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borishaifa [10]
4 years ago
7

Rosie is comparing √7 and 3.44444...she say that √7 >3.44444...because √7=3.5

Mathematics
2 answers:
lutik1710 [3]4 years ago
8 0
From what my calculator said, Radical 7 is 2.6457511311064591 (why not m8) and obviously 2.65 is not bigger than 3.5 or 3.444
ahrayia [7]4 years ago
4 0
Well

to compare we can squaer both

7 and11.86
11.86>7

therefor
3.444444>√7
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\displaystyle \frac{17}{2}.

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Let the x-coordinate of P be t. For P\! to be on the graph of the function y = \sqrt{x}, the y-coordinate of \! P would need to be \sqrt{t}. Therefore, the coordinate of P \! would be \left(t,\, \sqrt{t}\right).

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\begin{aligned} & d\left(\left(t,\, \sqrt{t}\right),\, (9,\, 0)\right) \\ &= \sqrt{(t - 9)^2 +\left(\sqrt{t}\right)^{2}} \\ &= \sqrt{t^2 - 18\, t + 81 + t} \\ &= \sqrt{t^2 - 17 \, t + 81}\end{aligned}.

The goal is to find the a t that minimizes this distance. However, \sqrt{t^2 - 17 \, t + 81} is non-negative for all real t\!. Hence, the \! t that minimizes the square of this expression, \left(t^2 - 17 \, t + 81\right), would also minimize \sqrt{t^2 - 17 \, t + 81}\!.

Differentiate \left(t^2 - 17 \, t + 81\right) with respect to t:

\displaystyle \frac{d}{dt}\left[t^2 - 17 \, t + 81\right] = 2\, t - 17.

\displaystyle \frac{d^{2}}{dt^{2}}\left[t^2 - 17 \, t + 81\right] = 2.

Set the first derivative, (2\, t - 17), to 0 and solve for t:

2\, t - 17 = 0.

\displaystyle t = \frac{17}{2}.

Notice that the second derivative is greater than 0 for this t. Hence, \displaystyle t = \frac{17}{2} would indeed minimize \left(t^2 - 17 \, t + 81\right). This t\! value would also minimize \sqrt{t^2 - 17 \, t + 81}\!, the distance between P \left(t,\, \sqrt{t}\right) and (9,\, 0).

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