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Question Number 175375    Answers: 1   Comments: 0

Question Number 175369    Answers: 1   Comments: 0

solve for x log_(∣sinx∣ ) (x^2 −8x+23) > (3/(log_2 ∣sinx∣ ))

$$\:\:\mathrm{solve}\:\mathrm{for}\:{x} \\ $$$$\:\:\:\mathrm{log}_{\mid\mathrm{sin}{x}\mid\:} \left({x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{23}\right)\:>\:\frac{\mathrm{3}}{\mathrm{log}_{\mathrm{2}} \mid\mathrm{sin}{x}\mid\:} \\ $$

Question Number 175362    Answers: 1   Comments: 2

lim_(x→(π/4)) ((sin x−cos x)/(tan ((π/8)−(x/2)))) =?

$$\:\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:{x}−\mathrm{cos}\:{x}}{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{8}}−\frac{{x}}{\mathrm{2}}\right)}\:=? \\ $$

Question Number 175353    Answers: 1   Comments: 0

ax^2 +bx+c = 0 x = ((−b±(√(b^2 −4ac)))/(2a)) Example: Find the values of x in the equation x^2 +5x+4 = 0 In order to solve for that, let′s first take a look on what are the values of a, b and c 1x^2 + 5x + 4 = 0 a = 1 ; b = 5 ; c = 4 Now, using the quadratic formula x = ((−5±(√(5^2 −4(1)(4))))/(2(1))) = ((−5±(√(25−16)))/2) = ((−5±3)/2)

$$\:\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}\:=\:\mathrm{0} \\ $$$$\: \\ $$$$\:\mathrm{x}\:=\:\frac{−\mathrm{b}\pm\sqrt{\mathrm{b}^{\mathrm{2}} −\mathrm{4ac}}}{\mathrm{2a}} \\ $$$$\: \\ $$$$\:\mathrm{Example}: \\ $$$$\: \\ $$$$\:\mathrm{Find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{x}\:\mathrm{in}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:\mathrm{x}^{\mathrm{2}} +\mathrm{5x}+\mathrm{4}\:=\:\mathrm{0} \\ $$$$\: \\ $$$$\:\mathrm{In}\:\mathrm{order}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{for}\:\mathrm{that},\:\mathrm{let}'\mathrm{s}\:\mathrm{first}\:\mathrm{take} \\ $$$$\:\mathrm{a}\:\mathrm{look}\:\mathrm{on}\:\mathrm{what}\:\mathrm{are}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:\mathrm{a},\:\mathrm{b}\:\mathrm{and}\:\mathrm{c}\: \\ $$$$\: \\ $$$$\:\mathrm{1x}^{\mathrm{2}} \:+\:\mathrm{5x}\:+\:\mathrm{4}\:=\:\mathrm{0} \\ $$$$\: \\ $$$$\:\mathrm{a}\:=\:\mathrm{1}\:;\:\mathrm{b}\:=\:\mathrm{5}\:;\:\mathrm{c}\:=\:\mathrm{4} \\ $$$$\: \\ $$$$\:\mathrm{Now},\:\mathrm{using}\:\mathrm{the}\:\mathrm{quadratic}\:\mathrm{formula} \\ $$$$\: \\ $$$$\:\mathrm{x}\:=\:\frac{−\mathrm{5}\pm\sqrt{\mathrm{5}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{4}\right)}}{\mathrm{2}\left(\mathrm{1}\right)}\: \\ $$$$\:\:\:\:=\:\frac{−\mathrm{5}\pm\sqrt{\mathrm{25}−\mathrm{16}}}{\mathrm{2}} \\ $$$$\:\:\:\:=\:\frac{−\mathrm{5}\pm\mathrm{3}}{\mathrm{2}} \\ $$

Question Number 175351    Answers: 1   Comments: 0

∫_0 ^(π/2) ((sin^3 x)/(sin x+cos x)) dx =?

$$\:\:\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\mathrm{sin}\:^{\mathrm{3}} {x}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\:{dx}\:=? \\ $$

Question Number 175374    Answers: 0   Comments: 0

lim_(n→∞) (Σ_(j=1) ^n Σ_(i = 1) ^n ((i+j)/(i^2 +j^2 ))−((π/2)+ ln 2)n + ln n) = ?

