The length of each side of a square is increased by 20%. Find the percentage increase in the area of the square.
can anyone help? thanks alot.
Hi Crucifiedx,
Let each side of the square be x
so after increasing by 20%, each side of the square is now 1.2x
Initial area is x * x = x²
Final area is 1.2x * 1.2x = 1.44x²
So, percentage increase = (1.44x² - x²)/x² * 100% = 44%
Don't be careless with the word 'increase' hor.
Take for example the length is 2cm, increase by 20% is equal to 2.4cm.
2 x 2 = 4cm^2
2.4 x 2.4 = 5.76cm^2
5.76 - 4 / 4 x 100 = 44%
Increase of % of area = 44%
I don't understand this part.
"so after increasing by 20%, each side of the square is now 1.2x?"
Originally posted by HelloKittyFan:Take for example the length is 2cm, increase by 20% is equal to 2.4cm.
i thought 0.4?
Ya, increase by 20%, so new length is equal to 2.4cm.
Apologies, grammatical error.
(",)
Originally posted by crucifiedx:I don't understand this part.
"so after increasing by 20%, each side of the square is now 1.2x?"
from length x,
increase by 20%, means multiply by 120%
you will get 1.2x
orh.. thanks. =)
Another ques:
Given this formula F=48/x^2, F=force and x=distance,
When the magnets are a certain distance apart, the force is 1.25 newtons. Write down the force when the distance is halved.
and this:
(2x-5)^2=49
my ans is 6 and -6. am i correct?
(2x - 5)^2 = 49
4x^2 - 20x + 25 = 49
4x^2 - 20x + 25 - 49 = 0
4x^2 - 20x - 24 = 0
(x - 6) (x + 1) = 0
ans = 6 or -1
Originally posted by crucifiedx:orh.. thanks. =)
Another ques:
Given this formula F=48/x^2, F=force and x=distance,
When the magnets are a certain distance apart, the force is 1.25 newtons. Write down the force when the distance is halved.
We do it in the simple way
When F = 1.25, 1.25 = 48/x²
so x² = 48/1.25
x = 6.197
when distance is halved, x is now 3.098
So, F = 48 / 3.098² = 5 N
Alternatively, a preferred method would be:
When x changes from x to x/2 (halved),
force = 48 / (x/2)²
force = 48 *4 / x²
force = 4F
Since F = 1.25 N originally,
force now = 4 * 1.25 = 5N
Originally posted by crucifiedx:and this:
(2x-5)^2=49
my ans is 6 and -6. am i correct?
One way could be as HelloKittyFan has done.
A shorter and simpler method would be
(2x-5)² = 49 (which is 7²)
so
2x - 5 = 7 or 2x - 5 = -7
2x = 12 or 2x = -2
x = 6 or x = -1
Originally posted by eagle:One way could be as HelloKittyFan has done.
A shorter and simpler method would be
(2x-5)² = 49 (which is 7²)
so
2x - 5 = 7 or 2x - 5 = -7
2x = 12 or 2x = -2
x = 6 or x = -1
i did it this way.. but careless again. -.-
Originally posted by crucifiedx:
i did it this way.. but careless again. -.-
I give you two more practice questions I just came up with. Try and see if you can get the answer
(x - 7)² = 36 => x = 13 or 1
(3x - 4)² = 121 => x = 5 or -7/3
Originally posted by eagle:We do it in the simple way
When F = 1.25, 1.25 = 48/x²
so x² = 48/1.25
x = 6.197
when distance is halved, x is now 3.098So, F = 48 / 3.098² = 5 N
Alternatively, a preferred method would be:
When x changes from x to x/2 (halved),
force = 48 / (x/2)²
force = 48 *4 / x²
force = 4FSince F = 1.25 N originally,
force now = 4 * 1.25 = 5N
thanks. =)
Originally posted by eagle:I give you two more practice questions I just came up with. Try and see if you can get the answer
(x - 7)² = 36 => x = 13 or 1
(3x - 4)² = 121 => x = 5 or -7/3
i got e ans. =)
this ques is abt vectors:
OP=2a-4/3c
BA=3a-2c
explain why OP is parallel to BA.
Originally posted by crucifiedx:this ques is abt vectors:
OP=2a-4/3c
BA=3a-2c
explain why OP is parallel to BA.
As taught earlier, you have to prove OP to be a multiple of BA, or vice versa.
Do some side calculations, and you can see that
OP = 2/3 BA
Hence, OP is parallel to BA.
oh ya.. ty
Given that (3X-1)(x+p)=3x^2+qx-2,
find the value of p and q.
y is proportional to x^n.
write down the value of n when
(i) ycm^2 is the area of a circle of radius xcm,
(ii) y hours is the time taken to travel a distance xkm at a constant speed.
(3x - 1)(x + p) = 3x^2 + 3xp - x - p = 3x^2 + qx - 2
p = 2
3xp - x = qx
3x(2) - x = qx
6x - x = qx
5x = qx
q = 5
Originally posted by HelloKittyFan:(3x - 1)(x + p) = 3x^2 + 3xp - x - p = 3x^2 + qx - 2
p = 2
3xp - x = qx
3x(2) - x = qx
6x - x = qx
5x = qx
q = 5
thanks alot. =)