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Question Number 170855    Answers: 1   Comments: 0

⌊x⌋= log_2 (4^( x) −2^( x) −1)⇒ ⌊ 4^( x) ⌋=?

$$ \\ $$$$\:\:\:\lfloor{x}\rfloor=\:{log}_{\mathrm{2}} \left(\mathrm{4}^{\:{x}} −\mathrm{2}^{\:{x}} −\mathrm{1}\right)\Rightarrow\:\lfloor\:\mathrm{4}^{\:{x}} \rfloor=? \\ $$$$ \\ $$

Question Number 181349    Answers: 0   Comments: 0

Question Number 170872    Answers: 1   Comments: 0

Question Number 170871    Answers: 1   Comments: 0

Why is it equal? (1/2)∫_0 ^π sin^(2p) udu=∫_0 ^(π/2) sin^(2p) udu

$${Why}\:{is}\:{it}\:{equal}? \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\pi} {\int}}{sin}^{\mathrm{2}{p}} {udu}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}{sin}^{\mathrm{2}{p}} {udu} \\ $$

Question Number 170849    Answers: 1   Comments: 0

2x+(√x)=(1/2) 8x+(1/( (√x)))=?

$$\mathrm{2}{x}+\sqrt{{x}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{8}{x}+\frac{\mathrm{1}}{\:\sqrt{{x}}}=? \\ $$

Question Number 170847    Answers: 0   Comments: 1

Question Number 170844    Answers: 2   Comments: 0

solve x^2 +(√(3−x))=3

$${solve}\:{x}^{\mathrm{2}} +\sqrt{\mathrm{3}−{x}}=\mathrm{3} \\ $$

Question Number 170843    Answers: 1   Comments: 0

Question Number 170838    Answers: 1   Comments: 0

if a<b, show that a<((mb+na)/(m+n))<b a,b,m,n are arbitrary constants

$$\mathrm{if}\:{a}<{b},\:\mathrm{show}\:\mathrm{that}\:{a}<\frac{{mb}+{na}}{{m}+{n}}<{b} \\ $$$${a},{b},{m},{n}\:\mathrm{are}\:\mathrm{arbitrary}\:\mathrm{constants} \\ $$

Question Number 170836    Answers: 2   Comments: 1

x^3 − 2x^2 − 5x + 6 = 0 α^3 + β^3 + γ^3 = ?

$${x}^{\mathrm{3}} \:−\:\mathrm{2}{x}^{\mathrm{2}} \:−\:\mathrm{5}{x}\:+\:\mathrm{6}\:=\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\alpha^{\mathrm{3}} \:+\:\beta^{\mathrm{3}} \:+\:\gamma^{\mathrm{3}} \:=\:? \\ $$

Question Number 170833    Answers: 1   Comments: 0

Solve { ((C_x ^y =C_x ^(y+1) )),((4C_x ^y =5C_x ^(y−1) )) :} or C_n ^k =((n!)/(k!(n−k)!))

$${Solve}\: \\ $$$$\begin{cases}{{C}_{{x}} ^{{y}} ={C}_{{x}} ^{{y}+\mathrm{1}} }\\{\mathrm{4}{C}_{{x}} ^{{y}} =\mathrm{5}{C}_{{x}} ^{{y}−\mathrm{1}} }\end{cases}\:{or}\:{C}_{{n}} ^{{k}} =\frac{{n}!}{{k}!\left({n}−{k}\right)!} \\ $$

Question Number 170832    Answers: 0   Comments: 9

Question Number 170831    Answers: 2   Comments: 0

Question Number 170958    Answers: 1   Comments: 0

log _5 (x+3)=log _6 (x+14)

$$\:\:\:\:\:\:\mathrm{log}\:_{\mathrm{5}} \left({x}+\mathrm{3}\right)=\mathrm{log}\:_{\mathrm{6}} \left({x}+\mathrm{14}\right) \\ $$

Question Number 170824    Answers: 1   Comments: 0

Question Number 170820    Answers: 0   Comments: 0

Solve y^((4)) +2y^((3)) +y^((2)) =xe^(−x)

$${Solve} \\ $$$${y}^{\left(\mathrm{4}\right)} +\mathrm{2}{y}^{\left(\mathrm{3}\right)} +{y}^{\left(\mathrm{2}\right)} ={xe}^{−{x}} \\ $$

