hielly's maths thread

[kamil9876]

You should be taking the sum of a geometric sequence rather than the 10th term since each bounce adds to the total distance, not just the 10th bounce. Also, be careful of the fact that the ball is travelling some distance when it drops as well as as when it ascends to the top. So you have to double the distance of each bounce and add on the distance travelled by the initial drop.

[TrueTears]

Hint:

Your variables are fine.

[GerrySly]

It seems about right, two simultaneous equations



Then using the quadratic formula we get x=18, y=6 (and x=-27/2 but x>0, so we discard that). Therefore he was traveling at 6km/h and it took 18 hrs. So if he traveled at 8km/h he would arrive 4 1/2 hours later. x=108/8=13 1/2 which is 4 1/2 hours faster

[kamil9876]

TT is on the right track. say distance=d, the actual average speed=v, time actually taken=t

d=vt (1)
108=vt

d=v't' (2)

v'- new average speed, v'=v+2
t'- new time, t'=t-4.5

The d is the same in both equations (1) and (2) since he travels the same distance, 108.

hence

(1) 108=vt
(2) 108=(v+2)(t-4.5)

Now solve simultaenously.

[Hielly]

a bus is due to reach its destination 30km away at a certain time. the bus usually travels with an average speed xkm/h. its start is delayed by 18 minutes but, by increasing its average speed by 5km/h, the driver arrives on time.

so
d=z-30
t=y
s=x

its start is delayed by 18 minutes but, by increasing its average speed by 5km/h, the driver arrives on time.

this wordiing is confusing me.
d=z-18 ?
t=y
s=x+5

ty

[kamil9876]

30=xt

it still travels that same distance, only time and velocity have changed, new velocity is x+5, new time is t-18

30=(x+5)(t-18)

now solve simultaenously.

Advice: try not to create too many variables as u can get lost. e.g: t=y adds nothing to the algebra.

Just write down the relevant formula: distance=velocity * time  

distance=30, but we don't know what velocity and time are and so we introduce variables here: x,y

Hence only introduce new unkowns when there is no information regarding that variable.

[Flaming_Arrow]

work out A by doing 180-(59+73), i think u can do it from there

[Hielly]

yerh i know that step, but you know when you use the sine rule, you use the angle that is opposite to the length right

[Flaming_Arrow]

yerh i know that step, but you know when you use the sine rule, you use the angle that is opposite to the length right

yep

so after u worked out A



and do it like that

[Hielly]

the bearing of point A from point B is 207 degrees.What is the bearing of B from A.

im not comfortable with bearings..... thanks

[TrueTears]

90 - [180 - [(207-180) + 90]]

[That thing near the right angle triangle is meant to be a S lol.]

[Hielly]

a person stands on level ground 60m from the nearest point of cylindrical tank of radius length 20m
a) the percentage of the circumference that is visible to the person.

i formed a right angled triangle, the base is 60 and the aim is to find the degree on of angle..?


thanks

[kamil9876]

What is visibile to him, is the arc length between A and B.

OB=20, OC=60+20=80

now use trig to find angle COB. The angle AOB is twice the angel COB by symmetry. Once you know angle AOB you can easily find the arclength AB.

[TrueTears]

a person stands on level ground 60m from the nearest point of cylindrical tank of radius length 20m
a) the percentage of the circumference that is visible to the person.

i formed a right angled triangle, the base is 60 and the aim is to find the degree on of angle..?


thanks

Or have a look here: http://vcenotes.com/forum/index.php/topic,4444.msg52717.html#msg52717

[kamil9876]

http://vcenotes.com/forum/index.php/topic,13687.msg150269.html#msg150269