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AllQuestion and Answers: Page 207

Question Number 199671    Answers: 1   Comments: 0

Question Number 199669    Answers: 0   Comments: 0

Question Number 199666    Answers: 3   Comments: 0

Question Number 199662    Answers: 1   Comments: 0

A point moves on the curve of the function f(x)=(√(x^2 +5)) such that it′s x−coordinate increases at a rate of 3(√(10)) cm/s. Find the rate of change of it′s distance from the point (1,0) when x = 2

$$ \\ $$$$\mathrm{A}\:\mathrm{point}\:\mathrm{moves}\:\mathrm{on}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{5}}\:\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{it}'\mathrm{s}\:\mathrm{x}−\mathrm{coordinate}\:\mathrm{increases}\: \\ $$$$\mathrm{at}\:\mathrm{a}\:\mathrm{rate}\:\mathrm{of}\:\mathrm{3}\sqrt{\mathrm{10}}\:\:\mathrm{cm}/\mathrm{s}.\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{rate}\:\mathrm{of}\:\mathrm{change}\:\mathrm{of}\:\mathrm{it}'\mathrm{s} \\ $$$$\mathrm{distance}\:\mathrm{from}\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{1},\mathrm{0}\right) \\ $$$$\mathrm{when}\:\mathrm{x}\:=\:\mathrm{2} \\ $$

Question Number 199659    Answers: 1   Comments: 0

Question Number 199658    Answers: 1   Comments: 0

Question Number 199654    Answers: 1   Comments: 0

x^4 + 4x^3 − 8x + 1 − m = 0 has 4 real roots find m=?

$$\mathrm{x}^{\mathrm{4}} \:+\:\mathrm{4x}^{\mathrm{3}} \:−\:\mathrm{8x}\:+\:\mathrm{1}\:−\:\mathrm{m}\:=\:\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{4}\:\mathrm{real}\:\mathrm{roots} \\ $$$$\mathrm{find}\:\:\mathrm{m}=? \\ $$

Question Number 199642    Answers: 1   Comments: 2

Question Number 199638    Answers: 0   Comments: 2

∫_0 ^1 ∫_0 ^1 cos(max(x^3 ,y^(3/2) ))dxdy=A old Quation By mr,univers x^3 =t,y^(3/2) =s A=(2/9)∫_0 ^1 ∫_0 ^1 cos(max(t,s))t^(−(2/3)) s^(−(1/3)) dtds ∫_0 ^1 ∫_0 ^1 cos(max(t,s))t^(−(2/3)) s^(−(1/3)) dtds=∫_0 ^1 t^(−(2/3)) ∫_t ^1 cos(s)s^(−(1/3)) dsdt_(=C) +∫_0 ^1 s^(−(1/3)) ∫_s ^1 cos(t)t^(−(2/3)) dtds_(=B) c=∫_0 ^1 t^(−(2/3)) ∫_t ^1 Σ_(n≥0) (((−1)^n )/(2n!)).s^(2n−(1/3)) dsdt=Σ_(n≥0) (((−1)^n )/(2n!))∫_0 ^1 t^(−(2/3)) ∫_t ^1 s^(2n−(1/3)) dsdt =Σ_(n≥0) (((−1)^n )/((2n)!))∫_0 ^1 ((1−t^(2n+(2/3)) )/(2n+(2/3))).t^(−(2/3)) dt =Σ_(n≥0) (((−1)^n )/((2n)!(2n+(2/3)))).(3−(1/(2n+1)))=3Σ_(n≥0) (((−1)^n )/((2n+1)!))=3sin(1) B=∫_0 ^1 s^(−(1/3)) ∫_s ^1 Σ_(n≥0) (((−1)^n )/((2n)!))t^(2n−(2/3)) dtds =Σ(((−1)^n )/((2n)!))∫_0 ^1 s^(−(1/3)) .(((1−s^(2n+(1/3)) )/(2n+(1/3))))ds =Σ_(n≥0) (((−1)^n )/((2n)!(2n+(1/3))))((3/2)−(1/(2n+1))) =(3/2)Σ_(n≥0) (((−1)^n )/((2n+1)!))=(3/2)sin(1) A=(2/9)(c+B)=(2/9)((3/2)sin(1)+3sin(1)) =sin(1) ∫_0 ^1 ∫_0 ^1 cos(Max(x^3 ,y^(3/2) ))dxdy=sin(1)

