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Question Number 120243    Answers: 1   Comments: 0

Question Number 120240    Answers: 1   Comments: 0

$$\rightarrow \\ $$

Question Number 120241    Answers: 0   Comments: 0

Correction to the last assignment (1)(1/(81^(x−2) )) = 27^(1−x) (1/3^(4(x−2)) )=3^(3(1−x)) (1/3^(4x−8) )=3^(3−3x) 3^(−4x+8) =3^(3−3x) −4x+8=3−3x C.L.T −4x+3x=3−8 −x=−5 x=5 (2)9^x =(1/(729)) 9^x =(1/3^6 ) 3^(2x) =3^(−6) 2x=−6 x=((−6)/2) x=−3 (3)16^x =0.125 2^(4x) =((125)/(1000)) 2^(4x) =(1/8) 2^(4x) =2^(−3) 4x=−3 x=((−3)/4) (4)a^(1/2) =4 a=

$$\boldsymbol{\mathrm{C}}{orrection}\:\boldsymbol{{to}}\:\boldsymbol{{the}}\:\boldsymbol{{last}}\:\boldsymbol{{assignment}} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\frac{\mathrm{1}}{\mathrm{81}^{\boldsymbol{{x}}−\mathrm{2}} }\:=\:\mathrm{27}^{\mathrm{1}−\boldsymbol{{x}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{4}\left(\boldsymbol{{x}}−\mathrm{2}\right)} }=\mathrm{3}^{\mathrm{3}\left(\mathrm{1}−\boldsymbol{{x}}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{4}\boldsymbol{{x}}−\mathrm{8}} }=\mathrm{3}^{\mathrm{3}−\mathrm{3}\boldsymbol{{x}}} \\ $$$$\mathrm{3}^{−\mathrm{4}\boldsymbol{{x}}+\mathrm{8}} =\mathrm{3}^{\mathrm{3}−\mathrm{3}\boldsymbol{{x}}} \\ $$$$−\mathrm{4}\boldsymbol{{x}}+\mathrm{8}=\mathrm{3}−\mathrm{3}\boldsymbol{{x}} \\ $$$$\boldsymbol{{C}}.{L}.{T} \\ $$$$−\mathrm{4}{x}+\mathrm{3}{x}=\mathrm{3}−\mathrm{8} \\ $$$$−{x}=−\mathrm{5} \\ $$$${x}=\mathrm{5} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\mathrm{9}^{{x}} =\frac{\mathrm{1}}{\mathrm{729}} \\ $$$$\mathrm{9}^{{x}} =\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{6}} } \\ $$$$\mathrm{3}^{\mathrm{2}{x}} =\mathrm{3}^{−\mathrm{6}} \\ $$$$\mathrm{2}{x}=−\mathrm{6} \\ $$$${x}=\frac{−\mathrm{6}}{\mathrm{2}} \\ $$$${x}=−\mathrm{3} \\ $$$$ \\ $$$$\left(\mathrm{3}\right)\mathrm{16}^{{x}} =\mathrm{0}.\mathrm{125} \\ $$$$\mathrm{2}^{\mathrm{4}{x}} =\frac{\mathrm{125}}{\mathrm{1000}} \\ $$$$\mathrm{2}^{\mathrm{4}{x}} =\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\mathrm{2}^{\mathrm{4}{x}} =\mathrm{2}^{−\mathrm{3}} \\ $$$$\mathrm{4}{x}=−\mathrm{3} \\ $$$${x}=\frac{−\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$$$\left(\mathrm{4}\right){a}^{\mathrm{1}/\mathrm{2}} =\mathrm{4} \\ $$$${a}= \\ $$

Question Number 120238    Answers: 0   Comments: 0

Question Number 120233    Answers: 0   Comments: 0

$$ \\ $$

Question Number 120237    Answers: 0   Comments: 1

Question Number 120230    Answers: 3   Comments: 0

∫ (x^3 /((x +2)^4 )) dx OR ∫ ((cos x)/(1 + cos x)) dx a^→ = i^ − j^ + 3k^ and b^(→ ) = 2i^ − 7j^ + k^ .

$$\:\:\:\:\int\:\frac{\mathrm{x}^{\mathrm{3}} }{\left(\mathrm{x}\:+\mathrm{2}\right)^{\mathrm{4}} }\:\mathrm{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{OR} \\ $$$$\:\:\:\int\:\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{1}\:+\:\mathrm{cos}\:\mathrm{x}}\:\mathrm{dx} \\ $$$$ \\ $$$$\:\:\:\:\:\overset{\rightarrow} {\mathrm{a}}\:=\:\hat {\mathrm{i}}\:−\:\hat {\mathrm{j}}\:+\:\mathrm{3}\hat {\mathrm{k}}\:\:\mathrm{and}\:\overset{\rightarrow\:} {\mathrm{b}}\:=\:\mathrm{2}\hat {\mathrm{i}}\:−\:\mathrm{7}\hat {\mathrm{j}}\:+\:\hat {\mathrm{k}}\:. \\ $$$$ \\ $$$$\:\:\:\: \\ $$

