Answer :
[tex]1.\ \sqrt{-25}-isn't\ real\ number\to\sqrt{-25}=\sqrt{25(-1)}=\sqrt{25}\cdot\sqrt{-1}=5i\\\\2.\ \sqrt{1.44}=\sqrt{1.2^2}=1.2\\\\3.\ \sqrt{\frac{16}{49}}=\frac{\sqrt{16}}{\sqrt{49}}=\frac{\sqrt{4^2}}{\sqrt{7^2}}=\frac{4}{7}\\\\\sqrt{-\frac{16}{49}}=\frac{4}{7}i\\\\4.\ \sqrt{361}=\sqrt{19^2}=19\\\\5.\ \sqrt{49}=\sqrt{7^2}=7\\\\6.\ \sqrt{0.64}=\sqrt{0.8^2}=0.8\\\\\sqrt{-0.64}=0.8i[/tex]
[tex]7.\ \sqrt{-6.25}=\sqrt{6.25(-1)}=\sqrt{6.25}\cdot\sqrt{-1}=\sqrt{2.5^2}\cdot\sqty{-1}=2.5i\\\\8.\ \sqrt\frac{169}{196}=\frac{\sqrt{169}}{\sqrt{196}}=\frac{\sqrt{13^2}}{\sqrt{14^2}}=\frac{13}{14}\\\\9.\ \sqrt\frac{25}{324}=\frac{\sqrt{25}}{\sqrt{324}}=\frac{\sqrt{5^2}}{\sqrt{18^2}}=\frac{5}{18}\\====================================[/tex]
[tex]i=\sqrt{-1}\ is\ the\ imaginary\ unit\\---------------------\\If\ a\geq0\ then\ \sqrt{a^2}=a[/tex]
[tex]7.\ \sqrt{-6.25}=\sqrt{6.25(-1)}=\sqrt{6.25}\cdot\sqrt{-1}=\sqrt{2.5^2}\cdot\sqty{-1}=2.5i\\\\8.\ \sqrt\frac{169}{196}=\frac{\sqrt{169}}{\sqrt{196}}=\frac{\sqrt{13^2}}{\sqrt{14^2}}=\frac{13}{14}\\\\9.\ \sqrt\frac{25}{324}=\frac{\sqrt{25}}{\sqrt{324}}=\frac{\sqrt{5^2}}{\sqrt{18^2}}=\frac{5}{18}\\====================================[/tex]
[tex]i=\sqrt{-1}\ is\ the\ imaginary\ unit\\---------------------\\If\ a\geq0\ then\ \sqrt{a^2}=a[/tex]
