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#101648 algebraic functions and graphs

Posted by saudiron on 13 December 2008 - 09:42 AM in Mathematics

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.

#101572 simple question

Posted by saudiron on 27 October 2008 - 04:09 AM in Problem Questions

Silver nitrate reacts with HCl to give a precipitate of Silver Chloride.

#101564 Differntiation

Posted by saudiron on 21 October 2008 - 07:35 AM in Problem Questions

QUOTE (Erin Anderson @ Oct 19 2008, 05:35 PM) <{POST_SNAPBACK}>
I have been trying this question for ages.
1. An yacht club is designed its new flag. the flag consists of a red triangle on a yellow rectangular background. In the yellow rectangle ABCD, AB mesures 8 units and AD mesures 6 units. E and F lie on BC and CD, x units from B and C as shown in the diagram.

a) Show that the area, H square units, of the red triangle AEF is
given by H(x) = 24-4x+1/2 x2 (x squared)


b) Hence find the greatest and least possible values of the area of triangle AEF

Its in one of your h/w excersise DIFFERENTIATION HOMEWOEK TM

Please help me unsure.gif

Calculate the area of the rectangle and then subtract the area of the three triangles formed between rectangle and flag ie triangles ABE, FCE and ADF. Area of a triangle is 1/2 base x altitude

Area of rectangle = 8 * 6 units squared = 48 units squared

Area of triangle ABE = 4X units squared (Base AB = 8 units)
Area of triangle FCE = X/2(6-X) units squared = (3X - X^2) units squared
Area of triangle ADF = 3(8-X) units squared = (24 3X) units squared

Total area of three triangles = 24 3X + 3X - X^2 + 4X
= 24 + 4X - X^2
Area of flag = Area of rectangle area of triangles
= 48 (24 + 4X - X^2)
H = 24 4X + X^2 units squared

Differentiating gives dH/dX = -4 + X
therefore have a turning point which is a minimum at X = 4
Substitute X = 4 in equation for H to give minimum area of 16
X can only have values between 0 and 6, and plot of function H will show maximum area of triangle of 24 square units occurs when X = 0.

Hope this helps