Question Number 120229    Answers: 2   Comments: 0

Question Number 120228    Answers: 0   Comments: 0

Please solve using special function like LambertW function or any other function (if possibe) ((8/7))^x +17^x =25x

$${Please}\:{solve}\:{using}\:{special}\:{function}\:{like}\:{LambertW} \\ $$$${function}\:{or}\:{any}\:{other}\:{function}\:\left({if}\:{possibe}\right) \\ $$$$\left(\frac{\mathrm{8}}{\mathrm{7}}\right)^{{x}} +\mathrm{17}^{{x}} =\mathrm{25}{x} \\ $$

Question Number 120225    Answers: 0   Comments: 0

Question Number 120215    Answers: 1   Comments: 0

Prove that (((((a+b)(b+c)(c+a))/8) ))^(1/3) ≥ (√((ab+bc+ca)/3)) for a,b,c > 0

$${Prove}\:{that}\:\sqrt[{\mathrm{3}}]{\frac{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)}{\mathrm{8}}\:}\:\geqslant\:\sqrt{\frac{{ab}+{bc}+{ca}}{\mathrm{3}}} \\ $$$${for}\:{a},{b},{c}\:>\:\mathrm{0} \\ $$

Question Number 120211    Answers: 3   Comments: 0

evaluate lim_(x→∞) f(x) and lim_(x→−∞) f(x) for f(x)=(x/( (√(x^2 +1)))).

$$\:{evaluate}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}{f}\left({x}\right)\:{and}\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}{f}\left({x}\right) \\ $$$${for}\:{f}\left({x}\right)=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}. \\ $$

Question Number 120209    Answers: 1   Comments: 0

Peter has 12 relatives (5 man & 7 woman) and his wife also has 12 relatives (5 woman &7 man). They do not have common relatives. They decided to invite 12 guests ,six each of their relatives, such that there are six man and six woman among the guests. How many ways can they choose 12 guests?

$${Peter}\:{has}\:\mathrm{12}\:{relatives}\:\left(\mathrm{5}\:{man}\:\&\:\mathrm{7}\:{woman}\right) \\ $$$${and}\:{his}\:{wife}\:{also}\:{has}\:\mathrm{12}\:{relatives} \\ $$$$\left(\mathrm{5}\:{woman}\:\&\mathrm{7}\:{man}\right).\:{They}\:{do}\:{not} \\ $$$${have}\:{common}\:{relatives}.\:{They}\:{decided} \\ $$$${to}\:{invite}\:\mathrm{12}\:{guests}\:,{six}\:{each}\:{of} \\ $$$${their}\:{relatives},\:{such}\:{that}\:{there} \\ $$$${are}\:{six}\:{man}\:{and}\:{six}\:{woman}\: \\ $$$${among}\:{the}\:{guests}.\:{How}\:{many} \\ $$$${ways}\:{can}\:{they}\:{choose}\:\mathrm{12}\:{guests}? \\ $$

Question Number 120207    Answers: 0   Comments: 0

Question Number 120206    Answers: 0   Comments: 0

Find the remainder when 1×3×5×7×…×2017×2019 is divided by 1000.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\: \\ $$$$\mathrm{1}×\mathrm{3}×\mathrm{5}×\mathrm{7}×\ldots×\mathrm{2017}×\mathrm{2019}\:\mathrm{is}\:\mathrm{divided} \\ $$$$\mathrm{by}\:\mathrm{1000}. \\ $$

Question Number 120204    Answers: 0   Comments: 2

Call a 7−digit telephone number d_1 d_2 d_3 −d_4 d_5 d_6 d_7 memorable if the prefix sequence d_1 d_2 d_3 is exactly the same as either of the sequences d_4 d_5 d_6 or d_5 d_6 d_7 (posibly both). Assuming that each d_i can be any of the ten decimal digits 0,1,2,...,9 . Find the number of different memorable telephone numbers

