maths help ~ (1 Viewer)

[xXmuffin0manXx]

Okay. I've forgotton how to do this one...

i) The average of 8 numbers is 15. When a 9th number is added, the average becomes 12. What is the 9th number.

And now for something a little harder..

ii)Write an expression for n

(1+ 1/2)(1+ 1/3)(1+ 1/4)...(1+ 1/n)

iii) A semicircle with points A, B, C, D all lie on the diameter and are equidistant. A and D both lie on the circumference.

(this is where you draw a diagram and label points A, B, C, D on the diamater and making sure they are equidistant A through to D)

Within the semicircle AD, a semicircle AB is drawn such that it takes away the part of the semicircle AD.

(So on AB, you draw a semicle that is corresponding with the larger semi circl)

The same is done for BD.

(So in a way, you get like a retarded umbrella without the handle)

What is the ratio of the new shape and the original semicircle AD?

_________

Thanks

 

[lyounamu]

Okay. I've forgotton how to do this one...

i) The average of 8 numbers is 15. When a 9th number is added, the average becomes 12. What is the 9th number.

And now for something a little harder..

ii)Write an expression for n

(1+ 1/2)(1+ 1/3)(1+ 1/4)...(1+ 1/n)

iii) A semicircle with points A, B, C, D all lie on the diameter and are equidistant. A and D both lie on the circumference.

(this is where you draw a diagram and label points A, B, C, D on the diamater and making sure they are equidistant A through to D)

Within the semicircle AD, a semicircle AB is drawn such that it takes away the part of the semicircle AD.

(So on AB, you draw a semicle that is corresponding with the larger semi circl)

The same is done for BD.

(So in a way, you get like a retarded umbrella without the handle)

What is the ratio of the new shape and the original semicircle AD?

_________

Thanks

Will just do i) as I desperately need sleep.
i) Average of 8 numbers is 15 so the total is 8 x 15 = 120
Average of 9 numbers is 12 so that total is 9 x 12 = 108
So the 9th number is -12.

ii) not sure what your 2nd one is asking

 

[tommykins]

Okay. I've forgotton how to do this one...

i) The average of 8 numbers is 15. When a 9th number is added, the average becomes 12. What is the 9th number.

x/8 = 15 -> x = 120
(x+a)/9 = 12
(120+a)/9 = 12
120+a = 108
a = -12

And now for something a little harder..

ii)Write an expression for n

(1+ 1/2)(1+ 1/3)(1+ 1/4)...(1+ 1/n)

n+1 where n = 1,2,3,4,5...

not really sure what it's asking though.

cbf circle geo.

 

[cHoke-meh]

ii)Write an expression for n

(1+ 1/2)(1+ 1/3)(1+ 1/4)...(1+ 1/n)

iii) A semicircle with points A, B, C, D all lie on the diameter and are equidistant. A and D both lie on the circumference.

(this is where you draw a diagram and label points A, B, C, D on the diamater and making sure they are equidistant A through to D)

Within the semicircle AD, a semicircle AB is drawn such that it takes away the part of the semicircle AD.

(So on AB, you draw a semicle that is corresponding with the larger semi circl)

The same is done for BD.

(So in a way, you get like a retarded umbrella without the handle)

What is the ratio of the new shape and the original semicircle AD?

_________

Thanks

ii) is weird, don't really understand what to do there.
iii) Say AD = 6
So then AB = 2, BC = 2, CD = 2
(they're all equidistant [didn't even know that was a word...])
Area of semicircle AD = 4.5 pi
Area of semicircle AB = ½ pi
Area of semicircle BD = 2 pi

Area of new shape = (Area of AD) - (Area of AB + Area of BD)
Area of new shape = 2pi
2pi : 4.5 pi
Therefore, ratio is 2.5 : 4.5 or as previously stated 4:9

??

edit: spellung
edit edit: forgot the ratio >.<
edit edit edit: woops.. semicircle area is 1/2 pi r^2

??

edit: spellung
edit edit: forgot the ratio >.<
edit edit edit: woops.. semicircle area is 1/2 pi r^2

 

