Question and Answers Forum

AllQuestion and Answers: Page 1013

Question Number 118369 Answers: 0 Comments: 1

657×10^4 =(5/7)+

$$\:\mathrm{657}×\mathrm{10}^{\mathrm{4}} =\frac{\mathrm{5}}{\mathrm{7}}+ \\ $$

Question Number 118366 Answers: 0 Comments: 4

4. Turunan fungsi f(x)=^5 (√((10x^2 −4)^8 )) adalah f^1 (x). Nilai f^1 (1)=... a. 1 b. 8 c. 14 d. 16 e. 20

$$\mathrm{4}.\:\mathrm{Turunan}\:\mathrm{fungsi}\:\mathrm{f}\left(\mathrm{x}\right)=^{\mathrm{5}} \sqrt{\left(\mathrm{10x}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{8}} \:\:} \\ $$$$\mathrm{adalah}\:\mathrm{f}^{\mathrm{1}} \left(\mathrm{x}\right).\:\mathrm{Nilai}\:\mathrm{f}^{\mathrm{1}} \left(\mathrm{1}\right)=... \\ $$$${a}.\:\mathrm{1} \\ $$$${b}.\:\mathrm{8} \\ $$$${c}.\:\mathrm{14} \\ $$$$\mathrm{d}.\:\mathrm{16} \\ $$$$\mathrm{e}.\:\mathrm{20} \\ $$

Question Number 118364 Answers: 1 Comments: 0

...ordinary differential equation... y′′ −xy′+y=0 general solution ::=?? .m.n.1970.

$$\:\:\:\:\:\:\:\:\:\:\:\:\:...{ordinary}\:{differential}\:{equation}... \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:{y}''\:−{xy}'+{y}=\mathrm{0} \\ $$$$\:\:\:\:{general}\:{solution}\:::=?? \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:.{m}.{n}.\mathrm{1970}. \\ $$$$ \\ $$

Question Number 118357 Answers: 2 Comments: 0

Find the sum of the prime factors of 20^5 +21.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{prime}\:\mathrm{factors}\:\mathrm{of} \\ $$$$\mathrm{20}^{\mathrm{5}} +\mathrm{21}. \\ $$

Question Number 118346 Answers: 0 Comments: 0

Let 30 furniture sets arrive at two city stations A and B ,15 sets for each station All funiture sets need to be delivered to two furniture stores C and D,and 10 sets must be delivered to store C,and to store D−20.It is known that delivery one furniture set from the station A to the stores C and D costs 1 and 3 monetary units,and from station B−2 and 5 units respectively,.It is necessary to draw out such a transportation plan that the cost of transportation is the lowest

$$\mathrm{Let}\:\mathrm{30}\:\mathrm{furniture}\:\mathrm{sets}\:\mathrm{arrive}\:\mathrm{at}\:\mathrm{two}\:\mathrm{city} \\ $$$$\mathrm{stations}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:,\mathrm{15}\:\mathrm{sets}\:\mathrm{for}\:\mathrm{each}\:\mathrm{station} \\ $$$$\mathrm{All}\:\mathrm{funiture}\:\mathrm{sets}\:\mathrm{need}\:\mathrm{to}\:\mathrm{be}\:\mathrm{delivered} \\ $$$$\mathrm{to}\:\mathrm{two}\:\mathrm{furniture}\:\mathrm{stores}\:\mathrm{C}\:\mathrm{and}\:\mathrm{D},\mathrm{and}\:\mathrm{10} \\ $$$$\mathrm{sets}\:\mathrm{must}\:\mathrm{be}\:\mathrm{delivered}\:\mathrm{to}\:\mathrm{store}\:\mathrm{C},\mathrm{and} \\ $$$$\mathrm{to}\:\mathrm{store}\:\mathrm{D}−\mathrm{20}.\mathrm{It}\:\mathrm{is}\:\mathrm{known}\:\mathrm{that}\:\mathrm{delivery} \\ $$$$\mathrm{one}\:\mathrm{furniture}\:\mathrm{set}\:\mathrm{from}\:\mathrm{the}\:\mathrm{station}\:\mathrm{A}\:\mathrm{to} \\ $$$$\mathrm{the}\:\mathrm{stores}\:\mathrm{C}\:\mathrm{and}\:\mathrm{D}\:\mathrm{costs}\:\mathrm{1}\:\mathrm{and}\:\mathrm{3}\:\mathrm{monetary} \\ $$$$\mathrm{units},\mathrm{and}\:\mathrm{from}\:\mathrm{station}\:\mathrm{B}−\mathrm{2}\:\mathrm{and}\:\mathrm{5}\:\mathrm{units} \\ $$$$\:\mathrm{respectively},.\mathrm{It}\:\mathrm{is}\:\mathrm{necessary}\:\mathrm{to}\:\mathrm{draw} \\ $$$$\mathrm{out}\:\mathrm{such}\:\mathrm{a}\:\mathrm{transportation}\:\mathrm{plan}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{cost}\:\mathrm{of}\:\mathrm{transportation}\:\mathrm{is}\:\mathrm{the}\:\mathrm{lowest} \\ $$

