Question

Caucule.

$a)3 + \sqrt{64}$

$b) {7}^{2} - \sqrt{25}$

$c) \sqrt{8 + 1}$

$d)\sqrt{ \frac{12}{3} }$
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Answer 1

a)

$3+\sqrt{64}\\\\=3+8\\\\=11$

b)

$7^2-\sqrt{25}\\\\=7^2-5\\\\7^2=49\\\\=49-5\\\\=44$

c)

$\sqrt{8+1}\\\\=\sqrt{9}\\\\=\sqrt{3^2}\\\\\sqrt{3^2}=3\\\\=3$

d)

$\sqrt{\frac{12}{3}}\\\\\frac{12}{3}=4\\\\=\sqrt{4}\\\\4=2^2\\\\=\sqrt{2^2}\\\\\sqrt{2^2}=2\\\\=2$