Jane drives at a constant rate of 45 miles per hour for 4 hours. How far will she have traveled in that time?

Mathematics

1 answer:

bearhunter [10]3 years ago

6 0

180 miles per hour. 45*4 to get the answer.<span />

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One more time!

CaHeK987 [17]

Since q(x) is inside p(x), find the x-value that results in q(x) = 1/4

$\frac{1}{4} = 5 - x^2\ \Rightarrow\ x^2 = 5 - \frac{1}{4}\ \Rightarrow\ x^2 = \frac{19}{4}\ \Rightarrow \\
x = \frac{\sqrt{19} }{2}$

so we conclude that
$q(\frac{\sqrt{19} }{2} ) = 1/4$

therefore

$p(1/4) = p\left( q\left(\frac{ \sqrt{19} }{2} \right) \right)$

plug $x=\sqrt{19}/2$ into p( q(x) ) to get answer

$p(1/4) = p\left( q\left( \frac{ \sqrt{19} }{2} \right) \right)\ \Rightarrow\ \dfrac{4 - \left( \frac{\sqrt{19} }{2}\right)^2 }{ \left( \frac{\sqrt{19} }{2}\right)^3 } \Rightarrow \\ \\ \dfrac{4 - \frac{19}{4} }{ \frac{19\sqrt{19} }{8}} \Rightarrow \dfrac{8\left(4 - \frac{19}{4}\right) }{ 8 \cdot \frac{19\sqrt{19} }{8}} \Rightarrow \dfrac{32 - 38}{19\sqrt{19}} \Rightarrow \dfrac{-6}{19\sqrt{19}} \cdot \frac{\sqrt{19}}{\sqrt{19}}\Rightarrow$

$\dfrac{-6\sqrt{19} }{19 \cdot 19} \\ \\ \Rightarrow -\dfrac{6\sqrt{19} }{361}$

$p(1/4) = -\dfrac{6\sqrt{19} }{361}$

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Step-by-step explanation:

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_____

If you keep the units with the numbers as I have done here, you find that the product of (10 ft) and (70 yd) does not have the units of ft². Rather, the units are ft·yd, which is <em>not what the problem is asking for</em>.

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