/sci/ - Science & Math » Thread #13674080

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Anonymous Thu 23 Sep 15:52:06 2021 No.13674080 View Reply Original Report

Quoted By: >>13674131 >>13675568

Why is it wrong when I use the base proportion of Width : Area to find a width with a given area ? The given ratio of width and length is 3 : 5 so the smallest area is 15. Thus the ratio of Width : Area is 3 : 15. When I multiply this proportion to 135 then I got a width of 27. This is however false. What is wrong with my logic ? :(

Anonymous

Anonymous Thu 23 Sep 2021 16:02:35 No.13674111 Report

Quoted By: >>13674147

just when i thought /sci/ was at the 'bottom' of intelligence.... conclude infinitely bottom well just as the infinity schizos love

Anonymous

Anonymous Thu 23 Sep 2021 16:07:13 No.13674127 Report

Quoted By:

nothing you said makes any sense at all to begin with so I dont even know what to say
AB=135
A/B=3/5
A=3B/5
3B^2/5=135
B^2=225
B=15
A=9

a proportion is literally just two numbers divided. you dont need to do any math to see that a length/area is definitely not a length under any circumstances.

Anonymous

Anonymous Thu 23 Sep 2021 16:07:35 No.13674131 Report

Quoted By: >>13674136 >>13674148

>>13674080
Anon, $\mathrm{area}=\mathrm{width}\cdot\mathrm{height}$ means that $k^2\cdot\mathrm{area}=(k\cdot\mathrm{width})\cdot(k\cdot\mathrm{height})$. To maintain the width-to-height ratio and multiply the area by $k$, you need to multiply each of the width and the height by the square root $\sqrt k$.

Anonymous

Anonymous Thu 23 Sep 2021 16:08:20 No.13674136 Report

Quoted By:

>>13674131
laughing at this retard rn

Anonymous

Anonymous Thu 23 Sep 2021 16:11:34 No.13674147 Report

Quoted By: >>13674166

>>13674111
I don't frequent this place a lot. I know I am wrong and I just want to understand the critical fault in my thinking, so I won't repeat the mistake. Whats wrong with you ? You must be very pathetic IRL to get off from some stranger's logical mistake on the internet.

Anonymous

Anonymous Thu 23 Sep 2021 16:11:40 No.13674148 Report

Quoted By: >>13674201

>>13674131
Sorry, let me re-type that:

Anon, $\mathrm{area}=\mathrm{width}\cdot\mathrm{height}$ means that $k^2\cdot\mathrm{area}=(k\cdot\mathrm{width})\cdot(k\cdot\mathrm{height})$ . To maintain the width-to-height ratio and multiply the area by $k$ , you need to multiply each of the width and the height by the square root $\sqrt k$ .

Anonymous

Anonymous Thu 23 Sep 2021 16:21:47 No.13674166 Report

Quoted By: >>13674206

>>13674147
seething

Anonymous

Anonymous Thu 23 Sep 2021 16:32:35 No.13674201 Report

Quoted By: >>13674258

>>13674148
Thank you. Now I see my mistake. The proportion 3:5 of width and length doesnt imply a fixed ratio 3:15 of width and area. When width is doubled (6), length is doubled (10), the area is 60 and not 30. So for n = 1,2,3... the equivalent width is 3n and length is 5n, area is 3x5x(n^2). The Width : Area ratio in each case is 3n/15n^2 = 1/5n. For Area = 135 I must find the equivalent scale number n from it to find width and length.

Anonymous

Anonymous Thu 23 Sep 2021 16:33:50 No.13674206 Report

Quoted By:

>>13674166
you must be at least 18yrs old to post here.

Anonymous

Anonymous Thu 23 Sep 2021 16:49:23 No.13674258 Report

Quoted By: >>13674282 >>13674836

>>13674201
Yeah, I guess if $\frac wh=\frac35$ then there's an $n$ such that $w=3n$ and $h=5n$ , and $wh=135=3n\cdot5n=15n^2$ means that $n=\sqrt{\frac{135}{15}}$ .

Anonymous

Anonymous Thu 23 Sep 2021 16:56:44 No.13674282 Report

Quoted By:

>>13674258
I don't know why that last one doesn't render properly, but anyway, what I was trying to say is $n=\sqrt{135/15}$ .

Anonymous

Anonymous Thu 23 Sep 2021 19:14:24 No.13674836 Report

Quoted By:

>>13674258
fucking schizo

Anonymous

Anonymous Thu 23 Sep 2021 23:11:33 No.13675568 Report

Quoted By:

>>13674080
>Width : Area is 3 : 15. When I multiply this proportion to 135 then I got a width of 27. This is however false.
It's right if you hold one of either width or height fixed. 3x45 and 27x5 both have 135 area.

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