Question Number 118349 Answers: 1 Comments: 0

solution xdy + (3y − e^x )dx with integration factor

$${solution}\:{xdy}\:+\:\left(\mathrm{3}{y}\:−\:{e}^{{x}} \right){dx}\: \\ $$$${with}\:{integration}\:{factor} \\ $$

Question Number 118347 Answers: 1 Comments: 0

∫ (dx/( (√x) +(x)^(1/(3 )) ))

$$\:\:\:\int\:\frac{{dx}}{\:\sqrt{{x}}\:+\sqrt[{\mathrm{3}\:}]{{x}}}\: \\ $$

Question Number 118362 Answers: 1 Comments: 0

Prove that 𝛑 is an irrational number

$$\mathrm{Prove}\:\mathrm{that}\:\boldsymbol{\pi}\:\mathrm{is}\:\mathrm{an}\:\mathrm{irrational}\:\mathrm{number} \\ $$

Question Number 118340 Answers: 1 Comments: 1

∫_(π/3) ^(π/2) (dx/(1+sin x−cos x))

$$\:\:\:\underset{\pi/\mathrm{3}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{{dx}}{\mathrm{1}+\mathrm{sin}\:{x}−\mathrm{cos}\:{x}} \\ $$

Question Number 118339 Answers: 0 Comments: 0

Question Number 118338 Answers: 3 Comments: 0

solve ∫ (dx/(3−5sin x))

$$\:\:\:\:{solve}\:\int\:\frac{{dx}}{\mathrm{3}−\mathrm{5sin}\:{x}}\: \\ $$

Question Number 118327 Answers: 4 Comments: 0

lim_(x→1) ((x−(√(2−x^2 )))/(2x−(√(2+2x^2 )))) ?

$$\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{x}−\sqrt{\mathrm{2}−{x}^{\mathrm{2}} }}{\mathrm{2}{x}−\sqrt{\mathrm{2}+\mathrm{2}{x}^{\mathrm{2}} }}\:? \\ $$

Question Number 118325 Answers: 0 Comments: 0

Question Number 118323 Answers: 0 Comments: 0

Prove that ζ(1−s)=2^(1−s) π^(−s) cos(((sπ)/2))Γ(s)ζ(s)

$${Prove}\:{that}\: \\ $$$$\zeta\left(\mathrm{1}−{s}\right)=\mathrm{2}^{\mathrm{1}−{s}} \pi^{−{s}} {cos}\left(\frac{{s}\pi}{\mathrm{2}}\right)\Gamma\left({s}\right)\zeta\left({s}\right) \\ $$

Question Number 118322 Answers: 2 Comments: 2

solve (1/x)+(1/y)+(1/z)=(3/4) with x,y,z∈N

$${solve} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${with}\:{x},{y},{z}\in\mathbb{N} \\ $$

Question Number 118318 Answers: 1 Comments: 0

Question Number 118308 Answers: 1 Comments: 1

Question Number 118307 Answers: 2 Comments: 1

∫^∞ _0 ((x^2 −2)/(x^4 +x^2 +1)) dx

$$\:\:\:\underset{\mathrm{0}} {\int}^{\infty} \:\frac{{x}^{\mathrm{2}} −\mathrm{2}}{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}\:{dx} \\ $$

Question Number 118304 Answers: 0 Comments: 0

... ⧫_▼ ^▼ Advanced calculus⧫_▼ ^▼ ... prove that :: Σ_(n=1) ^∞ (1/(3^n n^3 )) =^(???) (7/8)ζ( 3 )+(1/6)ln^3 (2)−(π^2 /(12)) ln(2) ✓ ...♠M.N.July 1970♠... .Good luck.

