1. ## equation

Hi Everyone,

I have an exam coming up the day after tomorrow and I am completely unable to get the correct answer for this question.

First off, I'd like to confirm I'm doing the last equation correctly. (apologies, but I'm not sure on how to use the math notation)

17 - Cr(sq) = 17 - 7Cr - 0.98Cr(sq)
17 = 17 - 7Cr + 0.02Cr(sq)
0 = -7Cr + 0.02Cr(sq)
-0.02Cr(sq) = -7Cr
-0.02Cr = -7
Cr = 350

Is this correct?

Thank you,
Andrew

2. Originally Posted by mcnandy
Hi Everyone,

I have an exam coming up the day after tomorrow and I am completely unable to get the correct answer for this question.

First off, I'd like to confirm I'm doing the last equation correctly. (apologies, but I'm not sure on how to use the math notation)

17 - Cr(sq) = 17 - 7Cr - 0.98Cr(sq)
17 = 17 - 7Cr + 0.02Cr(sq)
0 = -7Cr + 0.02Cr(sq)
-0.02Cr(sq) = -7Cr
-0.02Cr = -7
Cr = 350

Is this correct?

Thank you,
Andrew

$\displaystyle 17-C_{r}^{2}=17-7C_r-0.98C_r^2$
$\displaystyle \implies 0.02C_r^2-7C_r=0$
$\displaystyle \implies C_r(.02C_r-7)=0$
$\displaystyle \implies \color{red}\boxed{C_r=0}$ or $\displaystyle \color{red}\boxed{C_r=350}$

Remember that 0 may be a solution as well!

Hope this helped you out!

3. Thanks, It's confirmed that i'm doing something semi-right at least

Now, there's also the issue of the vector addition, which I must be wrong.

(4i - j) + Cr(5/5.66i + 4/5.66j)
= (4 - 0.7Cr)i + (-1 - 0.7Cr)j

Now, removing the i's and j's and squaring both of the sets of brackets, i.e

(4 - 0.7Cr)(4 - 0.7Cr) = 16 - 5.6Cr - 0.49Cr(sq)
(-1 - 0.7Cr)(-1 - 0.7Cr) = 1 - 1.4Cr - 0.49Cr(sq)

adding the two together and what not:

17 - 7Cr - 0.98Cr(sq) (the right hand side of the above equation)

Am I doing this correctly? If so, I can't see where I'm going wrong as the step before this is a simple subtraction and I hope to god I'm getting that right...

Thank you again,
Andrew