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tresset_1 [31]
2 years ago
5

Grogg typed the following $1000$ expressions into his calculator, one by one: \[\sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{4}, \dots, \

sqrt{999}, \sqrt{1000}. \]how many times will grogg's result be an integer?
Mathematics
1 answer:
Vanyuwa [196]2 years ago
8 0

Since 30² = 900 and 40² = 1600, between 30 and 40 of the numbers Grogg enters will reduce to integers. Now just find which count is correct. We don't have to look far:

31² = 961

32² = 1024

so there 31 of the outputs of √1, √2, √3, ..., √1000 are integers.

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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}
3 0
3 years ago
What are the numbers that are divisible by three
faust18 [17]
The numbers divisible by 3 are multiples of 3.
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