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Question Number 174807    Answers: 0   Comments: 0

We know that vertex form of parabola is given as y = a(x−h)^2 +k From the given diagram of bridge that resembles a parabola, we have a vertex points of (0, 30) and other points due to towers that supports the parabolic−shape cable, which is (200, 150). ∴ y = a(x−0)^2 +30 = ax^2 +30 To find the value of ′a′, let′s use the given points other than vertex 150 = a(200)^2 + 30 150−30 = 40000a a = ((120)/(40,000)) = (3/(1000)) ∴ y = ((3x^2 )/(1000)) +30 Also, since we′re asked to find a function that gives a length of metal rod needed relative to its distance from the midpoint of the bridge, with each rods have an equal distance to each other, then we must consider another variable ′d′ that represents the equal distance of metal rods relative to its decided quantity and variable ′n′ given as positive integer that divides the distance of midpoint to tower. d = ((200)/n) ⇒ nd = 200 Example: Engineers decided to use 8 metal rodus, then we have d = ((200)/8) = 25 To calculate the length of each rods, let′s use the formula above First rod: y = ((3(0∙25)^2 )/(1000)) +30 = 30 ft. Second rod: y = ((3(1∙25)^2 )/(1000)) +30 = 31.875 ft. Third rod: y = ((3(2∙25)^2 )/(1000)) +30 = 37.5 ft. Fourth rod: y = ((3(3∙25)^2 )/(1000)) +30 = 46.875 ft.

$$\:\mathrm{We}\:\mathrm{know}\:\mathrm{that}\:\mathrm{vertex}\:\mathrm{form}\:\mathrm{of}\:\mathrm{parabola}\:\mathrm{is}\:\mathrm{given}\:\mathrm{as} \\ $$$$\:\mathrm{y}\:=\:\mathrm{a}\left(\mathrm{x}−\mathrm{h}\right)^{\mathrm{2}} +\mathrm{k} \\ $$$$\: \\ $$$$\:\mathrm{From}\:\mathrm{the}\:\mathrm{given}\:\mathrm{diagram}\:\mathrm{of}\:\mathrm{bridge}\:\mathrm{that}\:\mathrm{resembles}\:\mathrm{a}\:\mathrm{parabola}, \\ $$$$\:\mathrm{we}\:\mathrm{have}\:\mathrm{a}\:\mathrm{vertex}\:\mathrm{points}\:\mathrm{of}\:\left(\mathrm{0},\:\mathrm{30}\right)\:\mathrm{and}\:\mathrm{other}\:\mathrm{points}\:\mathrm{due}\:\mathrm{to}\:\mathrm{towers} \\ $$$$\:\mathrm{that}\:\mathrm{supports}\:\mathrm{the}\:\mathrm{parabolic}−\mathrm{shape}\:\mathrm{cable},\:\mathrm{which}\:\mathrm{is}\:\left(\mathrm{200},\:\mathrm{150}\right). \\ $$$$\: \\ $$$$\:\therefore\:\:\mathrm{y}\:=\:\mathrm{a}\left(\mathrm{x}−\mathrm{0}\right)^{\mathrm{2}} +\mathrm{30}\:=\:\mathrm{ax}^{\mathrm{2}} +\mathrm{30} \\ $$$$\: \\ $$$$\:\mathrm{To}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:'\mathrm{a}',\:\mathrm{let}'\mathrm{s}\:\mathrm{use}\:\mathrm{the}\:\mathrm{given}\:\mathrm{points}\:\mathrm{other}\:\mathrm{than}\:\mathrm{vertex} \\ $$$$\:\mathrm{150}\:=\:\mathrm{a}\left(\mathrm{200}\right)^{\mathrm{2}} \:+\:\mathrm{30} \\ $$$$\:\mathrm{150}−\mathrm{30}\:=\:\mathrm{40000a} \\ $$$$\:\mathrm{a}\:=\:\frac{\mathrm{120}}{\mathrm{40},\mathrm{000}}\:=\:\frac{\mathrm{3}}{\mathrm{1000}} \\ $$$$\: \\ $$$$\:\therefore\:\:\mathrm{y}\:=\:\frac{\mathrm{3x}^{\mathrm{2}} }{\mathrm{1000}}\:+\mathrm{30} \\ $$$$\: \\ $$$$\:\mathrm{Also},\:\mathrm{since}\:\mathrm{we}'\mathrm{re}\:\mathrm{asked}\:\mathrm{to}\:\mathrm{find}\:\mathrm{a}\:\mathrm{function}\:\mathrm{that}\:\mathrm{gives}\:\mathrm{a}\:\mathrm{length}\:\mathrm{of}\:\mathrm{metal}\:\mathrm{rod} \\ $$$$\:\mathrm{needed}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{its}\:\mathrm{distance}\:\mathrm{from}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bridge},\:\mathrm{with}\:\mathrm{each}\:\mathrm{rods} \\ $$$$\:\mathrm{have}\:\mathrm{an}\:\mathrm{equal}\:\mathrm{distance}\:\mathrm{to}\:\mathrm{each}\:\mathrm{other},\:\mathrm{then}\:\mathrm{we}\:\mathrm{must}\:\mathrm{consider}\:\mathrm{another}\:\mathrm{variable}\:'\mathrm{d}' \\ $$$$\:\mathrm{that}\:\mathrm{represents}\:\mathrm{the}\:\mathrm{equal}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{metal}\:\mathrm{rods}\:\mathrm{relative}\:\mathrm{to}\:\mathrm{its}\:\mathrm{decided}\:\mathrm{quantity}\:\mathrm{and} \\ $$$$\:\mathrm{variable}\:'\mathrm{n}'\:\mathrm{given}\:\mathrm{as}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{that}\:\mathrm{divides}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{midpoint}\:\mathrm{to}\:\mathrm{tower}. \\ $$$$\: \\ $$$$\:\mathrm{d}\:=\:\frac{\mathrm{200}}{\mathrm{n}}\:\:\Rightarrow\:\:\mathrm{nd}\:=\:\mathrm{200} \\ $$$$\: \\ $$$$\:\mathrm{Example}: \\ $$$$\:\mathrm{Engineers}\:\mathrm{decided}\:\mathrm{to}\:\mathrm{use}\:\mathrm{8}\:\mathrm{metal}\:\mathrm{rodus},\:\mathrm{then}\:\mathrm{we}\:\mathrm{have}\:\mathrm{d}\:=\:\frac{\mathrm{200}}{\mathrm{8}}\:=\:\mathrm{25} \\ $$$$\:\mathrm{To}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{each}\:\mathrm{rods},\:\mathrm{let}'\mathrm{s}\:\mathrm{use}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{above} \\ $$$$\:\mathrm{First}\:\mathrm{rod}:\:\:\mathrm{y}\:=\:\frac{\mathrm{3}\left(\mathrm{0}\centerdot\mathrm{25}\right)^{\mathrm{2}} }{\mathrm{1000}}\:+\mathrm{30}\:=\:\mathrm{30}\:\mathrm{ft}. \\ $$$$\:\mathrm{Second}\:\mathrm{rod}:\:\:\mathrm{y}\:=\:\frac{\mathrm{3}\left(\mathrm{1}\centerdot\mathrm{25}\right)^{\mathrm{2}} }{\mathrm{1000}}\:+\mathrm{30}\:=\:\mathrm{31}.\mathrm{875}\:\mathrm{ft}. \\ $$$$\:\mathrm{Third}\:\mathrm{rod}:\:\:\mathrm{y}\:=\:\frac{\mathrm{3}\left(\mathrm{2}\centerdot\mathrm{25}\right)^{\mathrm{2}} }{\mathrm{1000}}\:+\mathrm{30}\:=\:\mathrm{37}.\mathrm{5}\:\mathrm{ft}. \\ $$$$\:\mathrm{Fourth}\:\mathrm{rod}:\:\:\mathrm{y}\:=\:\frac{\mathrm{3}\left(\mathrm{3}\centerdot\mathrm{25}\right)^{\mathrm{2}} }{\mathrm{1000}}\:+\mathrm{30}\:=\:\mathrm{46}.\mathrm{875}\:\mathrm{ft}. \\ $$$$\: \\ $$

