Respuesta :
ANSWER
[tex] \sqrt{9} \: and \: \sqrt{16} [/tex]
EXPLANATION
First, simplify the square roots.
[tex] \sqrt{4} = \sqrt{ {2}^{2} } = 2[/tex]
[tex] \sqrt{9} = \sqrt{ {3}^{2} } = 3[/tex]
[tex] \sqrt{16} = \sqrt{ {4}^{2} } = 4[/tex]
[tex] \sqrt{25} = \sqrt{ {5}^{2} } = 5[/tex]
Let us simplify t
[tex] \sqrt{12} [/tex]
Although 12 is not a perfect square, it contains, a perfect square.
[tex] \sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2 \sqrt{3} [/tex]
We can now observe that,
[tex]3 \: < \: 2 \sqrt{3} \: < \: 4[/tex]
Therefore,
[tex] \sqrt{9} \: < \: \sqrt{12} \: < \: \sqrt{16} [/tex]
Hence
[tex] \sqrt{12} [/tex]
is between,
[tex] \sqrt{9} [/tex]
and
[tex] \sqrt{16} [/tex]
[tex] \sqrt{9} \: and \: \sqrt{16} [/tex]
EXPLANATION
First, simplify the square roots.
[tex] \sqrt{4} = \sqrt{ {2}^{2} } = 2[/tex]
[tex] \sqrt{9} = \sqrt{ {3}^{2} } = 3[/tex]
[tex] \sqrt{16} = \sqrt{ {4}^{2} } = 4[/tex]
[tex] \sqrt{25} = \sqrt{ {5}^{2} } = 5[/tex]
Let us simplify t
[tex] \sqrt{12} [/tex]
Although 12 is not a perfect square, it contains, a perfect square.
[tex] \sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2 \sqrt{3} [/tex]
We can now observe that,
[tex]3 \: < \: 2 \sqrt{3} \: < \: 4[/tex]
Therefore,
[tex] \sqrt{9} \: < \: \sqrt{12} \: < \: \sqrt{16} [/tex]
Hence
[tex] \sqrt{12} [/tex]
is between,
[tex] \sqrt{9} [/tex]
and
[tex] \sqrt{16} [/tex]