Post by: spiderman on September 21, 2009, 02:44:17 pm
2)Rewrite these sentences using correct inequality signs
*) In an orchard the trees had between 300 to 400 pears on each tree.
im confused (should it be, 300<=p<400 or 300<=p<=400)
Post by: astarmathsandphysics on September 21, 2009, 02:57:55 pm
Post by: nid404 on September 21, 2009, 03:00:42 pm
1) A piece of wire 4 cm long ha sthe shape of a rectangle. Given that the width is w cm, show that the area A cm^2, of the rectangle is given by the function A=36-(6-w)^2. Find the greatest possible domain and the corresponding range of this function in this context.
2)Rewrite these sentences using correct inequality signs
*) In an orchard the trees had between 300 to 400 pears on each tree.
im confused (should it be, 300<=p<400 or 300<=p<=400)
spidey, which page in the textbook is this question from??
adi has answered the 2nd question right...
Post by: spiderman on September 21, 2009, 03:33:54 pm
qp 12
Post by: nid404 on September 21, 2009, 03:48:31 pm
Post by: nid404 on September 21, 2009, 04:12:42 pm
I'll put down the other part tho.....but u must wait till astar confirms the answer
We need to evaluate the equation for any real number....but area cannot have a negative value, so can't w
so
36-(6-w)^2>0
36-(6-w)^2<24w(it's 24m..u made a typo)
so this gives us...
0<w<12 domain
After that you try finding values of A between values of w 1-11
you'll notice
0<A<36 is the range....
if you didn't get it...i don't mind explaining it again....
I'll try figuring out the proving part too...
Post by: Ghost Of Highbury on September 21, 2009, 04:24:25 pm
lenght = x
width = w
wx = area
2w + 2x = 24
2x = 24 - 2w
x = 12 - w
area = wx = 12w - w2
----------
36 - (6-w)2 = 36 - (36 - 12w + w2)
==> 36 - 36 +12w -w2
= 12w - w2
-----------
done..
Post by: nid404 on September 21, 2009, 04:29:34 pm
2w+2x=24
this means twice width + twice length= length....can u explain this
Post by: Ghost Of Highbury on September 21, 2009, 04:31:15 pm
I got the part where you need to find the domain and the range but i don't get how to show the Area is 36-(6-w)^2
I'll put down the other part tho.....but u must wait till astar confirms the answer
We need to evaluate the equation for any real number....but area cannot have a negative value, so can't w
so
36-(6-w)^2>0
36-(6-w)^2<24w(it's 24m..u made a typo)
so this gives us...
0<w<12 domain
After that you try finding values of A between values of w 1-11
you'll notice
0<A<36 is the range....
if you didn't get it...i don't mind explaining it again....
I'll try figuring out the proving part too...
the domain part can also be explained this way
first simplify 36-(6-w)^2 ==> 12w - w2
the area has to be greater than 0
thus, 12w - w2 > 0
12w > w2
w2 < 12w
w < 12
therefore..........0<w<12
-------------------------------------
Post by: Ghost Of Highbury on September 21, 2009, 04:32:29 pm
why did u take
2w+2x=24
this means twice width + twice length= length....can u explain this
the length of the wire = 24 ...
therefore the permieter of the rectangle is also = 24
perimeter of a rectangle = 2(length) + 2(width) = 2x + 2w = 2w + 2x
thus, 2w + 2x =24
Post by: master786 on September 21, 2009, 04:35:16 pm
its not 4cm long its 24cm long ?
Post by: nid404 on September 21, 2009, 04:36:06 pm
the length of the wire = 24 ...
therefore the permieter of the rectangle is also = 24
perimeter of a rectangle = 2(length) + 2(width) = 2x + 2w = 2w + 2x
thus, 2w + 2x =24
how can the length of the wire be equal to the perimeter??