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{i}\:=\:\mathrm{1}} {\overset{{n}} {\sum}}\frac{{i}+{j}}{{i}^{\mathrm{2}} +{j}^{\mathrm{2}} }−\left(\frac{\pi}{\mathrm{2}}+\:\mathrm{ln}\:\mathrm{2}\right){n}\:+\:\mathrm{ln}\:{n}\right)\:=\:? \\ $$

Question Number 175373    Answers: 4   Comments: 0

if x^6 + y^6 = 9 x^4 + y^4 = 5 then x^2 +y^2 =?

$$\:\:\mathrm{if}\:\:\:\:{x}^{\mathrm{6}} \:+\:{y}^{\mathrm{6}} \:=\:\mathrm{9} \\ $$$$\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{4}} \:+\:{y}^{\mathrm{4}} \:=\:\mathrm{5} \\ $$$$\:\mathrm{then}\:\:\:\:{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:=? \\ $$

Question Number 175346    Answers: 1   Comments: 0

Solve it by horner′s method. (2x^3 y^2 +3x^2 y−4x+5y−12)÷(x−3)

$${Solve}\:{it}\:{by}\:{horner}'{s}\:{method}. \\ $$$$\left(\mathrm{2}{x}^{\mathrm{3}} {y}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{2}} {y}−\mathrm{4}{x}+\mathrm{5}{y}−\mathrm{12}\right)\boldsymbol{\div}\left({x}−\mathrm{3}\right) \\ $$

Question Number 175343    Answers: 3   Comments: 1

∫_0 ^∞ (x^5 /( (√(3−x))))dx

$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{5}} }{\:\sqrt{\mathrm{3}−{x}}}{dx} \\ $$

Question Number 175335    Answers: 2   Comments: 0

solve the inequalities Q.(1) ((1+log_a ^2 x)/(1+log_a x)) > 1 , 0<a<1 Q.(2) log_x ((4x+5)/(6−5x)) < −1

$$\:\:\:\:\:\mathrm{solve}\:\mathrm{the}\:\mathrm{inequalities} \\ $$$$\:{Q}.\left(\mathrm{1}\right)\:\:\frac{\mathrm{1}+\mathrm{log}_{{a}} ^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{log}_{{a}} {x}}\:\:\:>\:\mathrm{1}\:\:\:,\:\:\mathrm{0}<{a}<\mathrm{1} \\ $$$$\:\:{Q}.\left(\mathrm{2}\right)\:\:\:\:\:\:\:\mathrm{log}_{{x}} \:\frac{\mathrm{4}{x}+\mathrm{5}}{\mathrm{6}−\mathrm{5}{x}}\:\:<\:\:−\mathrm{1} \\ $$

Question Number 175333    Answers: 1   Comments: 0

solve (x ∈ R ). ⌊ (( 2^( x) +1)/3) ⌋ + ⌊ 4^( (x/2)) ⌋ = 4 −−−−−−

$$ \\ $$$$\:\:\:{solve}\:\left({x}\:\in\:\mathbb{R}\:\right). \\ $$$$ \\ $$$$\:\:\lfloor\:\frac{\:\mathrm{2}^{\:{x}} \:+\mathrm{1}}{\mathrm{3}}\:\rfloor\:+\:\lfloor\:\mathrm{4}^{\:\frac{{x}}{\mathrm{2}}} \:\rfloor\:=\:\mathrm{4} \\ $$$$\:\:\:\:−−−−−−\: \\ $$

Question Number 175320    Answers: 3   Comments: 3

{ ((4x=2(mod 9))),((7x=2 (mod 13))) :}

$$\:\:\begin{cases}{\mathrm{4}{x}=\mathrm{2}\left({mod}\:\mathrm{9}\right)}\\{\mathrm{7}{x}=\mathrm{2}\:\left({mod}\:\mathrm{13}\right)}\end{cases} \\ $$

Question Number 181489    Answers: 1   Comments: 0

a and b are mutuallly prime numbers If (3/8) < (a/b) < (2/5) Find (b−a)_(min)

$$\boldsymbol{\mathrm{a}}\:\:\mathrm{and}\:\:\boldsymbol{\mathrm{b}}\:\mathrm{are}\:\mathrm{mutuallly}\:\mathrm{prime}\:\mathrm{numbers} \\ $$$$\mathrm{If}\:\:\:\frac{\mathrm{3}}{\mathrm{8}}\:<\:\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{b}}}\:<\:\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\mathrm{Find}\:\:\:\left(\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{a}}\right)_{\boldsymbol{\mathrm{min}}} \\ $$