$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{cos}\left(\mathrm{max}\left(\mathrm{x}^{\mathrm{3}} ,\mathrm{y}^{\frac{\mathrm{3}}{\mathrm{2}}} \right)\right)\mathrm{dxdy}=\mathrm{A} \\ $$$$\mathrm{old}\:\mathrm{Quation}\:\mathrm{By}\:\mathrm{mr},\mathrm{univers} \\ $$$$\mathrm{x}^{\mathrm{3}} =\mathrm{t},\mathrm{y}^{\frac{\mathrm{3}}{\mathrm{2}}} =\mathrm{s} \\ $$$$\mathrm{A}=\frac{\mathrm{2}}{\mathrm{9}}\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{cos}\left(\mathrm{max}\left(\mathrm{t},\mathrm{s}\right)\right)\mathrm{t}^{−\frac{\mathrm{2}}{\mathrm{3}}} \mathrm{s}^{−\frac{\mathrm{1}}{\mathrm{3}}} \mathrm{dtds} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{cos}\left(\mathrm{max}\left(\mathrm{t},\mathrm{s}\right)\right)\mathrm{t}^{−\frac{\mathrm{2}}{\mathrm{3}}} \mathrm{s}^{−\frac{\mathrm{1}}{\mathrm{3}}} \mathrm{dtds}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{t}^{−\frac{\mathrm{2}}{\mathrm{3}}} \int_{\mathrm{t}} ^{\mathrm{1}} \mathrm{cos}\left(\mathrm{s}\right)\mathrm{s}^{−\frac{\mathrm{1}}{\mathrm{3}}} \mathrm{dsdt}_{=\mathrm{C}} \\ $$$$+\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{s}^{−\frac{\mathrm{1}}{\mathrm{3}}} \int_{\mathrm{s}} ^{\mathrm{1}} \mathrm{cos}\left(\mathrm{t}\right)\mathrm{t}^{−\frac{\mathrm{2}}{\mathrm{3}}} \mathrm{dtds}_{=\mathrm{B}} \\ $$$$\mathrm{c}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{t}^{−\frac{\mathrm{2}}{\mathrm{3}}} \int_{\mathrm{t}} ^{\mathrm{1}} \underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{2n}!}.\mathrm{s}^{\mathrm{2n}−\frac{\mathrm{1}}{\mathrm{3}}} \mathrm{dsdt}=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{2n}!}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{t}^{−\frac{\mathrm{2}}{\mathrm{3}}} \int_{\boldsymbol{\mathrm{t}}} ^{\mathrm{1}} \mathrm{s}^{\mathrm{2n}−\frac{\mathrm{1}}{\mathrm{3}}} \boldsymbol{\mathrm{dsdt}} \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}\right)!}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−\mathrm{t}^{\mathrm{2n}+\frac{\mathrm{2}}{\mathrm{3}}} }{\mathrm{2n}+\frac{\mathrm{2}}{\mathrm{3}}}.\mathrm{t}^{−\frac{\mathrm{2}}{\mathrm{3}}} \mathrm{dt} \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}\right)!\left(\mathrm{2n}+\frac{\mathrm{2}}{\mathrm{3}}\right)}.\left(\mathrm{3}−\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{1}}\right)=\mathrm{3}\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}+\mathrm{1}\right)!}=\mathrm{3sin}\left(\mathrm{1}\right) \\ $$$$\mathrm{B}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{s}^{−\frac{\mathrm{1}}{\mathrm{3}}} \int_{\mathrm{s}} ^{\mathrm{1}} \underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}\right)!}\mathrm{t}^{\mathrm{2n}−\frac{\mathrm{2}}{\mathrm{3}}} \mathrm{dtds} \\ $$$$=\Sigma\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}\right)!}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{s}^{−\frac{\mathrm{1}}{\mathrm{3}}} .\left(\frac{\mathrm{1}−\mathrm{s}^{\mathrm{2n}+\frac{\mathrm{1}}{\mathrm{3}}} }{\mathrm{2n}+\frac{\mathrm{1}}{\mathrm{3}}}\right)\mathrm{ds} \\ $$$$=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}\right)!\left(\mathrm{2n}+\frac{\mathrm{1}}{\mathrm{3}}\right)}\left(\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2n}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}+\mathrm{1}\right)!}=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{sin}\left(\mathrm{1}\right) \\ $$$$\mathrm{A}=\frac{\mathrm{2}}{\mathrm{9}}\left(\mathrm{c}+\mathrm{B}\right)=\frac{\mathrm{2}}{\mathrm{9}}\left(\frac{\mathrm{3}}{\mathrm{2}}\mathrm{sin}\left(\mathrm{1}\right)+\mathrm{3sin}\left(\mathrm{1}\right)\right) \\ $$$$=\mathrm{sin}\left(\mathrm{1}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{cos}\left(\mathrm{Max}\left(\mathrm{x}^{\mathrm{3}} ,\mathrm{y}^{\frac{\mathrm{3}}{\mathrm{2}}} \right)\right)\mathrm{dxdy}=\mathrm{sin}\left(\mathrm{1}\right) \\ $$$$ \\ $$