Post by: nid404 on September 21, 2009, 04:36:33 pm
ok wait so spidermans questiontyped is rong ?
its not 4cm long its 24cm long ?
yup
Post by: master786 on September 21, 2009, 04:38:15 pm
its how long the wire is totally nd this 24cm when madeinto a rectangle has a length nd width which is unkown to us :)
Post by: Ghost Of Highbury on September 21, 2009, 04:38:46 pm
how can the length of the wire be equal to the perimeter??
ok..suppose...u take a 30cm wire...fold it to make a circle...a triangle...a square...whatever,,.,doesnt matter
u havent added any extra wire nor have u removed a part of the wire....the perimeter will be equal to the length of wire taken rite?
dont confuse it with the area....u'm talking abt the perimeter....
Post by: nid404 on September 21, 2009, 04:40:11 pm
got it...it's a 24cm wire, made into a rectangle
then ok...sorry my bad
Post by: master786 on September 21, 2009, 04:41:43 pm
Post by: Ghost Of Highbury on September 21, 2009, 04:47:31 pm
but..i strongly think that it shud be --> 300<p<400
because it says its BETWEEN the range...
Post by: nid404 on September 21, 2009, 04:49:38 pm
i'm not pretty sure of the second part..
but..i strongly think that it shud be --> 300<p<400
because it says its BETWEEN the range...
that's what I know after doing inequalities from the text....u don't include the upper and lower bound when it says between
Post by: astarmathsandphysics on September 21, 2009, 07:25:54 pm
Post by: spiderman on September 22, 2009, 04:15:18 am
Post by: spiderman on September 22, 2009, 12:18:09 pm
(x-3)^2 -3=0
Solve this quadratic function leaving surds in your answer.....
Post by: Ghost Of Highbury on September 22, 2009, 12:32:42 pm
------------------------------
first step - expand...
x2 - 6x + 9 -3 = 0
x2 -6x + 6 = 0
use the formula x = -b (+)<b2 - 4ac
-----------------------------
2a
and.. x = -b (-)<b2 - 4ac
-----------------------------
2a
simplify...
6 +<(36-24)
------------ = 6 +<12
2 -------
2
=> 3 + (<12/2)
and 3 - (<12/2)
------------------
<12 represents the answer in surd form...
------
<12/2 can be written as (<3*4)/2 = (<3 * <4)/2 = (<3 *2)/2 = <3
-----
so the final answer is 3(+/-) <3
Post by: Ghost Of Highbury on September 22, 2009, 12:35:18 pm
Post by: spiderman on September 22, 2009, 12:44:06 pm
Post by: Ghost Of Highbury on September 22, 2009, 12:46:24 pm
but the answer in the text is 3+-square root of 3....its the same dude....<12/2 is further simplified as +-<3
Post by: astarmathsandphysics on September 22, 2009, 12:47:23 pm
please can anyone help me with this sum
(x-3)^2 -3=0
Solve this quadratic function leaving surds in your answer.....
Make x the subject
1. Add 3 to both sides
(x-3)^2 =3
2. Square root
(x-3)=
3. Add 3 to both sides.
==3
Post by: Ghost Of Highbury on September 22, 2009, 12:49:06 pm
its the same dude....<12/2 is further simplified as +-<3
modified the answer....check again,...
simplified <12/2 to <3
:)
Post by: spiderman on September 22, 2009, 12:49:54 pm
Post by: spiderman on September 22, 2009, 01:09:34 pm
Post by: Ghost Of Highbury on September 22, 2009, 01:12:32 pm
u mean...the one before IGCSEs....lolz..
my juniors have been troubling me all day for those papers...its not dere in FEP....the CIE site also has 2005..
difficult to get them..
Post by: astarmathsandphysics on September 22, 2009, 01:35:58 pm
Post by: astarmathsandphysics on September 24, 2009, 10:40:50 pm
2 there are some values of a, b and c for which the equation ax+by+c=0 does not represent a straight line. give an example of such values
1)
3two lines have equations y=m1x+c1 and y=m2x+c2 and m1m2=-1. prove that the lines are perpendicular.
(p,q) on y=mx+c so q=mp+c (1)
(r,s) on y=mx+c so s=mr+c (2)
(1)-(2) gives q-s=m(p-r) so m=(q-s)/(p-r)
2) a=b=c=0. Any value of x and y satisfies the equation so the while x,y plane is the solution