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MATHS - more help needed, plox
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IllusoryDeath Away
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Post: #1
MATHS - more help needed, plox

algebra/geometry mixed....and seeing as how I had constant Cs & Bs in geometry class last year, you can tell how confusing stuff like this is to me, no matter how basic.

1)[Image: MXLVIII.jpg?t=1255562223]

2)[Image: MXLVII.jpg?t=1255562223]

3)[Image: MXLVI.jpg?t=1255562223]

4)[Image: MXLV.jpg?t=1255562227]

5)[Image: MXLLL.jpg?t=1255562227]

6)[Image: MXLL.jpg?t=1255562228]
Lovely picture - ain't it just? 8D ^
10-15-2009 09:21 AM
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IllusoryDeath Away
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Post: #2
Re: MATHS - more help needed, plox

bump for necessity D:
10-16-2009 11:50 AM
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magikarp Offline
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Post: #3
Re: MATHS - more help needed, plox

Hopefully these are right, but it's been a long time since I took geometry.
IllusoryDeath Wrote:1)[Image: MXLVIII.jpg?t=1255562223]
The surface area of a sphere is 4*pi*radius^2. The circumference is 2*pi*radius.

Find r.

4 = 2*pi*r
r = 4/(2*pi)

Substitute that value into the equation for surface area.

SA = 4*pi*r^2
= 4*pi*(4/(2*pi))^2
=4*pi*(16/(4*pi^2))
=(16*4*pi)/(4*pi^2)
=16/pi
=5cm^2

Hidden stuff:
You can also do it by calculating 4/(2*pi) first (it equals about 0.6336), but your teacher may not want you to do it that way.

SA = 4*pi*r^2
=4*pi*0.6336^2
=5cm^2




Quote:2)[Image: MXLVII.jpg?t=1255562223]
surface area of the 4 triangular sides:
-First, you need to find the height of the triangle.
height = sqrt(3^2 + 6^2)
=sqrt(45)

Then,

4 * ((sqrt(45)*6)/2)
=12(sqrt(45)

4 rectangular sides:
4 * (6*8)
=192

square bottom:

6*6
=36

total = 12(sqrt(45)) + 192 + 36
=308m^2
K, I'll post these for now. I'll try to do the rest soon.

"Do we treat straight public sex differently than we do gay public sex? Of course. Straight people are so proud of their public sex that they named a cocktail after it."
10-16-2009 12:57 PM
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magikarp Offline
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Post: #4
Re: MATHS - more help needed, plox

IllusoryDeath Wrote:3)[Image: MXLVI.jpg?t=1255562223]

area of the 6 triangular sides:

6 * ((base*height)/2)
=6 * ((4*8)/2)
=96

area of the base:

0.5*number of sides*apothem*side length
=0.5*6*4(sqrt(3))*4
=48(sqrt(3)

total:

96 + 48(sqrt(3))
=179m^2

Quote:4)[Image: MXLV.jpg?t=1255562227]
I don't know how to do this, sorry.

Quote:5)[Image: MXLLL.jpg?t=1255562227]
The easiest way to do this is to imagine the path as a rectangle with another rectangle (the lawn) taken out of the middle. The length of the first rectangle is 32m (1m on each side of the lawn), and the width is 27m.

area of outer rectangle:

length*width
=32*27
=864

area of lawn:

25*30
=750

total:

864-750
=114m^2

Quote:6)[Image: MXLL.jpg?t=1255562228]

Find the area of the field:

120*80
=9600m^2

Find how much the strips cost per m^2:

$15/(1/2m*10m)
=$3 per m^2

Find the cost:

$3 * 9600
=$28800

"Do we treat straight public sex differently than we do gay public sex? Of course. Straight people are so proud of their public sex that they named a cocktail after it."
10-16-2009 01:22 PM
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IllusoryDeath Away
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Post: #5
Re: MATHS - more help needed, plox

Thank you magikarp. You've helped immensely. Smile
10-18-2009 04:45 AM
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Amortisatie Offline
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Post: #6
Re: MATHS - more help needed, plox

I have a feeling i am going to hate geometry immensely, oh wait...

i already do and i haven't even gotten to this stuff yet, ughhh

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“The good thing about science is that it's true whether or not you believe in it.”
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10-18-2009 06:52 AM
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IllusoryDeath Away
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Post: #7
Re: MATHS - more help needed, plox

Oh yes. Geometry was a special kind of hell for me sophomore year.... No
10-18-2009 07:24 AM
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