Question Number 118298 Answers: 0 Comments: 1

b, x, y, c are consecutive terms of a G.P. ∴ x = br, y = br^2 , c = br^3 A.M. between b and c is a ∴ ((b+c)/2) = a ⇒ 2a = b+c 2abc = (b+c)bc = b^2 c + bc^2 = b^2 (br^3 ) + b(br^3 )^2 = b^3 r^3 + b^3 r^6 = (br)^3 + (br^2 )^3 = x^3 + y^3 ∴ x^3 + y^3 = 2abc ←

Question Number 118292 Answers: 2 Comments: 0

Prove that: ∫_0 ^( 1) ((x^n −1)/(lnx)) = ln∣n+1∣

$$\boldsymbol{\mathrm{P}}\mathrm{rove}\:\mathrm{that}: \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{x}^{\mathrm{n}} −\mathrm{1}}{\mathrm{lnx}}\:=\:\boldsymbol{\mathrm{ln}}\mid\boldsymbol{\mathrm{n}}+\mathrm{1}\mid \\ $$

Question Number 118286 Answers: 2 Comments: 0

What condition should be satisfied by the vectors a and b for the following relations to hold true :(a)∣a+b∣=∣a−b∣ ;(b)∣ a+b∣>∣ a−b∣;(c)∣a+ b∣< ∣a−b∣

Question Number 118280 Answers: 0 Comments: 1

(26) OA^(→) = a^→ = (((4.8)),((3.6)) ) , OB^(→) = b^→ = ((( 8)),((15)) ) a^→ . b^→ = (4.8)(8) + (3.6)(15) = 92.4^(→ ) a = (√(4.8^2 +3.6^2 )) = (√(23.04+12.96)) = (√(36)) = 6 b = (√(8^2 +15^2 )) = (√(64+225)) = (√(289)) = 17 a.b = 6 . 17 = 102 cos AOB = ((a^→ .b^→ )/(a.b)) = ((92.4)/(102)) = 0.9059 = cos 25.06° ∴ ∠AOB = 25.06°

$$\left(\mathrm{26}\right)\:\:\:\overset{\rightarrow} {\mathrm{OA}}\:=\:\overset{\rightarrow} {\mathrm{a}}\:=\:\begin{pmatrix}{\mathrm{4}.\mathrm{8}}\\{\mathrm{3}.\mathrm{6}}\end{pmatrix}\:\:,\:\:\overset{\rightarrow} {\mathrm{OB}}\:=\:\overset{\rightarrow} {\mathrm{b}}\:=\:\begin{pmatrix}{\:\:\mathrm{8}}\\{\mathrm{15}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\overset{\rightarrow} {\mathrm{a}}.\:\overset{\rightarrow} {\mathrm{b}}\:=\:\left(\mathrm{4}.\mathrm{8}\right)\left(\mathrm{8}\right)\:+\:\left(\mathrm{3}.\mathrm{6}\right)\left(\mathrm{15}\right)\:=\:\mathrm{92}.\overset{\rightarrow\:} {\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}\:=\:\sqrt{\mathrm{4}.\mathrm{8}^{\mathrm{2}} +\mathrm{3}.\mathrm{6}^{\mathrm{2}} }\:=\:\:\sqrt{\mathrm{23}.\mathrm{04}+\mathrm{12}.\mathrm{96}}\:=\:\sqrt{\mathrm{36}}\:=\:\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}\:=\:\sqrt{\mathrm{8}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} }\:=\:\sqrt{\mathrm{64}+\mathrm{225}}\:=\:\sqrt{\mathrm{289}}\:=\:\mathrm{17} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}.\mathrm{b}\:=\:\mathrm{6}\:.\:\mathrm{17}\:=\:\mathrm{102} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{cos}\:\mathrm{AOB}\:=\:\frac{\overset{\rightarrow} {\mathrm{a}}.\overset{\rightarrow} {\mathrm{b}}}{\mathrm{a}.\mathrm{b}}\:=\:\frac{\mathrm{92}.\mathrm{4}}{\mathrm{102}}\:=\:\mathrm{0}.\mathrm{9059}\:=\:\mathrm{cos}\:\mathrm{25}.\mathrm{06}° \\ $$$$\:\:\:\:\:\:\:\:\:\:\therefore\:\:\angle\mathrm{AOB}\:=\:\mathrm{25}.\mathrm{06}° \\ $$