Question Number 199632    Answers: 2   Comments: 0

In a swimming pool there are four pipes that fill it with water. When pipes 1, 2, 3 work, the pool fills in 12 minutes. When pipes 2, 3, 4 work, the pool fills in 15 minutes. When pipes 1 and 4 work, the pool fills in 20 minutes. Find how many minutes the pool fills if all four pipes work simultaneously.

In a swimming pool there are four pipes that fill it with water. When pipes 1, 2, 3 work, the pool fills in 12 minutes. When pipes 2, 3, 4 work, the pool fills in 15 minutes. When pipes 1 and 4 work, the pool fills in 20 minutes. Find how many minutes the pool fills if all four pipes work simultaneously.

Question Number 199606    Answers: 1   Comments: 0

Question Number 199602    Answers: 3   Comments: 0

2 + 7 + 12 + ... + x = 270 Find: x = ?

$$\mathrm{2}\:+\:\mathrm{7}\:+\:\mathrm{12}\:+\:...\:+\:\boldsymbol{\mathrm{x}}\:=\:\mathrm{270} \\ $$$$\mathrm{Find}:\:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$

Question Number 199598    Answers: 0   Comments: 0

Question Number 199597    Answers: 2   Comments: 0

calculate... Q : lim_( x→ (π/4)) ( 1 + sin(x) −cos(x) )^( tan(2x)) = ? • Nice − Mathematics •

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{calculate}... \\ $$$$ \\ $$$$\:\:\:\:\:\:\mathrm{Q}\::\:\:\:\:\:\:\mathrm{lim}_{\:{x}\rightarrow\:\frac{\pi}{\mathrm{4}}} \:\:\left(\:\:\mathrm{1}\:+\:\:{sin}\left({x}\right)\:−{cos}\left({x}\right)\:\right)^{\:{tan}\left(\mathrm{2}{x}\right)} \:=\:\:?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\bullet\:\mathscr{N}{ice}\:−\:\mathscr{M}{athematics}\:\bullet\: \\ $$$$\:\:\:\:\:\:\: \\ $$

Question Number 199592    Answers: 0   Comments: 9

Question Number 199593    Answers: 0   Comments: 0

if {a_n } is geometric sequence and we know , p=((n+m)/2) ⇒ S_p = f(a_n ,a_m ,n,m)=? ■Attention: The purpose of obtaining a relationship without the need for direct calculation is the common ratio of the sequence. For example, in a geometric sequence with a negative common ratio, if the sixth and tenth terms are equal to 5 and 12, the sum of the first 7 terms of this sequence?

$${if}\:\left\{{a}_{{n}} \right\}\:{is}\:{geometric}\:{sequence}\:{and}\: \\ $$$$ \\ $$$${we}\:{know}\:,\:{p}=\frac{{n}+{m}}{\mathrm{2}}\:\:\Rightarrow \\ $$$$\:{S}_{{p}} \:=\:{f}\left({a}_{{n}} ,{a}_{{m}} ,{n},{m}\right)=? \\ $$$$\blacksquare{Attention}:\: \\ $$The purpose of obtaining a relationship without the need for direct calculation is the common ratio of the sequence. For example, in a geometric sequence with a negative common ratio, if the sixth and tenth terms are equal to 5 and 12, the sum of the first 7 terms of this sequence?

