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Question Number 174811 Answers: 0 Comments: 2

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.

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

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).

$G i v e n f (x) = x^{4} + {a x}^{3} + {b x}^{2} + c x + d$ $w h e r e a, b, c a n d d a r e r e a l n u m b e r$ $s u p p o s e t h e g r a p h f (x) i n t e r s e c t s$ $t h e g r a p h o f y = 2 x - 1 a t x = 1, 2, 3 .$ $F i n d t h e v a l u e o f f (0) + f (4) .$

Question Number 174804 Answers: 0 Comments: 4

Please solve .

$P l e a s e s o l v e .$

Question Number 174800 Answers: 1 Comments: 0

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

$For which value of x this cubic equation will$ $be 0 ? a^{3} - 16 a - 3$

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

$\int (\frac{sinx}{\sqrt{x}}) dx$ $Mastermind$

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)) −−−−−−

$Q : b y u s i n g f o u r i e r s e r i e s,$ $p r o v e t h a t :$ $Ω = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1}}{{(2 n - 1)}^{3}} = \frac{π^{3}}{32}$ $- - - - - -$

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 ?

$w h e r e a m i w r o n g ?$

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! 👎👎👎

$≪_{∙}^{∙} \underline{\overset{―}{I THINK . . .^{}}}_{∙}^{∙} ≫$ $One question per post, IDEAL$ 👍👍👍 $Two questions per post, OK (BEARABLE)$ (👎+👍)/2 $Three or more questions : NO, NO, NO!$ 👎👎👎

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?

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

Question Number 174735 Answers: 0 Comments: 0

Question Number 174732 Answers: 1 Comments: 0

Question Number 174728 Answers: 1 Comments: 3

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