Question Number 118277 Answers: 0 Comments: 0

(23) Let a^→ = ((( 7)),((24)) ) and b^→ = ((( 3)),((−4)) ) a^→ .b^→ = (7)(3) + (24)(−4) = 21−96 = −75 a = (√(7^2 +24^2 )) = (√(49+576)) = (√(625)) = 25 b = (√(3^2 +(−4)^2 )) = (√(9+16)) = (√(25)) = 5 a.b = 25 . 5 = 125 Let θ be the angle between a^→ and b^→ . cosθ = ((a^→ .b^→ )/(a.b)) = ((−75)/(125)) = −0.6 = cos126.87° ∴ θ = 126.87°

$$\left(\mathrm{23}\right)\:\mathrm{Let}\:\overset{\rightarrow} {\mathrm{a}}\:=\:\begin{pmatrix}{\:\:\mathrm{7}}\\{\mathrm{24}}\end{pmatrix}\:\:\:\mathrm{and}\:\:\overset{\rightarrow} {\mathrm{b}}\:=\:\begin{pmatrix}{\:\:\:\mathrm{3}}\\{−\mathrm{4}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\overset{\rightarrow} {\mathrm{a}}.\overset{\rightarrow} {\mathrm{b}}\:=\:\left(\mathrm{7}\right)\left(\mathrm{3}\right)\:+\:\left(\mathrm{24}\right)\left(−\mathrm{4}\right)\:=\:\mathrm{21}−\mathrm{96}\:=\:−\mathrm{75} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}\:=\:\:\sqrt{\mathrm{7}^{\mathrm{2}} +\mathrm{24}^{\mathrm{2}} }\:=\:\sqrt{\mathrm{49}+\mathrm{576}}\:=\:\sqrt{\mathrm{625}}\:=\:\mathrm{25} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}\:=\:\:\sqrt{\mathrm{3}^{\mathrm{2}} +\left(−\mathrm{4}\right)^{\mathrm{2}} }\:=\:\sqrt{\mathrm{9}+\mathrm{16}}\:=\:\sqrt{\mathrm{25}}\:=\:\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}.\mathrm{b}\:=\:\mathrm{25}\:.\:\mathrm{5}\:=\:\:\mathrm{125}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{Let}\:\theta\:\mathrm{be}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{between}\:\overset{\rightarrow} {\mathrm{a}}\:\mathrm{and}\:\overset{\rightarrow} {\mathrm{b}}. \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{cos}\theta\:=\:\frac{\overset{\rightarrow} {\mathrm{a}}.\overset{\rightarrow} {\mathrm{b}}}{\mathrm{a}.\mathrm{b}}\:=\:\frac{−\mathrm{75}}{\mathrm{125}}\:=\:−\mathrm{0}.\mathrm{6}\:=\:\mathrm{cos126}.\mathrm{87}° \\ $$$$\:\:\:\:\:\:\:\:\:\:\therefore\:\:\theta\:=\:\mathrm{126}.\mathrm{87}°\:\:\:\: \\ $$

Question Number 118275 Answers: 1 Comments: 1

Question Number 118274 Answers: 2 Comments: 0

y′′−3y′+2y = (1/(1+e^(−x) ))

$$\:\:{y}''−\mathrm{3}{y}'+\mathrm{2}{y}\:=\:\frac{\mathrm{1}}{\mathrm{1}+{e}^{−{x}} }\: \\ $$

Pg 1008 Pg 1009 Pg 1010 Pg 1011 Pg 1012 Pg 1013 Pg 1014 Pg 1015 Pg 1016 Pg 1017

Contact: info@tinkutara.com