$${Call}\:{a}\:\mathrm{7}−{digit}\:{telephone}\:{number}\:{d}_{\mathrm{1}} {d}_{\mathrm{2}} {d}_{\mathrm{3}} −{d}_{\mathrm{4}} {d}_{\mathrm{5}} {d}_{\mathrm{6}} {d}_{\mathrm{7}} \\ $$$${memorable}\:{if}\:{the}\:{prefix}\:{sequence}\:{d}_{\mathrm{1}} {d}_{\mathrm{2}} {d}_{\mathrm{3}} \\ $$$${is}\:{exactly}\:{the}\:{same}\:{as}\:{either}\:{of}\:{the}\:{sequences} \\ $$$${d}_{\mathrm{4}} {d}_{\mathrm{5}} {d}_{\mathrm{6}} \:{or}\:{d}_{\mathrm{5}} {d}_{\mathrm{6}} {d}_{\mathrm{7}} \:\left({posibly}\:{both}\right). \\ $$$${Assuming}\:{that}\:{each}\:{d}_{{i}} \:{can}\:{be}\:{any}\:{of}\:{the}\:{ten} \\ $$$${decimal}\:{digits}\:\mathrm{0},\mathrm{1},\mathrm{2},...,\mathrm{9}\:.\:{Find}\:{the}\:{number} \\ $$$${of}\:{different}\:{memorable}\:{telephone} \\ $$$${numbers} \\ $$

Question Number 120202    Answers: 0   Comments: 0

Show for the equation a^n = b^2 −1 where n>1 and a>2 are any natural numbers , there are no positive integer solutions for a and b ?

$${Show}\:{for}\:{the}\:{equation}\:{a}^{{n}} \:=\:{b}^{\mathrm{2}} −\mathrm{1}\: \\ $$$${where}\:{n}>\mathrm{1}\:{and}\:{a}>\mathrm{2}\:{are}\:{any}\:{natural} \\ $$$${numbers}\:,\:{there}\:{are}\:{no}\:{positive}\:{integer} \\ $$$${solutions}\:{for}\:{a}\:{and}\:{b}\:? \\ $$

Question Number 120201    Answers: 2   Comments: 0

lim_(x→∞) ((ln (x+e^x +e^(2x) ))/x)=?

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\left({x}+{e}^{{x}} +{e}^{\mathrm{2}{x}} \right)}{{x}}=? \\ $$

Question Number 120196    Answers: 1   Comments: 0

The ratio of male and female participants in a competition is 3:2. Given that 15% of participants are prize winner, and among the prize winner, the ratio of male to female is 2:1. Find the ratio of male and female among the participants that are not prize winner.

$$\mathrm{The}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{male}\:\mathrm{and}\:\mathrm{female}\:\mathrm{participants} \\ $$$$\mathrm{in}\:\mathrm{a}\:\mathrm{competition}\:\mathrm{is}\:\mathrm{3}:\mathrm{2}.\:\mathrm{Given}\:\mathrm{that}\:\mathrm{15\%}\: \\ $$$$\mathrm{of}\:\mathrm{participants}\:\mathrm{are}\:\mathrm{prize}\:\mathrm{winner},\:\mathrm{and}\:\mathrm{among} \\ $$$$\mathrm{the}\:\mathrm{prize}\:\mathrm{winner},\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{male}\:\mathrm{to}\:\mathrm{female} \\ $$$$\mathrm{is}\:\mathrm{2}:\mathrm{1}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{male}\:\mathrm{and}\:\mathrm{female} \\ $$$$\mathrm{among}\:\mathrm{the}\:\mathrm{participants}\:\mathrm{that}\:\mathrm{are}\:\mathrm{not}\:\mathrm{prize} \\ $$$$\mathrm{winner}. \\ $$

Question Number 120195    Answers: 3   Comments: 0

C .lim_(z→0) ((Re(z).Im(z))/(Re(z)+Im(z))) =?

$$\mathbb{C}\:.\underset{\mathrm{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{Re}\left(\mathrm{z}\right).\mathrm{Im}\left(\mathrm{z}\right)}{\mathrm{Re}\left(\mathrm{z}\right)+\mathrm{Im}\left(\mathrm{z}\right)}\:=? \\ $$

Question Number 120194    Answers: 1   Comments: 0

Question Number 120188    Answers: 0   Comments: 0

Let n be a positive integer . Prove that Σ_(k=0) ^n 2^k ((n),(k) ) ((( n−k)),((⌊((n−k)/2)⌋)) ) = (((2n+1)),(( n)) )

$${Let}\:{n}\:{be}\:{a}\:{positive}\:{integer}\:. \\ $$$${Prove}\:{that}\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{2}^{{k}} \:\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\:\begin{pmatrix}{\:\:{n}−{k}}\\{\lfloor\frac{{n}−{k}}{\mathrm{2}}\rfloor}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:{n}}\end{pmatrix} \\ $$

Question Number 120186    Answers: 1   Comments: 0

Question Number 120185    Answers: 1   Comments: 0

solve for x, y∈Z (√x)+(√y)=(√(2020))

$${solve}\:{for}\:{x},\:{y}\in{Z} \\ $$$$\sqrt{{x}}+\sqrt{{y}}=\sqrt{\mathrm{2020}} \\ $$

Question Number 120183    Answers: 0   Comments: 0

Question Number 123960    Answers: 0   Comments: 0

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