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algebraic functions and graphs


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#1 sammy w

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Posted 12 December 2008 - 11:31 PM

Pg 97, Q12, maths in action higher textbook

hey, i was wondering if anyone could help me out with this question as no one can seem to help me get the correct answer for part c! the diagram should be attached!

12)the gradient of the tightrope must not be more than 1/10 or less than 0.

a) find a formula for L(x), the length of the rope> answer:L= square root of 2500 +(x-2) squared

b) find an expression for the gradient of the rope and use it to dertermine the domain of L(x)>answer:2<=x<=7, for the gradient to be more than 1/10

c) calcualte the range of L(x) correct to 2 decimal place> answer ????????? the answer is 50<=L(x)<= 50.25


All i can guess is that it is if they have rotated the diagram 90 degrees to the right for the range of L(x) and that the 2 rectangles at either end of the diagram have something to do with the extra 0.25, but there are no other no.'s to give this indication?????

if anyone can help that would be great!!!!

many thanks

ps..let me no if the diagram does not upload!

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#2 saudiron

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Posted 13 December 2008 - 09:42 AM

QUOTE (sammy w @ Dec 13 2008, 02:31 AM) <{POST_SNAPBACK}>
Pg 97, Q12, maths in action higher textbook

hey, i was wondering if anyone could help me out with this question as no one can seem to help me get the correct answer for part c! the diagram should be attached!

12)the gradient of the tightrope must not be more than 1/10 or less than 0.

a) find a formula for L(x), the length of the rope> answer:L= square root of 2500 +(x-2) squared

b) find an expression for the gradient of the rope and use it to dertermine the domain of L(x)>answer:2<=x<=7, for the gradient to be more than 1/10

c) calcualte the range of L(x) correct to 2 decimal place> answer ????????? the answer is 50<=L(x)<= 50.25


All i can guess is that it is if they have rotated the diagram 90 degrees to the right for the range of L(x) and that the 2 rectangles at either end of the diagram have something to do with the extra 0.25, but there are no other no.'s to give this indication?????

if anyone can help that would be great!!!!

many thanks

ps..let me no if the diagram does not upload!


From the answer in (b) the domain of X is 2<=X<=7
The formula for the length of L(X) is square root of 2500 + (X-2)squared
Substituting the values for the domain of X into this formula we have:-
When X=2, L(X) = sqare root of 2500 + (2-2) squared
= square root of 2500 + 0
= 50

When X=7, L(X) = square root of 2500 + (7-2)squared
= square root of 2500 + 5squared
= square root of 2500 +25
= square root of 2525
= 50.25 (to 2 decimal places)
Hence the range of L(X) is 50<=L(X)<=50.25

Hope this is clear.



#3 sammy w

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Posted 13 December 2008 - 12:20 PM

QUOTE (saudiron @ Dec 13 2008, 09:42 AM) <{POST_SNAPBACK}>
QUOTE (sammy w @ Dec 13 2008, 02:31 AM) <{POST_SNAPBACK}>
Pg 97, Q12, maths in action higher textbook

hey, i was wondering if anyone could help me out with this question as no one can seem to help me get the correct answer for part c! the diagram should be attached!

12)the gradient of the tightrope must not be more than 1/10 or less than 0.

a) find a formula for L(x), the length of the rope> answer:L= square root of 2500 +(x-2) squared

b) find an expression for the gradient of the rope and use it to dertermine the domain of L(x)>answer:2<=x<=7, for the gradient to be more than 1/10

c) calcualte the range of L(x) correct to 2 decimal place> answer ????????? the answer is 50<=L(x)<= 50.25


All i can guess is that it is if they have rotated the diagram 90 degrees to the right for the range of L(x) and that the 2 rectangles at either end of the diagram have something to do with the extra 0.25, but there are no other no.'s to give this indication?????

if anyone can help that would be great!!!!

many thanks

ps..let me no if the diagram does not upload!


From the answer in (b) the domain of X is 2<=X<=7
The formula for the length of L(X) is square root of 2500 + (X-2)squared
Substituting the values for the domain of X into this formula we have:-
When X=2, L(X) = sqare root of 2500 + (2-2) squared
= square root of 2500 + 0
= 50

When X=7, L(X) = square root of 2500 + (7-2)squared
= square root of 2500 + 5squared
= square root of 2500 +25
= square root of 2525
= 50.25 (to 2 decimal places)
Hence the range of L(X) is 50<=L(X)<=50.25

Hope this is clear.



yeh helps alot! my teacher could not even get it, so thanks!!!!!





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