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Maths, Addmaths and Statistics: Post your doubts here!

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Without using a calculator evaluate
(lg 5)^2+lg2 lg50

(lg 5)^2 + lg2 lg 50
(lg 5)^2 + lg2 lg(2 * 25)
(lg 5)^2 + lg2 (lg 2 + lg25)
(lg 5)^2 + lg2 (lg 2 + lg 5^2)
(lg 5)^2 + lg 2 ( lg 2 + 2log 5)
(lg 5)^2 + lg2*log2 + 2*log2*log5
(lg 5)^2 +2*lg2*lg5 + (log2)^2 [Now you can notice the form a^2 + 2ab + b^2 = (a+b)^2
[lg5 +lg2]^2
(lg(5*2))^2
(lg10)^2
1^2
1
 
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Anyone can post the Nov 2012 Maths paper for O levels please....? secondly, make x the subject of formula.. x(x-2)=t-2
Thanks.
 
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hey! i want some help...
i need past papers/links from before 2000 along with there er and ms
plz help!
thnx.
 

tdk

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Thnx a lot. I have a doubt here:
If a measurement is reported as 7 mm rounded to nearest mm, then its lower limit is obviously 6.50 mm but what is its upper limit? Is it 7.49 or 7.50?
 
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Thnx a lot. I have a doubt here:
If a measurement is reported as 7 mm rounded to nearest mm, then its lower limit is obviously 6.50 mm but what is its upper limit? Is it 7.49 or 7.50?
Asslamu Alikum Wa Rahmatullah Wa Barakatoho

7.5
because the actual lower limit is 7.499999999999....so when you round you get 7.5
But it is usually written in the form
7.5>x>=6.5 ( read as: xis less than 7.5 and more than or equal to 6.5)
 
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Thnx a lot. I have a doubt here:
If a measurement is reported as 7 mm rounded to nearest mm, then its lower limit is obviously 6.50 mm but what is its upper limit? Is it 7.49 or 7.50?
let me tell u an easy method. they say nearest to 'a' millimeter. mean 1 millimeter. divide 1 by 2. u get 0.5 millimeter. add it to the real value u get upper bound and subtract it u get lower bound.
for example is they say 1245m to the nearest 5 m. divide by 2 u get 2.5 add it u get 1247.5 cm as upper bound. subtract u get 1242.5 as lower bound.
if u have gud concept of it u can just do it by looking at it but sometimes u have confusion so use this method rather than guess.
 
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Please I need full explanation of this relative velocity question:
A coastguard station receives a distress call from a ship which is travelling at 15 km/h on a bearing of 150°. A lifeboat leaves the coastguard station at 15 00 hours; at this time the ship is at a distance of 30 km on a bearing of 270°. The lifeboat travels in a straight line at constant speed and reaches the ship at 15 40 hours.
(i) Find the speed of the lifeboat.
(ii) Find the bearing on which the lifeboat travelled.

PhyZac , scouserlfc , FAHMEED
 
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Please I need full explanation of this relative velocity question:
A coastguard station receives a distress call from a ship which is travelling at 15 km/h on a bearing of 150°. A lifeboat leaves the coastguard station at 15 00 hours; at this time the ship is at a distance of 30 km on a bearing of 270°. The lifeboat travels in a straight line at constant speed and reaches the ship at 15 40 hours.
(i) Find the speed of the lifeboat.
(ii) Find the bearing on which the lifeboat travelled.

PhyZac , scouserlfc , FAHMEED
Assalamu Alikum
What are the answers?

Is the speed 60 km/h and bearing 310?
 
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Given that lg 2=a and lg3=b and that lg x=3a-4b+1 find x.
Express lg cube root 972 in terms of a and b.
 
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