Mathematics


 

**একটি নিরেট** **লম্ব বৃত্তাকার লোহার দন্ডের ভূমি তলের ব্যাসার্ধের দৈর্ঘ্য 32 সেমি এবং দৈর্ঘ্য 35 সেমি। দন্ডটি গলিয়ে 8 সেমি দৈর্ঘ্যের ব্যাসার্ধ ও 28 সেমি উচ্চতা বিশিষ্ট কতগুলি নিরেট শঙ্কু তৈরি করা যাবে?**




[![]()](https://modelactivitytask.com/question/6512 "একটি লম্ব বৃত্তাকার শঙ্কুর ভূমিতলের ব্যাসার্ধের দৈর্ঘ্য একে রেখে উচ্চতা অর্ধেক করলে শঙ্কুটির আয়তন কত হ্রাস পাবে??")


[![]()](https://modelactivitytask.com/question/6606 "Mathematics")
উত্তর।
Δl =l αΔt =2 •10⁻²•14•10⁻⁶•70=1.96•10⁻⁵m=1.96•10⁻² mm 圖片參考：http://imgcld.yimg.com/8/n/ab00135862/o/161006040753513872668990.jpg
 
首先請先 複習下列公式
 
設扇形的中心角θ弳，半徑為r，弧長為s，面積為a，則
s=rθ；a= r2θ= rs
sol: 設: ∠aob=θ 弳 (弧度) od=x
(x+2.5) θ =6 ……………………………..(1)
求黃色部分面積= (x+5)2 θ - x2θ
= θ ( x2+10x+25-x2 )
= θ (10x+25)
=5θ( x + 2.5)=30



圖片參考：http://imgcld.yimg.com/8/n/ab00135862/o/161006040753513872669011.jpg
 

圖片參考：http://imgcld.yimg.com/8/n/ab00135862/o/161006040753513872669012.jpg
 

圖片參考：http://imgcld.yimg.com/8/n/ab00135862/o/161006040753513872669033.jpg
 



 2010-06-04 23:25:06 補充：
y= |cosx|

恕不再多畫 將以文字說明

1. 當cosx >0 時 y=cosx
2. 當cosx <0 時 y = -cosx

 2010-06-04 23:26:21 補充：
注意

y=3sin2(x-π)+2

之 a , b 步驟為同一圖形

 2010-06-05 13:06:49 補充：
奇怪 1/2 通通不見了

更正

s=rθ；a=0.5 r2θ=0.5 rs

求黃色部分面積=0.5(x+5)2 θ - 0.5x2 θ
=0.5θ ( x2+10x+25-x2 )
=θ (10x+25)
=5θ( x + 2.5)=30

p.s. x2 即為 x平方

 2010-08-28 11:35:07 補充：
staff
crusty
dough
yeast
ferment
form
trap
raise
rise
flat
slab
acorn
slope
pound
protein
wheat
rye
grind
coarse
gluten
elastic
expand
carbon
dioxide
technique
wear away
grit
swiss
crush
steel
roller
rid of The given case is SAS (two sides and included angle) so use the Law of Cosines:
  

\(\displaystyle{b}^{{2}}={a}^{{2}}+{c}^{{2}}-{2}{a}{c}{\cos{{B}}}\)
  

\(\displaystyle{b}^{{2}}={5}^{{2}}+{3}^{{2}}-{2}{\left({5}\right)}{\left({3}\right)}{\cos{{92}}}°\)
  

\(\displaystyle{b}^{{2}}={34}-{30}{\cos{{92}}}°\)
  

\(\displaystyle{b}=\sqrt{{{34}-{30}{\cos{{92}}}°}}\)
  

b ~ 5.92 d = 21 m
r =![](https://img-nm%20mnimgs%20com/img/study_content/curr/1/7/3/41/1541/chapter%2011_html_96a2a55%20gif)
circumference = 2πr =![](https://img-nm%20mnimgs%20com/img/study_content/curr/1/7/3/41/1541/chapter%2011_html_521eed1b%20gif)= 66 m
length of rope required for fencing = 2 × 66 m = 132 m
cost of 1 m rope = rs 4
cost of 132 m rope = 4 × 132 = rs 528 hello, red.since the triangle is a right triangle, you can use the pythagorean theorem.let x be the length of the unknown side.then,92 + x2 = (9√2)281 + x2 = 81·2 = 162x2 = 162 - 81x2 = 81take the square root of both sides.x = ±9since the length cannot be a negative number, x = 9 m sin(α β) = sinαcosβ cosαsinβ 
sin(α-β) = sinαcosβ - cosαsinβ


cos(α β) = cosαcosβ - sinαsinβ
cos(α-β) = cosαcosβ sinαsinβ


tan(α β) = tanα tanβ / 1-tanα tanβ
tan(α-β) = tanα-tanβ / 1 tanα tanβ


sin2θ = 2sinθcosθ = 2tanθ /1 tanθtanθ


cos2θ = cos平方θ−sin平方θ = 2cos平方θ−1 = 1−2sin平方θ


△abc=海龍公式 √s(s-a)(s-b)(s-c) 
[s=a b c/2 ，就是周長/2 ]



△abc=1/2 ａ.b*sinc = axbxc / 4r = rxs 

 [r=外接圓半徑，r=內接圓半徑]




應該是這樣吧!希望對你有幫助。 (a) material q with resistivity 2.63×10-8 ohm-m can be used for making electric wires because it has very low resistivity.(b) material r with resistivity 1.0 ×1015 ohm-m can be used for making handle of soldering iron because it has very high resistivity.(c) material p with resistivity 2.3 × 103 ohm-m can be used for making solar cell because it is a semiconductor. 18 sq.units explanation: x2 = 12y ∴ 4b = 12 ∴ b = 3 area of triangle = `1/2 xx "ab" xx "os"` = `1/2 xx 4"a" xx "a"` = `1/2(12)(3)` = 18 sq. units ![](https://s3mn%20mnimgs%20com/img/shared/discuss_editlive/4110270/2012_11_30_15_28_05/vvnzofse6383734697724592802%20png)


let the length of path be a m


total area ![](https://s3mn%20mnimgs%20com/img/shared/discuss_editlive/4110270/2012_11_30_15_28_05/ceaxujzx7923478049123545928%20png)


area of lawn = total area â€“ area of
path

![](https://s3mn%20mnimgs%20com/img/shared/discuss_editlive/1800204/2012_12_04_21_09_56/mathmlequation2002748073868866733%20png)


area of lawn


![](https://s3mn%20mnimgs%20com/img/shared/discuss_editlive/4110270/2012_11_30_15_28_05/dwqusyms3311632565229211681%20png) [readmore]typed answera) δg° = -rtlnksp δg° = -(8.314 jmol-k)(298k)ln(4.6×10-2 )= 7628.72 j = 7.63 kjb) ∆g° = nfe°e° = ∆g°-nfe° = 7.63-(2)(96,000)(11000)e° =-0.0397 v