Question Number 203650    Answers: 2   Comments: 0

lim_(x→0) (((2^x +3^x )/2))^(2/x) =?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{2}^{{x}} +\mathrm{3}^{{x}} }{\mathrm{2}}\right)^{\frac{\mathrm{2}}{{x}}} =? \\ $$

Question Number 203675    Answers: 1   Comments: 0

Question Number 203648    Answers: 1   Comments: 0

Question Number 199570    Answers: 1   Comments: 0

I = ∫_0 ^(π/2) tan^(−1) (((sinx )/2))dx

$$\:\:\:\:\:\:\:\:\mathrm{I}\:\:\:\:\:=\:\:\:\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{sin}{x}\:}{\mathrm{2}}\right){dx} \\ $$

Question Number 199568    Answers: 1   Comments: 1

Question Number 199553    Answers: 1   Comments: 0

lim_(x→0) ((tan ((x/2))−sin ((x/2)))/(x^2 ((√(x^2 +x−2))−(√(x^2 +2x−2)) ))) =?

$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)−\mathrm{sin}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)}{\mathrm{x}^{\mathrm{2}} \:\left(\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{2}}−\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}−\mathrm{2}}\:\right)}\:=? \\ $$

Question Number 199550    Answers: 2   Comments: 0

If x = (√(7 + 4(√3))) Find: ((x^4 − x^3 − 9x^2 − 2x + 5)/(x^2 − 4x + 3)) = ?

$$\mathrm{If} \\ $$$$\mathrm{x}\:=\:\sqrt{\mathrm{7}\:+\:\mathrm{4}\sqrt{\mathrm{3}}} \\ $$$$\mathrm{Find}: \\ $$$$\frac{\mathrm{x}^{\mathrm{4}} \:−\:\mathrm{x}^{\mathrm{3}} \:−\:\mathrm{9x}^{\mathrm{2}} \:−\:\mathrm{2x}\:+\:\mathrm{5}}{\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{4x}\:+\:\mathrm{3}}\:\:=\:\:? \\ $$

Question Number 199547    Answers: 3   Comments: 0

Question Number 199544    Answers: 1   Comments: 2

((Area(ydllow))/(Area(Blue)))=?

$$\:\:\frac{\boldsymbol{\mathrm{Area}}\left(\boldsymbol{{ydllow}}\right)}{\boldsymbol{\mathrm{Area}}\left(\boldsymbol{{Blue}}\right)}=? \\ $$

Question Number 199587    Answers: 0   Comments: 1

I konw that Bosons are with even number of proton and neutron and eletron but I am not sure of that fermions. Can someone help answer the question below? Which of this belongs to bosons or fermions? 1. 12C (Carbon−12) 2. 12C^+ ion 3. 4He^+ 4. H^− 5. 13C 6. Positron atom

$$ \\ $$$${I}\:{konw}\:{that}\:{Bosons}\:{are}\:{with}\:{even}\:{number}\:{of}\: \\ $$$${proton}\:{and}\:{neutron}\:{and}\:{eletron}\:{but}\:{I}\:{am}\:{not}\:{sure} \\ $$$${of}\:{that}\:{fermions}.\: \\ $$$${Can}\:{someone}\:{help}\:{answer}\:{the}\:{question}\:{below}? \\ $$$$ \\ $$$${Which}\:{of}\:{this}\:{belongs}\:{to}\:{bosons}\:{or}\:{fermions}? \\ $$$$\mathrm{1}.\:\mathrm{12}{C}\:\left({Carbon}−\mathrm{12}\right) \\ $$$$\mathrm{2}.\:\mathrm{12}{C}^{+} \:{ion} \\ $$$$\mathrm{3}.\:\mathrm{4}{He}^{+} \\ $$$$\mathrm{4}.\:{H}^{−} \\ $$$$\mathrm{5}.\:\mathrm{13}{C} \\ $$$$\mathrm{6}.\:{Positron}\:{atom} \\ $$

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