Question Number 175315    Answers: 0   Comments: 0

Question Number 175311    Answers: 1   Comments: 1

Question Number 175305    Answers: 1   Comments: 0

Question Number 175304    Answers: 1   Comments: 0

Question Number 175302    Answers: 2   Comments: 0

Question Number 175301    Answers: 0   Comments: 0

Question Number 175297    Answers: 1   Comments: 0

Question Number 175294    Answers: 1   Comments: 0

Question Number 175285    Answers: 2   Comments: 0

simplfy sinh(log2)

$$\mathrm{simplfy} \\ $$$$\:\boldsymbol{\mathrm{sinh}}\left(\boldsymbol{\mathrm{log}}\mathrm{2}\right) \\ $$

Question Number 175284    Answers: 1   Comments: 0

Question Number 175280    Answers: 2   Comments: 0

A biology class at central high school predicted that a local population of animals will double in size every 12 years. The population at the beginning of 2014 was estimated to be 50 animals. If P represents the population n years after 2014. Find the equations represents the class model of the population over time?

$$\mathrm{A}\:\mathrm{biology}\:\:\mathrm{class}\:\mathrm{at}\:\mathrm{central}\:\:\mathrm{high}\:\:\mathrm{school} \\ $$$$\mathrm{predicted}\:\:\mathrm{that}\:\:\mathrm{a}\:\:\mathrm{local}\:\:\mathrm{population}\:\:\mathrm{of}\:\:\:\mathrm{animals} \\ $$$$\mathrm{will}\:\:\mathrm{double}\:\:\mathrm{in}\:\:\mathrm{size}\:\:\mathrm{every}\:\:\mathrm{12}\:\:\mathrm{years}. \\ $$$$\mathrm{The}\:\:\mathrm{population}\:\:\mathrm{at}\:\:\mathrm{the}\:\:\mathrm{beginning}\:\:\mathrm{of}\:\:\mathrm{2014} \\ $$$$\mathrm{was}\:\:\mathrm{estimated}\:\:\mathrm{to}\:\mathrm{be}\:\:\mathrm{50}\:\:\mathrm{animals}. \\ $$$$\mathrm{If}\:\:{P}\:\:\mathrm{represents}\:\:\mathrm{the}\:\:\mathrm{population}\:\:{n}\:\:\mathrm{years}\:\:\mathrm{after}\:\:\mathrm{2014}. \\ $$$$\mathrm{Find}\:\:\mathrm{the}\:\:\mathrm{equations}\:\:\mathrm{represents}\:\:\mathrm{the}\:\:\mathrm{class}\:\:\mathrm{model} \\ $$$$\mathrm{of}\:\:\mathrm{the}\:\:\mathrm{population}\:\:\mathrm{over}\:\:\mathrm{time}? \\ $$

Question Number 175277    Answers: 0   Comments: 0

Question Number 175279    Answers: 3   Comments: 0

Find the angle between the lines r = ((1),(0),(4) ) + λ ((2),((−1)),((−1)) ) r = ((1),(0),(4) ) + λ ((3),(0),(1) )

$$\mathrm{Find}\:\:\mathrm{the}\:\:\mathrm{angle}\:\:\mathrm{between}\:\:\mathrm{the}\:\:\mathrm{lines} \\ $$$${r}\:=\:\begin{pmatrix}{\mathrm{1}}\\{\mathrm{0}}\\{\mathrm{4}}\end{pmatrix}\:\:+\:\lambda\begin{pmatrix}{\mathrm{2}}\\{−\mathrm{1}}\\{−\mathrm{1}}\end{pmatrix} \\ $$$${r}\:=\:\begin{pmatrix}{\mathrm{1}}\\{\mathrm{0}}\\{\mathrm{4}}\end{pmatrix}\:\:+\:\lambda\begin{pmatrix}{\mathrm{3}}\\{\mathrm{0}}\\{\mathrm{1}}\end{pmatrix} \\ $$$$ \\ $$

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