maths again!!!

[spiderman]

1) A piece of wire 24 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)

[astarmathsandphysics]

I am not in for the next 4  hours. Will answer it when i get back.

[nid404]

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...

[spiderman]

page no 36
qp 12

[nid404]

gimme a few minutes...i'll do it

[nid404]

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...

[Ghost Of Highbury]

proving part

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..

[nid404]

why did u take
2w+2x=24
this means twice width + twice length= length....can u explain this

[Ghost Of Highbury]

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

-------------------------------------

[Ghost Of Highbury]

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

[master786]

ok wait so spidermans questiontyped is rong ?
its not 4cm long its 24cm long ?

[nid404]

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??

[nid404]

ok wait so spidermans questiontyped is rong ?
its not 4cm long its 24cm long ?

yup

[master786]

its not the lenghth of the wire ..
its how long the wire is totally nd this 24cm when madeinto a rectangle has a length nd width which is unkown to us

[Ghost Of Highbury]

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....