[Aznmichael92]

ii) is weird, don't really understand what to do there.
iii) Humm, my guess was weird - and i think it's wrong...
Say AD = 6
So then AB = 2, BC = 2, CD = 2
(they're all equidistant [didn't even know that was a word...])
Area of semicircle AD = 9pi
Area of semicircle AB = pi (1pi)
Area of semicircle BD = 4pi

Area of new shape = (Area of AD) - (Area of AB + Area of BD)
Area of new shape = 4pi
4pi:9pi = 4/9
Therefore, ratio is 4:9

??

edit: spellung
edit edit: forgot the ratio >.<

i am no good either but isnt the area of the semi circle AD = 4.5 pi as the area of circle = pi x r^2
semi circle is half that?

using chokes method,

let distance AD = 12, therefore AB, BC, CD = 4
Area of semi circle AD = 18pi (26pi/2)
Area of semi circle AB = 2pi (4pi/2)
Area of Semi circle DB = 8pi (16pi/2)

Area of new shape = 18pi - (2pi+8pi)
Area of new shape = 8 pi

Therfore ratio is 4:9

OMG Chokes correct !

 

[cHoke-meh]

Ye lol, i kept the values constant so it shouldn't have mattered anyway. Still though, whoops

 

[victoria10]

this is confusing me as well!
i dont even understand this. my school must be behind or something.
i'm not even surprised.

 

[Forbidden.]

Okay. I've forgotton how to do this one...

i) The average of 8 numbers is 15. When a 9th number is added, the average becomes 12. What is the 9th number.

And now for something a little harder..

ii)Write an expression for n

(1+ 1/2)(1+ 1/3)(1+ 1/4)...(1+ 1/n)

iii) A semicircle with points A, B, C, D all lie on the diameter and are equidistant. A and D both lie on the circumference.

(this is where you draw a diagram and label points A, B, C, D on the diamater and making sure they are equidistant A through to D)

Within the semicircle AD, a semicircle AB is drawn such that it takes away the part of the semicircle AD.

(So on AB, you draw a semicle that is corresponding with the larger semi circl)

The same is done for BD.

(So in a way, you get like a retarded umbrella without the handle)

What is the ratio of the new shape and the original semicircle AD?

_________

Thanks

∑ 1 + (1/n)

on top of that sigma symbol write the infinity symbol or something but below it you should write a 2

 

[kurt.physics]

∑ 1 + (1/n)

on top of that sigma symbol write the infinity symbol or something but below it you should write a 2

Sorry, but i dont think that is right, see, in the expression it is (1 + 1/2) multiplied by (1 + 1/3) and so on to (1 + 1/n). So this is a product series and therefore you would write it as π (<-- product sign look alike) between i = 2 and n of (1 + 1/i)

 

[tommykins]

i am no good either but isnt the area of the semi circle AD = 4.5 pi as the area of circle = pi x r^2
semi circle is half that?

using chokes method,

let distance AD = 12, therefore AB, BC, CD = 4
Area of semi circle AD = 18pi (26pi/2)
Area of semi circle AB = 2pi (4pi/2)
Area of Semi circle DB = 8pi (16pi/2)

Area of new shape = 18pi - (2pi+8pi)
Area of new shape = 8 pi

Therfore ratio is 4:9

OMG Chokes correct !

You can't randomly let the distance be a constant, it has to be x or a variable.
Basically you're saying, let AD = 12, well - does the ratio still work if AD is 13? 14? 15? 16?

 

[cHoke-meh]

You can't randomly let the distance be a constant, it has to be x or a variable.
Basically you're saying, let AD = 12, well - does the ratio still work if AD is 13? 14? 15? 16?

...

Would it matter anyway? No matter what variables you use, you'd end up cancelling out the variables (e.g. 2x/x = 2) to get a constant ratio anyway..

edit: but for the pedantic...

