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Flura [38]
3 years ago
10

The French mathematician Pierre de Fermat (1601-1665) considered the sequence 5, 17, 257, 65537,

Mathematics
1 answer:
Semenov [28]3 years ago
3 0
These are prime numbers.

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10. Complete the table below based on the following function:f(x)=-6x-1
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<span>f(x)=-6x-1

</span><span> X    Y
 0    -1
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 6    -37
 5     -31</span>
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The point (17,−9) is translated to the point (6,2). Select the description of the translation.
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3 years ago
Simplify the number into simplest radical form. Use the factor tree to help determine the factors. StartRoot 96 Endroot StartRoo
NikAS [45]

Answer:

[tex]4608\sqrt{3}[/tex]

Step-by-step explanation:

1. \sqrt{96} *\sqrt{6}* 2*\sqrt{6}*4 *\sqrt{6} *4*\sqrt{3}

2. \sqrt{2^{5} *3} \sqrt{6} *2\sqrt{6}*4\sqrt{6}*4\sqrt{3}   <em>factoring 96</em>

<em>since \sqrt{2^{5}*3 } = \sqrt{2^{5} } \sqrt{3}</em>

3. \sqrt{2^{5} } \sqrt{3}\sqrt{6} *2\sqrt{6}*4\sqrt{6}*4\sqrt{3}

<em>using exponent rule - (a^{b}) ^{c} = a^{bc}</em>

<em> \sqrt{2^{5} } = 2^{5/2}</em>

4. 2^{5/2}\sqrt{3}\sqrt{2*3} *2\sqrt{6}*4\sqrt{6}*4\sqrt{3}

<em>doing some simple simplification and 4=2^{2}  and 6=2*3</em>

5. 2^{5/2} \sqrt{3} \sqrt{2} \sqrt{3} *2\sqrt{2} \sqrt{3} *2^{2}\sqrt{2} \sqrt{3}*4\sqrt{3}

<em>collecting the roots on one side and applying exponent rule</em>

6. \sqrt{3} \sqrt{3}\sqrt{3} \sqrt{3}\sqrt{2} \sqrt{2}\sqrt{2} *2^{5/2+1+2+2} \sqrt{3}

<em>Applying exponents rule on all \sqrt{3} and \sqrt{2}</em>

<em>7. 2^{1/2+1/2+1/2} *2^{5/2+1+2+2}*3^{1/2+1/2+1/2+1/2+1/2}</em>

<em>combining all powers of 2</em>

8. 2^{1/2+1/2+1/2+5/2+1+2+2}*3^{1/2+1/2+1/2+1/2+1/2}

<em>Simplifying</em>

9. 2^{9} *3^{2}\sqrt{3}

10. 512*9\sqrt{3}

11. 4608\sqrt{3}

3 0
3 years ago
Read 2 more answers
How many millions to get a billion? <br><br>A. 100<br>B.4000<br>C.1000<br>D.10​
hoa [83]

Answer:

c

Step-by-step explanation:

3 0
2 years ago
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