Another Q (1 Viewer)

[Paj20]

Yachts in the Sydney to Hobart Race need to satisfy a set criteria to be placed in the 2
various divisions. If one of the divisions required a yacht to weigh 28 tonnes ± 20 kg,
calculate the percentage error allowed in this measurement to 1 significant figure.

was the ± 20 part about??... i just done 20/28000 which is 0.0007


the answer is 0.07%

Also --- A gardener has 16 metres of fencing with which to enclose a rectangular vegetable patch.
(i) If the length of the rectangular vegetable patch is x metres, show that the area of 2
the rectangular vegetable patch is given by A = 8x – x2.

How do you 'show' it?

 

[bored of sc]

Yachts in the Sydney to Hobart Race need to satisfy a set criteria to be placed in the 2
various divisions. If one of the divisions required a yacht to weigh 28 tonnes ± 20 kg,
calculate the percentage error allowed in this measurement to 1 significant figure.

was the ± 20 part about??... i just done 20/28000 which is 0.0007
the answer is 0.07%

That's correct. For your decimal, convert it to a percentage by multiplying by 100.

0.0007 x 100 = 0.07%

The +or-20 part simply means the weight of the yacht can be either 20kg heavier or 20kg lighter than 28 tonnes. Thus maximum weight allowed would be 28000+20 = 28020kg while the minimum weight would be 28000 - 20 = 27080kg.

 

[bored of sc]

A gardener has 16 metres of fencing with which to enclose a rectangular vegetable patch.
(i) If the length of the rectangular vegetable patch is x metres, show that the area of 2
the rectangular vegetable patch is given by A = 8x – x2.

How do you 'show' it?

Can you reword the question? It doesn't make sense to me.

 

[Paj20]

Can you reword the question? It doesn't make sense to me.

Thats what is says in the Trial Paper

 

[PC]

A gardener has 16 metres of fencing with which to enclose a rectangular vegetable patch.
(i) If the length of the rectangular vegetable patch is x metres, show that the area of 2
the rectangular vegetable patch is given by A = 8x – x2.

How do you 'show' it?

"Show" just means that you should set out your working as you normally would, working towards the answer. Pretend that you don't know the answer.

The perimeter of the rectangle must be 16 metres.
Let x be the length and y be the width.
P = 2(x + y) = 16
2x + 2y = 16
2y = 16 – 2x
y = 8 – x

So the width is 8 - x.

Now Area = length x width
= xy
= x(8 – x) (since y = 8 – x)
= 8x – x²