Say AD = 6x
So then AB = 2x, BC = 2x, CD = 2x
(they're all equidistant [didn't even know that was a word...])
Area of semicircle AD = 4.5 pi x^2
Area of semicircle AB = ½ pi x^2
Area of semicircle BD = 2 pi x^2

Area of new shape = (Area of AD) - (Area of AB + Area of BD)
Area of new shape = 2pi x^2
2pi x^2 : 4.5 pi x^2 or (2 pi x^2)/(4.5 pi x^2)
Cancel each pi x^2 and you're left with 2:4.5 or 2/4.5
Therefore, ratio is 2.5 : 4.5 or as previously stated 4:9

 

[tommykins]

回复: Re: maths help ~

...

Would it matter anyway? No matter what values you use, you'd end up cancelling out the variables (e.g. 4x/9x = 4/9) to get a constant ratio anyway..

Obviously it wouldn't matter, but for future reference it's better off to let it = x than actually give it a value.

 

[xXmuffin0manXx]

...

Would it matter anyway? No matter what values you use, you'd end up cancelling out the variables (e.g. 4x/9x = 4/9) to get a constant ratio anyway..

edit: but for the pedantic...

Say AD = 6x
So then AB = 2x, BC = 2x, CD = 2x
(they're all equidistant [didn't even know that was a word...])
Area of semicircle AD = 4.5 pi x^2
Area of semicircle AB = ½ pi x^2
Area of semicircle BD = 2 pi x^2

Area of new shape = (Area of AD) - (Area of AB + Area of BD)
Area of new shape = 2pi x^2
2pi x^2 : 4.5 pi x^2 or (2 pi x^2)/(4.5 pi x^2)
Cancel each pi x^2 and you're left with 2:4.5 or 2/4.5
Therefore, ratio is 2.5 : 4.5 or as previously stated 4:9

yeah.it doesnt matter which distance you use because the points are equidistant anyway.so all you have to do is make sure that they are in proportion..

 

[tommykins]

回复: Re: maths help ~

yeah.it doesnt matter which distance you use because the points are equidistant anyway.so all you have to do is make sure that they are in proportion..

I'm just saying, use x, as x = any number.

 

[xXmuffin0manXx]

Re: 回复: Re: maths help ~

I'm just saying, use x, as x = any number.

yeah i know what you mean..but for this question it would be easier to just give them distances..for me anyway

 

[tommykins]

回复: Re: 回复: Re: maths help ~

yeah i know what you mean..but for this question it would be easier to just give them distances..for me anyway

Sorry for being anal over this, but it'll only help you later on.

For example, let's say prove x^2 + y^2 ≥ 2xy for x,y ≥ 0


Your method is basically saying, let x = 4, y = 2

16 + 4 ≥ 2.4.2
20 ≥ 16

True, .'. x^2 + y^2 ≥ 2xy
But what about x = 2 and y = 3? how about x = 0.4 and y = 0.0000001 ?
See the problem here? you need cover all variables.

PS. you can attempt the question if you wish, ill postup the solution soon.

 

[xXmuffin0manXx]

Re: 回复: Re: 回复: Re: maths help ~

Sorry for being anal over this, but it'll only help you later on.

For example, let's say prove x^2 + y^2 ≥ 2xy for x,y ≥ 0


Your method is basically saying, let x = 4, y = 2

16 + 4 ≥ 2.4.2
20 ≥ 16

True, .'. x^2 + y^2 ≥ 2xy
But what about x = 2 and y = 3? how about x = 0.4 and y = 0.0000001 ?
See the problem here? you need cover all variables.

PS. you can attempt the question if you wish, ill postup the solution soon.

no matter what measurements you put in to the geo question..the answer will always turn out the same..

 

[tommykins]

回复: Re: 回复: Re: 回复: Re: maths help ~

Sigh...don't worry, whatever.

 

[xXmuffin0manXx]

Re: 回复: Re: 回复: Re: 回复: Re: maths help ~

Sigh...don't worry, whatever.

haha..dw dw i know what you mean

i was just playing around with you

couldnt resist^^

 

[tommykins]

回复: Re: 回复: Re: 回复: Re: 回复: Re: maths help ~

haha..dw dw i know what you mean

i was just playing around with you

couldnt resist^^

Sif take advantage of my willingness to help people with math.