Question Number 170812    Answers: 1   Comments: 0

f(x)=((x+1)/( (√(x^2 +9)))) find the horizontal asymptote

$$\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\frac{\boldsymbol{\mathrm{x}}+\mathrm{1}}{\:\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{9}}} \\ $$$$\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{horizontal}}\:\boldsymbol{\mathrm{asymptote}} \\ $$

Question Number 170811    Answers: 0   Comments: 0

find minimum and maximum z = x^3 −3x^2 −4y^2 +2 ?

$${find}\:{minimum}\:{and}\:{maximum}\:{z}\:=\:{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} +\mathrm{2}\:? \\ $$

Question Number 170810    Answers: 0   Comments: 0

The mean height of a population of girls aged 15 to 19 years in a northern province in Ghana was found to be 165 cm with a standard deviation of 15 cm. Assuming that the heights are normally distributed, find the heights in centimetres that correspond to the following percentiles: a. Between the 20th and 50th percentiles.

$$ \\ $$The mean height of a population of girls aged 15 to 19 years in a northern province in Ghana was found to be 165 cm with a standard deviation of 15 cm. Assuming that the heights are normally distributed, find the heights in centimetres that correspond to the following percentiles: a. Between the 20th and 50th percentiles.

Question Number 170809    Answers: 2   Comments: 0

Question Number 170802    Answers: 2   Comments: 0

Question Number 170800    Answers: 1   Comments: 0

Prove that, for any real number x and odd positive integer n, cos^n x=(1/2^(n+1) )Σ_(k=0) ^((n−1)/2) C_k ^n cos (n−2k)x

$$\mathrm{Prove}\:\mathrm{that},\:\mathrm{for}\:\mathrm{any}\:\mathrm{real}\:\mathrm{number}\:{x}\:\mathrm{and}\:\mathrm{odd}\:\mathrm{positive}\:\mathrm{integer}\:{n}, \\ $$$$\:\:\:\:\:\:\:\mathrm{cos}^{{n}} {x}=\frac{\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} }\underset{{k}=\mathrm{0}} {\overset{\left({n}−\mathrm{1}\right)/\mathrm{2}} {\sum}}{C}_{{k}} ^{{n}} \mathrm{cos}\:\left({n}−\mathrm{2}{k}\right){x} \\ $$

Question Number 170796    Answers: 2   Comments: 0

Question Number 170795    Answers: 2   Comments: 0

solve (x^2 +y^2 +1)dx+x(x−2y)dy=0

$${solve} \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{1}\right){dx}+{x}\left({x}−\mathrm{2}{y}\right){dy}=\mathrm{0} \\ $$

Question Number 170794    Answers: 2   Comments: 0

2n objects of each of three kinds are given to two persons, so that each person gets 3n objects. Prove that this can be done in 3n^2 + 3n + 1 ways.

$$\mathrm{2}{n}\:\mathrm{objects}\:\mathrm{of}\:\mathrm{each}\:\mathrm{of}\:\mathrm{three}\:\mathrm{kinds}\:\mathrm{are} \\ $$$$\mathrm{given}\:\mathrm{to}\:\mathrm{two}\:\mathrm{persons},\:\mathrm{so}\:\mathrm{that}\:\mathrm{each} \\ $$$$\mathrm{person}\:\mathrm{gets}\:\mathrm{3}{n}\:\mathrm{objects}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{this}\:\mathrm{can}\:\mathrm{be}\:\mathrm{done}\:\mathrm{in}\:\mathrm{3}{n}^{\mathrm{2}} \:+\:\mathrm{3}{n}\:+\:\mathrm{1}\:\mathrm{ways}. \\ $$

Question Number 170781    Answers: 2   Comments: 0

(√((2−x))) = (2−x)^2 solve for x

$$\sqrt{\left(\mathrm{2}−\mathrm{x}\right)}\:\:\:=\:\:\left(\mathrm{2}−\mathrm{x}\right)^{\mathrm{2}} \:\:\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\:\:\:\:\:\boldsymbol{\mathrm{x}} \\ $$

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