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anonymous
 4 years ago
I AM CERTAIN THAT NO ONE WILL ANSWER THIS QUESTION...
Suppose that l be a line through the points (0,3)and (5,2) and tangent to the graph of the function \[F(x)=\frac{ x^2p }{ 3x+1 }\] where ‘p’ is real number. What is the equation of a line parallel to l and tangent to the graph of F.
A) 4x+4y+7 = 0 B) 9x+9y+37 = 0 C) x+y3 = 0 D) 4x+6y37 = 0
anonymous
 4 years ago
I AM CERTAIN THAT NO ONE WILL ANSWER THIS QUESTION... Suppose that l be a line through the points (0,3)and (5,2) and tangent to the graph of the function \[F(x)=\frac{ x^2p }{ 3x+1 }\] where ‘p’ is real number. What is the equation of a line parallel to l and tangent to the graph of F. A) 4x+4y+7 = 0 B) 9x+9y+37 = 0 C) x+y3 = 0 D) 4x+6y37 = 0

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I will post the answer exactly after one hour.

mathslover
 4 years ago
Best ResponseYou've already chosen the best response.2Hint: Never challenge OpenStudy

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0m= 23/50 = 1 the equation will be y3= 1(x0) x+y3 =0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@hartnn that means???

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.3that means answer can be a,b or c

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yep.... but the answer hasn't yet finished... @hartnn

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I will the slope of the line through the two points

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.3solve y=3x=F(x)=.... u get two roots of x, equate them because the line is tangent and can have only one intersection point. from this i got p=7. now answer can only be either A or B so solving both of them with F(x), gave only B) option with one intersection point, A) doesn't even intersect.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0PLZ DONT POST THE SOLUTION

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am here to post the answer.

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.3whats wrong with my answer? and method ??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0correct but the approach not.

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.3didn't u get p=7, the same way as i got ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you use the options to answer the question, isn't it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Time is up @sauravshakya

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0First of all what you have to do is finding l so as to get p \[l:\frac{ y3 }{ x0 }=\frac{ 23 }{ 50 }\] \[l:y=x+3\]thus \[F(x)=\frac{ x^2p }{ 3x+1} =x+3=y\]simplifying the above expression \[4x^{2}8xp3=0\]Now since l is tangent to F(x), the quadratic equation above has only one solution. That means \[b^{2}4ac=0\] \[(8)^{2}4(4)(p3)=0\]thus \[(8)^{2}4(4)(p3)=0\]\[p=7\]Now we get F(x) to be \[F(x)=\frac{ x^2+7 }{ 3x+1 }\]Therefore lets solve F’(x)=1 so as to get the coordinate of the unknown line. \[F'(x)=\frac{ 2x(3x+1)3(x^{2}+7) }{ (3x+1)^2 }=1\] \[3x^{2}+2x21=9x^{2}6x1\] \[12x^{2}+8x20=0\]thus \[x=1 or x=\frac{ 5 }{ 3 }\]and \[F(1)=2 and F(\frac{ 5 }{ 3 })=\frac{ 22 }{ 9 }\]Thus we got this two points (1,2) and (5/3,22/9). The two possible lines therefore are \[\frac{ y2 }{ x1 }=1{\rightarrow}y=x+3\]w/c is the given line l. And \[\frac{ y(\frac{ 22 }{9 }) }{ x(\frac{ 5 }{ 3 })}=1{\rightarrow}9y+9x+37=0\]w/c is the answer.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0b/c the slope at that point is 1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Because the line l and the required line are parallel

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So, both of their slope is 1

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.3it wasn't that difficult.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes but need smart approach

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0YEP....... Actually I did a algebra mistake, as usual

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.3lol! checking options(when given) is the smartest approach!! :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that saves time in exam

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.3atleast i got the answer...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok thanks for your help....... closing
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