Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
[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]
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.