Respuesta :
Answer:
1) [tex]\sqrt{\frac{-9}{16}}=\pm\frac{3}{4}i[/tex]
2) [tex]\sqrt[3]{-0.000125}=-0.05[/tex]
3) [tex]\sqrt{(36g^6)}=6(g^3)[/tex]
4) [tex]\sqrt[3]{125x^{21}y^{24}}=5(x^{7})(y^{8})[/tex]
5) [tex]\sqrt{\frac{27x^4}{75y^2}}=\frac{3(x^2)}{5y}[/tex]
Step-by-step explanation:
To find : The following expression
1) Find all the real square roots of -9/16.
[tex]\sqrt{\frac{-9}{16}}[/tex]
There is no real roots because [tex]\sqrt{-1}=i[/tex] is an imaginary number
Therefore, [tex]\sqrt{\frac{-9}{16}}=\pm\frac{3}{4}i[/tex]
2) Find all the real cube roots of -0.000125.
[tex]\sqrt[3]{-0.000125}[/tex]
There is a real roots.
Therefore, [tex]\sqrt[3]{-0.000125}=-0.05[/tex]
3) Simplify [tex]\sqrt{(36g^6)}[/tex]
[tex]=\sqrt{(36g^6)}[/tex]
[tex]=\sqrt{6^2(g^3)^2}[/tex]
[tex]=6(g^3)[/tex]
Therefore, simplified form is [tex]\sqrt{(36g^6)}=6(g^3)[/tex]
4) Simplify [tex]\sqrt[3]{125x^{21}y^{24}}[/tex]
[tex]=\sqrt[3]{125x^{21}y^{24}}[/tex]
[tex]=\sqrt[3]{5^3(x^{7})^3(y^{8})^3}[/tex]
[tex]=5(x^{7})(y^{8})[/tex]
Therefore, simplified form is [tex]\sqrt[3]{125x^{21}y^{24}}=5(x^{7})(y^{8})[/tex]
5) Simplify [tex]\sqrt{\frac{27x^4}{75y^2}[/tex]
[tex]=\sqrt{\frac{27x^4}{75y^2}[/tex]
[tex]=\sqrt{\frac{9(x^2)^2}{25y^2}[/tex] (Divide nr. and dr. by 3)
[tex]=\sqrt{\frac{3^2(x^2)^2}{5^2y^2}[/tex]
[tex]=\frac{3(x^2)}{5y}[/tex]
Therefore, simplified form is [tex]\sqrt{\frac{27x^4}{75y^2}}=\frac{3(x^2)}{5y}[/tex]
A real square root of number is the value which is when multiplicand by itself gives the same value as the number posses.
- a) All the real square roots of -9/16 are [tex]\dfrac{3i}{4}[/tex] and [tex]-\dfrac{3i}{4}[/tex].
- b) All the real square roots of -9/16 are 0.05[tex]i[/tex] and -0.05 [tex]i[/tex].
- c) The simplified value of [tex]\sqrt{36g^6}[/tex] is [tex]6g^3[/tex].
- e) The Simplified value of [tex]\sqrt[]{125x^{21}y^{24}}[/tex] is [tex]5x^7y^8[/tex]..
- d) The simplified value of [tex]\sqrt{\dfrac{27x^4}{75y^2}}[/tex] is [tex]{\dfrac{9x^2}{25y} }[/tex].
What is real square root?
A real square root of number is the value which is when multiplicand by itself gives the same value as the number posses.
Given information-
- a) All the real square roots of -9/16-
Put the square root on the number to find the real square roots,
[tex]\sqrt{\dfrac{-9}{16} } =\sqrt{\dfrac{(-1)3^2}{4^2} } \\\sqrt{\dfrac{-9}{16} } =\pm \dfrac{3}{4} \times\sqrt{-1} \\\sqrt{\dfrac{-9}{16} } =\pm \dfrac{3}{4} (i)\\\sqrt{\dfrac{-9}{16} } =\pm\dfrac{3i}{4}[/tex]
Thus all the real square roots of -9/16 are [tex]\dfrac{3i}{4}[/tex] and [tex]-\dfrac{3i}{4}[/tex].
- b) All the real cube roots of -0.000125.
Put the square root on the number to find the real square roots,
[tex]\sqrt{-0.000125} =\sqrt{(-1)\times(0.05)^2} \\\sqrt{-0.000125} =\pm (0.05)\sqrt{(-1)} \\\sqrt{-0.000125} =\pm (0.05)(i)\\\sqrt{-0.000125} =\pm (0.05i)[/tex]
Thus all the real square roots of -9/16 are 0.05[tex]i[/tex] and-0.05 [tex]i[/tex].
- c) Simplify [tex]\sqrt{36g^6}[/tex]
Simplify above square root as,
[tex]\sqrt{36g^6}=\sqrt{6^2\times g^6} \\\sqrt{36g^6}=(6^2\times g^6} )^\dfrac{1}{2}\\\sqrt{36g^6}=(6^2\times g^{(3)^2}} )^\dfrac{1}{2} \\[/tex]
[tex]\sqrt{36g^6}=(6\times g^3} )^\dfrac{2}{2}\\[/tex]
[tex]\sqrt{36g^6}=6 g^3}}[/tex]
Hence the Simplified value of [tex]\sqrt{36g^6}[/tex] is [tex]6g^3[/tex].
- d) Simplify [tex]\sqrt[]{125x^{21}y^{24}}[/tex]
Simplify above square root as,
[tex]\sqrt[]{125x^{21}y^{24}} ={(5^3)(x^7)^3(y^8)^3} )^\dfrac{1}{3}}\\\sqrt[]{125x^{21}y^{24}} ={(5^3)(x^7)^3(y^8)^3} )^\dfrac{3}{3}}\\\sqrt[]{125x^{21}y^{24}} =5^3x^7y^8}[/tex]
Hence the Simplified value of [tex]\sqrt[]{125x^{21}y^{24}}[/tex] is [tex]5x^7y^8[/tex]..
- e) Simplify [tex]\sqrt{\dfrac{27x^4}{75y^2}}[/tex]
Simplify above square root as,
[tex]\sqrt{\dfrac{27x^4}{75y^2} } =\sqrt{\dfrac{9\times3\times x^4}{15\times3\times y^2} }\\\sqrt{\dfrac{27x^4}{75y^2} } =({\dfrac{9\times (x^2)^2}{25\times y^2} } )^\dfrac{1}{2}} \\\sqrt{\dfrac{27x^4}{75y^2} } ={\dfrac{9x^2}{25y} }[/tex]
Hence the Simplified value of [tex]\sqrt{\dfrac{27x^4}{75y^2}}[/tex] is [tex]{\dfrac{9x^2}{25y} }[/tex].
Thus,
- a) All the real square roots of -9/16 are [tex]\dfrac{3i}{4}[/tex] and [tex]-\dfrac{3i}{4}[/tex].
- b) All the real square roots of -9/16 are 0.05[tex]i[/tex] and -0.05 [tex]i[/tex].
- c) The simplified value of [tex]\sqrt{36g^6}[/tex] is [tex]6g^3[/tex].
- d) The Simplified value of [tex]\sqrt[]{125x^{21}y^{24}}[/tex] is [tex]5x^7y^8[/tex]..
- e) The simplified value of [tex]\sqrt{\dfrac{27x^4}{75y^2}}[/tex] is [tex]{\dfrac{9x^2}{25y} }[/tex].
Learn more about the real square root here;
https://brainly.com/question/664132
