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Question-180744




Question Number 180744 by harckinwunmy last updated on 16/Nov/22
Commented by cyathokoza last updated on 17/Nov/22
I TRY   THIS THEN SUBMIT. THE ANSWERS
$$\boldsymbol{{I}}\:\boldsymbol{{TRY}}\:\:\:\boldsymbol{{THIS}}\:\boldsymbol{{THEN}}\:\boldsymbol{{SUBMIT}}.\:\boldsymbol{{THE}}\:\boldsymbol{{ANSWERS}} \\ $$
Commented by Acem last updated on 17/Nov/22
Cool
$${Cool} \\ $$
Answered by Acem last updated on 16/Nov/22
Commented by Acem last updated on 17/Nov/22
a•   ⇒ R= ((d^( 2) + b^( 2) )/(2b))    Or  R= ((ℓ^( 2) + 4b^( 2) )/(8 b))    ; ℓ= ∣xy∣    The questions “b, c” are easy, am sorry that am   very busy right now. When being free, would   complete your question.  Good luck!
$${a}\bullet \\ $$$$\:\Rightarrow\:{R}=\:\frac{{d}^{\:\mathrm{2}} +\:{b}^{\:\mathrm{2}} }{\mathrm{2}{b}}\:\:\:\:{Or}\:\:\boldsymbol{{R}}=\:\frac{\ell^{\:\mathrm{2}} +\:\mathrm{4}\boldsymbol{{b}}^{\:\mathrm{2}} }{\mathrm{8}\:\boldsymbol{{b}}}\:\:\:\:;\:\ell=\:\mid{xy}\mid \\ $$$$ \\ $$$${The}\:{questions}\:“{b},\:{c}''\:{are}\:{easy},\:{am}\:{sorry}\:{that}\:{am} \\ $$$$\:{very}\:{busy}\:{right}\:{now}.\:{When}\:{being}\:{free},\:{would} \\ $$$$\:{complete}\:{your}\:{question}. \\ $$$${Good}\:{luck}! \\ $$
Answered by Acem last updated on 16/Nov/22
Commented by Acem last updated on 17/Nov/22
b•    S_1 = ((11 (d^( 2) + b^( 2) )^2  arctan ((d/L)))/(2520 b^( 2) )) − d.L    Or    S_1 = ((11(ℓ^( 2) + 4b^( 2) )^2  arctan ((ℓ/(2L))))/(40 320 b^( 2) )) − (1/2) ℓ.L      ; L= ((ℓ^( 2) + 4b^( 2) )/(8 b)) − k  ; ℓ= Seg. wide  ,  b: Seg. height  ,  k: arch dome height            S_1 = 405.88 Cm^2  ≈ 0.0406 m^2     c• S= S_1  + d.a= 5 860.236 Cm^2  ≈ 0.58602 m^2
$${b}\bullet\:\:\:\:{S}_{\mathrm{1}} =\:\frac{\mathrm{11}\:\left({d}^{\:\mathrm{2}} +\:{b}^{\:\mathrm{2}} \right)^{\mathrm{2}} \:\mathrm{arctan}\:\left(\frac{{d}}{{L}}\right)}{\mathrm{2520}\:{b}^{\:\mathrm{2}} }\:−\:{d}.{L} \\ $$$$ \\ $$$${Or}\:\:\:\:\boldsymbol{{S}}_{\mathrm{1}} =\:\frac{\mathrm{11}\left(\ell^{\:\mathrm{2}} +\:\mathrm{4}\boldsymbol{{b}}^{\:\mathrm{2}} \right)^{\mathrm{2}} \:\boldsymbol{\mathrm{arctan}}\:\left(\frac{\ell}{\mathrm{2}\boldsymbol{{L}}}\right)}{\mathrm{40}\:\mathrm{320}\:\boldsymbol{{b}}^{\:\mathrm{2}} }\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\ell.\boldsymbol{{L}}\:\: \\ $$$$ \\ $$$$;\:{L}=\:\frac{\ell^{\:\mathrm{2}} +\:\mathrm{4}{b}^{\:\mathrm{2}} }{\mathrm{8}\:{b}}\:−\:{k} \\ $$$$;\:\ell=\:{Seg}.\:{wide}\:\:,\:\:{b}:\:{Seg}.\:{height}\:\:,\:\:{k}:\:{arch}\:{dome}\:{height} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:{S}_{\mathrm{1}} =\:\mathrm{405}.\mathrm{88}\:{Cm}^{\mathrm{2}} \:\approx\:\mathrm{0}.\mathrm{0406}\:{m}^{\mathrm{2}} \\ $$$$ \\ $$$${c}\bullet\:{S}=\:{S}_{\mathrm{1}} \:+\:{d}.{a}=\:\mathrm{5}\:\mathrm{860}.\mathrm{236}\:{Cm}^{\mathrm{2}} \:\approx\:\mathrm{0}.\mathrm{58602}\:{m}^{\mathrm{2}} \\ $$$$ \\ $$

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