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Maths year 11

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Original post by z_o_e
Got it!! I got 71% on that work!!

Now this...



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Sorry I put my answers in terms of Pi because I'm lazy.

Original post by z_o_e
I got the top ones. I think they are right. What about these?



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Cross section of the first one is quarter of a circle.
For the second youre finding the volume of the solid outer bit of the pipe. So the cross sectional area you're concerned with will be the area of the larger circle - area of the smaller circle.
(edited 7 years ago)
Reply 282
Original post by 34908seikj
Sorry I put my answers in terms of Pi because I'm lazy.



So it's Csa*L

These are tricker ones do I use same method :/



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Original post by z_o_e
So it's Csa*L

These are tricker ones do I use same method :/



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Reply 284
Original post by 34908seikj


I got them all wrong "(

The answer was 7.3
2) 2
3) 7.2

I think it's working out isn't right.


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Original post by z_o_e
I got them all wrong "(

The answer was 7.3
2) 2
3) 7.2

I think it's working out isn't right.


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Ugh, my answers are incorrect, are you sure?

I rounded to two 2dp, so my answers are correct for questions 1 and 3, lemme figure out where I went wrong on 2.
(edited 7 years ago)
Reply 286
Original post by 34908seikj
Ugh, my answers are incorrect, are you sure?


According to the website. :frown:


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Original post by z_o_e
According to the website. :frown:


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ohhhh for question two, I didn't multiply by 1/2. Sorry, that's my fault.
Original post by z_o_e
I got them all wrong "(

The answer was 7.3
2) 2
3) 7.2

I think it's working out isn't right.


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You're supposed to do it to 1 decimal place.
Reply 289
Original post by Ano123
You're supposed to do it to 1 decimal place.


I did.

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Did you round it correctly?
Reply 291
Original post by Ano123
You're supposed to do it to 1 decimal place.


Heya. I'm going to start from today xx

I needed explanation on this question please.

I would like to workout the answers and stuff all by myself :smile: as I wouldn't learn if I got them x



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Reply 292
Original post by 34908seikj
Did you round it correctly?




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Original post by z_o_e
Heya. I'm going to start from today xx

I needed explanation on this question please.

I would like to workout the answers and stuff all by myself :smile: as I wouldn't learn if I got them x



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If you let the length BD be denoted y y , then using the triangle on the right hand side we get
sin42=12.8y \displaystyle \sin 42^{\circ} =\frac{12.8}{y} .


If you let BCA be denoted x x , then tanx=12.86.8 \displaystyle \tan x =\frac{12.8}{6.8} . So you just solve for x x .
Reply 295
Original post by B_9710
If you let BCA be denoted x x , then tanx=12.86.8 \displaystyle \tan x =\frac{12.8}{6.8} . So you just solve for x x .


I did this


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You're working out the wrong angle.
Original post by z_o_e
Heya. I'm going to start from today xx

I needed explanation on this question please.

I would like to workout the answers and stuff all by myself :smile: as I wouldn't learn if I got them x



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For part b

Use Sin to find BD, you can do this because you have the opposite side to an angle. - SOHCAHTOAH - Sin opp/hyp Cos adj/hyp Tan opp/adj


So set it up like this: 12.8/x = Sin(42)

Now try rearranging the formulae to make x the subject.


Long time no see Zoe, have you been revising :biggrin:
Reply 298
Original post by B_9710
You're working out the wrong angle.


Oh?
Is this fine?

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Original post by z_o_e
Oh?
Is this fine?

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That's right.

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