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hugomimihu is right, we can't answer most of this question for certain. In order to solve it, we would have to assume many things from the diagram due to lack of information.

It said it was wrong...

The points on the graph tend to have a downward motion. This downward motion means that there is a negative correlation. Te age of the collector classic doesn't affect the cost. This means that the car can still be old and cost a lot. There is only one outlier on the graph like that with coordinates of (8 yrs, 8200 pounds).

By the quadratic formula, the roots are (-2 - \pm 2i sqrt(7))/3, and so r^2 + s^2 = 62/9.

The probability is 11/81.

Great! Thanks!

Almost right guest, but a+ b = -5 not +5

Using that your final formula becomes  -11 - 3*5 + 9 = -17

Yeah, it really helps!

Can you please help me find the value of the repeated root? Thanks!

I interpret this as if you want faster answers, share your question link on social media which will verify your question and also be in more places (i.e.) social media and this forum.

It said it wasn't correct. Thank you for trying though!

While zero divided by anything is zero, there is a slight error in this problem. You CANNOT divide by zero. So the answer is undefined.

My reason for Lucky to not like his own posts is because if there is a wrong answer but it is marked with a +1, people can take away the wrong information from a question / post.

*Note: ​I'm not saying that Lucky has a lot wrong answers or anything...

Compute the sum:
\(101^2-97^2+93^2-89^2+\ldots+5^2-1^2\)

\(\begin{array}{|rcll|} \hline && \mathbf{ 101^2-97^2+93^2-89^2+\ldots+5^2-1^2} \\\\ &=& 101^2 + 93^2 + 85^2 +\ldots+ 13^2+5^2 \\ && -97^2 -89^2 -81^2 -\ldots- 9^2-1^2 \\\\ &=& \mathbf{\sum \limits_{k=1}^{13}\Big(~ 8k-3~\Big)^2 -\sum \limits_{k=1}^{13}\Big(~ 8k-7~\Big)^2} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\sum \limits_{k=1}^{13}\Big(~ 8k-3~\Big)^2 -\sum \limits_{k=1}^{13}\Big(~ 8k-7~\Big)^2} \\\\ &=& \sum \limits_{k=1}^{13}\Big(~ 8k-3~\Big)^2 - \Big(~ 8k-7~\Big)^2 \\\\ &=& \sum \limits_{k=1}^{13}64k^2-48k+9-(64k^2-112k+49) \\\\ &=& \sum \limits_{k=1}^{13} {\color{red}64k^2}-48k+9{\color{red}-64k^2}+112k-49 \\\\ &=& \sum \limits_{k=1}^{13} -48k+9+112k-49 \\\\ &=& \sum \limits_{k=1}^{13} 64k-40 \\\\ &=& 64\sum \limits_{k=1}^{13}k -\sum \limits_{k=1}^{13}40 \quad | \quad \sum \limits_{k=1}^{13}40 = 40*13 \\\\ &=& 64\sum \limits_{k=1}^{13}k - 40*13 \quad | \quad \sum \limits_{k=1}^{13}k = \dfrac{(1+13)}{2}*13 \\\\ &=& 64\dfrac{(1+13)}{2}*13 - 40*13 \\\\ &=& 32*14*13 - 40*13 \\\\ &=& 5824 - 520 \\\\ &=& \mathbf{5304} \\ \hline \end{array}\)

\(\mathbf{ 101^2-97^2+93^2-89^2+\ldots+5^2-1^2} = \mathbf{5304}\)

4x + 2 = 2x + 10   | -2x

2x + 2 = 10          | -2

      2x = 8            | :2

        x = 4

-(x⁹) + x⁹ = 0

and  

(-x)⁹ + x⁹ = 0

Points \(A\), \(B\), and \(C\) are on sides \(\overline{WX}\), \(\overline{YZ}\), and \(\overline{XY}\) of rectangle \(WXYZ\) as shown below such that
\(XA = 4\), \(YB = 18\),  \(\angle ACB= 90^\circ\), and \(YC = 2XC\).
Find \(AB\).

\(\begin{array}{|rcll|} \hline \tan(A) = \dfrac{4}{XC} &=& \dfrac{2XC}{18} \\\\ \dfrac{4}{XC} &=& \dfrac{2XC}{18} \\\\ \dfrac{4}{XC} &=& \dfrac{XC}{9} \\\\ 36 &=& XC^2 \\\\ 6 &=& XC \\ \mathbf{XC} &=& \mathbf{6} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline x^2 &=& (3XC)^2 + BD^2 \quad | \quad BD =YB-XA \\ x^2 &=& (3XC)^2 + (YB-XA)^2 \\ x^2 &=& (3*6)^2 + (18-4)^2 \\ x^2 &=& 18^2 + 14^2 \\ x^2 &=& 324 + 196 \\ x^2 &=& 520 \\ x^2 &=& 4*130 \\ x &=& 2\sqrt{130} \\ \mathbf{x} &=& \mathbf{22.8035085020} \\ \hline \end{array} \)

\(\mathbf{AB \approx 22.8}\)

df/dx ≈ (f(9.01) - f(9))/(9.01 - 9)  = (3 - 3)/0.01  = 0