Question Number 174806    Answers: 1   Comments: 1

Given f(x)=x^4 +ax^3 +bx^2 +cx+d where a,b,c and d are real number suppose the graph f(x) intersects the graph of y=2x−1 at x=1,2,3. Find the value of f(0)+f(4).

$${Given}\:{f}\left({x}\right)={x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d} \\ $$$${where}\:{a},{b},{c}\:{and}\:{d}\:{are}\:{real}\:{number} \\ $$$${suppose}\:{the}\:{graph}\:{f}\left({x}\right)\:{intersects} \\ $$$${the}\:{graph}\:{of}\:{y}=\mathrm{2}{x}−\mathrm{1}\:{at}\:{x}=\mathrm{1},\mathrm{2},\mathrm{3}. \\ $$$$\:{Find}\:{the}\:{value}\:{of}\:\:{f}\left(\mathrm{0}\right)+{f}\left(\mathrm{4}\right). \\ $$

Question Number 174804    Answers: 0   Comments: 4

Please solve .

$${Please}\:{solve}\:. \\ $$

Question Number 174800    Answers: 1   Comments: 0

For which value of x this cubic equation will be 0 ? a^3 − 16a − 3

$$\mathrm{For}\:\mathrm{which}\:\mathrm{value}\:\mathrm{of}\:{x}\:\mathrm{this}\:\mathrm{cubic}\:\mathrm{equation}\:\mathrm{will} \\ $$$$\mathrm{be}\:\mathrm{0}\:?\:{a}^{\mathrm{3}} \:−\:\mathrm{16}{a}\:−\:\mathrm{3} \\ $$

