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1. Core maths help needed

How would you work out part 8 : IV

and for question 9 part iii) why is it MT^2 = radius^2 - 2^2

and not MT^2 = radius^2 + 2^2

because doesn't MT = the hypotenuse so it would be a^2 + b^2 = c^2 ?

Thank you
2. Re: Core maths help needed
(Original post by Ineedhelp998)

How would you work out part 8 : IV

and for question 9 part iii) why is it MT^2 = radius^2 - 2^2

and not MT^2 = radius^2 + 2^2

because doesn't MT = the hypotenuse so it would be a^2 + b^2 = c^2 ?

Thank you
Question 8 (iv),

Differentiate,

It is an increasing function when,
3. Re: Core maths help needed
8iv. If a function is increasing, what does that tell you about it's gradient while it's increasing?

9iii. MT is not the hypotenuse but it's hard for me to show you via text. Try to notice that CM is perpendicular to MT. So CT must be the hypotenuse.
4. Re: Core maths help needed
(Original post by notnek)
8iv. If a function is increasing, what does that tell you about it's gradient while it's increasing?

9iii. MT is not the hypotenuse but it's hard for me to show you via text. Try to notice that CM is perpendicular to MT. So CT must be the hypotenuse.
Doesn't it look like this?
5. Re: Core maths help needed
(Original post by Ineedhelp998)
Doesn't it look like this?
ST is a chord i.e. a straight line connecting S and T. Your ST is not a chord.

And what makes you think angle MCT is a right-angle?
Last edited by notnek; 12-05-2012 at 13:17.
6. Re: Core maths help needed
(Original post by notnek)
ST is a chord i.e. a straight line connecting S and T. Your ST is not a chord.
Yes it is supposed to be a straight line, but hard to draw
7. Re: Core maths help needed
(Original post by notnek)
ST is a chord i.e. a straight line connecting S and T. Your ST is not a chord.

And what makes you think angle MCT is a right-angle?
hmm come to think of it I don't know

actually because mc and tc are perpendicular
Last edited by Ineedhelp998; 12-05-2012 at 13:19.
8. Re: Core maths help needed
(Original post by Ineedhelp998)

How would you work out part 8 : IV

and for question 9 part iii) why is it MT^2 = radius^2 - 2^2

and not MT^2 = radius^2 + 2^2

because doesn't MT = the hypotenuse so it would be a^2 + b^2 = c^2 ?

Thank you
for 8:iv

differentiate y=7+6x-x^2
so dy/dx= 6-2x
increasing function is where dy/dx>0
so 6-2x>0 just treat this like an equals sign and factorise
we have 2 which goes into both 6 and 2x so we put this outside the bracket
2(3-x)>0 so x=0 and x=3
because it is a cubic, you draw the graph of it and it crosses the x-axis at 0 and 3. Then shade in where the line is above 0 and write down this inequality

you should get 2 answers
9. Re: Core maths help needed
(Original post by Ineedhelp998)
hmm come to think of it I don't know

actually because mc and tc are perpendicular
Why do you think that? Even in your drawing, angle MCT doesn't look like a right-angle.

Consider the triangle CST. This triangle is isosceles since it's made up of two radii. And by joining a line from the midpoint of the chord to the centre of the circle, you're splitting the triangle into two.

Can you see why angle CMT is a right-angle now?
10. Re: Core maths help needed
(Original post by notnek)
Why do you think that? Even in your drawing, angle MCT doesn't look like a right-angle.

Consider the triangle CST. This triangle is isosceles since it's made up of two radii. And by joining a line from the midpoint of the chord to the centre of the circle, you're splitting the triangle into two.

Can you see why angle CMT is a right-angle now?
May be my diagram will help you explaining him,

Last edited by raheem94; 12-05-2012 at 13:28.
11. Re: Core maths help needed
(Original post by notnek)
Why do you think that? Even in your drawing, angle MCT doesn't look like a right-angle.

Consider the triangle CST. This triangle is isosceles since it's made up of two radii. And by joining a line from the midpoint of the chord to the centre of the circle, you're splitting the triangle into two.

Can you see why angle CMT is a right-angle now?
thanks now I get it, don't know where I got my diagram from :P
12. Re: Core maths help needed
(Original post by dongonaeatu)
for 8:iv

differentiate y=7+6x-x^2
so dy/dx= 6-2x
increasing function is where dy/dx>0
so 6-2x>0 just treat this like an equals sign and factorise
we have 2 which goes into both 6 and 2x so we put this outside the bracket
2(3-x)>0 so x=0 and x=3
because it is a cubic, you draw the graph of it and it crosses the x-axis at 0 and 3. Then shade in where the line is above 0 and write down this inequality

you should get 2 answers
Firstly, full solutions are not permitted in this forum.

Secondly, the line in bold above is wrong. Why do you have two solutions for x in a linear equation? I think you're getting your inequality confused with

Last edited by notnek; 12-05-2012 at 13:36.
13. Re: Core maths help needed
(Original post by raheem94)
May be my diagram will help you explaining him,

Ohh thats how it looks
14. Re: Core maths help needed
(Original post by notnek)
Firstly, full solutions are not permitted in this forum.

Secondly, the line in bold above is wrong. Why do you have two solutions for x in a linear equation? I think you're getting your inequality confused with

it is not wrong, and i didnt post full solutions. He has to draw the graph to get the answer. And i meant 2 solutions as in x> than a number or x< than a number is where its an increasing function.
15. Re: Core maths help needed
(Original post by notnek)
Firstly, full solutions are not permitted in this forum.

Secondly, the line in bold above is wrong. Why do you have two solutions for x in a linear equation? I think you're getting your inequality confused with

I have said him multiple times to not answer questions on the maths forum, i have still not seen a single correctly answered question from him.

This is the response he gave me a few minutes ago:

(Original post by dongonaeatu)
ok, sorry. I wont answer any questions from now on.
Now why did you answered again?

I think the OP had done this question, he didn't needed any more help. You should leave the questions for the more experienced mathematicians to answer.
16. Re: Core maths help needed
(Original post by dongonaeatu)
it is not wrong, and i didnt post full solutions. He has to draw the graph to get the answer. And i meant 2 solutions as in x> than a number or x< than a number is where its an increasing function.
You aren't clear with this concept yourself mate.
17. Re: Core maths help needed
(Original post by raheem94)
You aren't clear with this concept yourself mate.
What did i do wrong? increasing function is where dy/dx>0
so you get the 2 x values from factorising and draw the graph of it and shade in where the graph is above 0 and this is when its an increasing function
18. Re: Core maths help needed
(Original post by dongonaeatu)
it is not wrong, and i didnt post full solutions. He has to draw the graph to get the answer. And i meant 2 solutions as in x> than a number or x< than a number is where its an increasing function.
I was too quick to say that you'd posted a full solution when you hadn't actually given the solution. Sorry.

But your working is still incorrect. What would your answer be?
Last edited by notnek; 12-05-2012 at 13:45.
19. Re: Core maths help needed
(Original post by dongonaeatu)
What did i do wrong? increasing function is where dy/dx>0
so you get the 2 x values from factorising and draw the graph of it and shade in where the graph is above 0 and this is when its an increasing function

Done

This is what we needed to do.
20. Re: Core maths help needed
(Original post by notnek)
I was too quick to say that you'd posted a full solution when you hadn't actually given the solution. Sorry.

But your working is still incorrect. What would your answer be?
okay, the answer would be
increasing function when x<0 and x>3
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