Answer :
Answer:
The greatest value of [tex]x[/tex] is in equation
(2), [tex]x=\sqrt{26}\approx5.09[/tex]
Step-by-step explanation:
Equation (1) :
[tex]x^2=17[/tex]
Take square root of both side
[tex]\sqrt{x^2}=\sqrt{17}\\\\(x^2)^{\frac{1}{2}}=\sqrt{17}\\\\x\approx4.12[/tex]
Equation (2) :
[tex]x^2=26[/tex]
Take square root of both side
[tex]\sqrt{x^2}=\sqrt{26}\\\\(x^2)^{\frac{1}{2}}=\sqrt{26}\\\\x\approx5.09[/tex]
Equation (3) :
[tex]x^3=64\\\\x^3=4\times4\times4\\\\x^3=4^3[/tex]
Take cube root both side
[tex]\sqrt[3]{x^3}=\sqrt[3]{4^3} \\\\(x^3)^{\frac{1}{3}}=(4^3)^{\frac{1}{3}}\\\\x=4[/tex]
Equation (4) :
[tex]x^3=27\\\\x^3=3\times3\times3\\\\x^3=3^3[/tex]
[tex]Take\ cube\ root\ both\ side\\\\(x^3)^{\frac{1}{3}}=(3^3)^{\frac{1}{3}}\\\\x=3[/tex]
Answer:
Equation 2! (or B)
Step-by-step explanation:
[tex]\sqrt{17}[/tex] = 4.12
[tex]\sqrt{26}[/tex] = 5.10
we know [tex]\sqrt{26}[/tex] is larger than [tex]\sqrt{17}[/tex]
[tex]\sqrt[3]{64}[/tex] = 4 = 4 x 4 x 4
And we know this will have to be larger than [tex]\sqrt[3]{27}[/tex]
5.10 is greater than 4, so equation 2 is larger!