Question Number 174799    Answers: 1   Comments: 0

Question Number 174790    Answers: 3   Comments: 1

Question Number 174785    Answers: 0   Comments: 0

Question Number 174779    Answers: 1   Comments: 2

Question Number 174776    Answers: 1   Comments: 0

Question Number 174775    Answers: 1   Comments: 0

Question Number 174770    Answers: 1   Comments: 0

∫(((sinx)/( (√x))))dx Mastermind

$$\int\left(\frac{\mathrm{sinx}}{\:\sqrt{\mathrm{x}}}\right)\mathrm{dx} \\ $$$$ \\ $$$$\mathrm{Mastermind} \\ $$

Question Number 174769    Answers: 2   Comments: 0

Question Number 174767    Answers: 1   Comments: 0

Question Number 174766    Answers: 0   Comments: 0

Q: by using fourier series, prove that : 𝛀=Σ_(n=1) ^∞ (((−1)^( n−1) )/((2n−1 )^( 3) )) = (𝛑^( 3) /(32)) −−−−−−

$$ \\ $$$$\:\:\:\:{Q}:\:\:{by}\:{using}\:{fourier}\:{series}, \\ $$$$\:\:\:\:\:\:{prove}\:\:{that}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\Omega}=\underset{\boldsymbol{{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\:\boldsymbol{{n}}−\mathrm{1}} }{\left(\mathrm{2}\boldsymbol{{n}}−\mathrm{1}\:\right)^{\:\mathrm{3}} }\:=\:\frac{\boldsymbol{\pi}^{\:\mathrm{3}} }{\mathrm{32}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−−−−−− \\ $$

Question Number 174765    Answers: 0   Comments: 0

Question Number 174763    Answers: 2   Comments: 0

Question Number 174758    Answers: 1   Comments: 3

where am i wrong ?

$${where}\:{am}\:{i}\:{wrong}\:?\: \\ $$$$ \\ $$

Question Number 174755    Answers: 0   Comments: 1

Question Number 174754    Answers: 0   Comments: 0

Question Number 174751    Answers: 0   Comments: 0

Question Number 174742    Answers: 0   Comments: 0

≪_• ^• I THINK...^ ^(−) _•^• _(−) ≫ One question per post, IDEAL 👍👍👍 Two questions per post,OK (BEARABLE) (👎+👍)/2 Three or more questions:NO, NO, NO! 👎👎👎

$$\:\:\:\:\:\ll_{\bullet} ^{\bullet} \underset{−} {\overline {\boldsymbol{\mathrm{I}}\:\boldsymbol{\mathrm{THINK}}...^{} }}\:_{\bullet} ^{\bullet} \gg \\ $$$$\boldsymbol{\mathrm{One}}\:\boldsymbol{\mathrm{question}}\:\boldsymbol{\mathrm{per}}\:\boldsymbol{\mathrm{post}},\:\boldsymbol{\mathrm{IDEAL}} \\ $$👍👍👍 $$\boldsymbol{\mathrm{Two}}\:\boldsymbol{\mathrm{questions}}\:\boldsymbol{\mathrm{per}}\:\boldsymbol{\mathrm{post}},\boldsymbol{\mathrm{OK}}\:\left(\boldsymbol{\mathrm{BEARABLE}}\right) \\ $$(👎+👍)/2 $$\boldsymbol{\mathrm{Three}}\:\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{more}}\:\boldsymbol{\mathrm{questions}}:\boldsymbol{\mathrm{NO}},\:\boldsymbol{\mathrm{NO}},\:\boldsymbol{\mathrm{NO}}! \\ $$👎👎👎

Question Number 174737    Answers: 1   Comments: 0

A graduating student keeps applying for a job until she gets an offer. The probability of getting an offer at any trial is 0.35. What is the expected number of applications?

$$\mathrm{A}\:\mathrm{graduating}\:\mathrm{student}\:\mathrm{keeps}\:\mathrm{applying} \\ $$$$\mathrm{for}\:\mathrm{a}\:\mathrm{job}\:\mathrm{until}\:\mathrm{she}\:\mathrm{gets}\:\mathrm{an}\:\mathrm{offer}.\:\mathrm{The} \\ $$$$\mathrm{probability}\:\mathrm{of}\:\mathrm{getting}\:\mathrm{an}\:\mathrm{offer}\:\mathrm{at}\:\mathrm{any} \\ $$$$\mathrm{trial}\:\mathrm{is}\:\mathrm{0}.\mathrm{35}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{expected}\: \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{applications}? \\ $$

Question Number 174735    Answers: 0   Comments: 0

Question Number 174732    Answers: 1   Comments: 0

Question Number 174728    Answers: 1   Comments: 3

Question Number 174727    Answers: 1   Comments: 0

∫_0 ^∞ (e^(−x^2 ) /((x^2 +a^2 )^2 ))dx

$$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}^{\mathrm{2}} } }{\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$

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