 2cos(x/4), different from txtbook's.
http://i56.tinypic.com/jrdxxs.jpg\">
Thanks in advance,
Xiao Hu

sqrt[2+sqrt(2+2cosx)]
=sqrt{2+sqrt[2+2(2cos^2(x/2)-1)]}
=sqrt[2+sqrt(4cos^2(x/2))]
Note here, 135<x/2<180 (2nd), so cos(x/2) is negative.
=sqrt[2-2cos(x/2)]<--- I suppose your mistake is because you have +2cos(x/2) in this step, I will explain this at the end.
=sqrt[2-2(1-2(sin^2(x/4))]
=sqrt[4sin^2(x/4)]
Note here, 67.5<x/4<90 (1st), so sin(x/4) is positive.
=2sin(x/4)

So the common mistake here is that some students will assume sqrt(x^2)=x, which is not true.

In fact, sqrt(x^2)=|x|, which is equal to x if x is positive and is equal to -x if x is negative. (e.g. if x = -2, we have sqrt(x^2)=2 which is -x).

Therefore, whenever you need to pull a 'perfect square' from the square root, make sure you determine whether you are pulling the square of a positive or negative number. In this question, cos(x/2) is negative, that is why when you pull cos^(x/2) out of the square root, you have to add a negative sign in front.

Hope you could understand. Cheers