Recent content by Casio1

Thanks, I was getting there but was struggling to understand initially how to get rid of the denominator 9, which I multiplied through like you and got the results, but I must admit at the point of 34x^2 + 34x - 68 = 0, I had difficulties because I was getting solutions saying I had no roots and...

Just a query that I don't understand how you got from; $\displaystyle (x + 3)^2 + ( - 5x + 14 - 4)^2 = 17$ To $\displaystyle (x + 3)^2 + 25( x - 2)^2 = 17$ Thanks Mark you asked; Why are you changing -5x + 14 - 4 to -5x + 2/3? Because the circle intersects the line 3y = - 5x + 14...

Continued from; Originally Posted by Jameson http://www.mathhelpboards.com/f2/understanding-how-deal-fractions-using-brackets-2596/#post11674 What is the full problem you are trying to solve? I can't make sense of your post until I know that. I have a circle problem and am trying to find...

OK from this point onwards some people could now say that this is going to be in the wrong thread if I post as this is a circle problem, so I guess I should ask if I am permitted to continue or start another thread in geometry?

I can't see that being the end of the problem? X^2 + 6x + 9 + (1\9)(25X^2 - 20x + 4) Something is not quite finished as I require to find the coordinates of any points of my circle to finish, therefore the fraction (1\9) must go so that I can add and subtract terms to end up with an expression...

I have followed through all the above and can see what you have done with the denominator 3, where you have ended up with my original solution, but I didn't know what to do with the denominators. If then \[\frac{1}{9}\left(25x^2 - 20x + 4\right)\] is equal to what I had originally, from this...

(- 5x/3 + 2/3) (- 5x/3 + 2/3) If the above example was; (-5x + 2) (-5x + 2) = 25x^2 - 10x - 10x + 4 = 25x^2 - 20x + 4 The problem is I don't know how to deal with the denominators in this form? Anyone help

$\begin{align}r^2=(x + 5)^2 + (y - 3)^2\end{align}$ $\begin{align}r^2=(x - 1)^2 + (y + 1)^2\end{align}$ $\begin{align}r^2=(x - 0)^2 + (y + 2)^2\end{align}$ ${2}/{3} + {4}/{6}=$ How is this? Better

$y=2x+1$ $y=2x^2+3x+5$ $ a+3b=1$ $2a - b=1$ Multiply through by 3 $6a - 3b = 3$ Subtract equation (3) from (1) $6a - 3b = 3$ $ a + 3b = 1$ $7a - 0b = 4$ a =$\frac{4}{7}$ Just learning latex on here and trying to understand how I can keep everything in line, which seems difficult with...

Sorry that is a typo error on my behalf now I have looked back.:o The midpoints are; [4/2, - 2/2 ], [-1/2, 3/2 ] With reference to the parametric equation I give the solution and not the original parametric equation. So y = (3x - 8) / (2) = (3(4/2) - 8)) / (2) = -2/2 Kind regards...

OK, its going to be a long day. I have sone points which represent a circle. A(5, -3), B(-1,1) and C(0,2) I am asked to find the slope AB, this is - 2/3 I am asked to find the coordinates of the midpoint of the line segment AB, these are [4/2, - 2/2] using my answers above I am asked to...

OK I feel that I am stumped with this now. I have a circle problem. Given some coordinates I have calculated the gradient at -2/3. Then I am looking to find the midpoints, of the line, which are [4/2, -2/2]. I have then calculated the perpendicular of the line AB, which had a gradient - 2/3. The...

I have not experienced this one before and would appreciate a little help. 2/3x - 3 = - x + 2 2/3x = - x + 5 Now how do I add x to the LHS, is it; 2/4x or 3/3x or am I completely off the mark? Kind regards Casio

Thanks for setting me on the right line of thought there. OK let me take this one step at a time so I get the proper understanding of what is actually going on with these sequences. First, please explain what this part refers to;\text{Given:}\:U_{n+1} Casio

U1 = 2, Un+1 =-0.3Un + 3 (n = 1,2,3...) U2= U2.4+1 = -0.3(3.4) + 3 = 1.98 U3 = U1.98+1 = -0.3(2.98)+3 = 2.11 U4 = U2.11+1 = -0.3(3.11)+3 = 2.07 Four terms are;2, 1.98, 2.11, 2.07,... Not sure whether the first method I used to work out the terms was correct, or whether the method I used...