Answer:
Part 5) Option b [tex]2x\sqrt{3}\ ft[/tex]
Part 6) Option d. [tex]4y\sqrt[3]{2}\ mm[/tex]
Step-by-step explanation:
Part 5) we know that
The area of a square is equal to
[tex]A=b^2[/tex]
where
b is the length side of the square
we have
[tex]A=12x^2\ ft^2[/tex]
substitute
[tex]12x^2=b^2[/tex]
Solve for b
take square root both sides
[tex]b=\sqrt{12x^{2}}[/tex]
Remember that
[tex]12=(2^2)(3)[/tex]
substitute
[tex]b=\sqrt{(2^2)(3)x^{2}}[/tex]
Applying property of exponents
[tex]b=\sqrt{(2^2)(3)x^{2}}=[(2^2)(3)x^{2}]^{\frac{1}{2}}=[2^2x^2]^{\frac{1}{2}}3^{\frac{1}{2}}=2x\sqrt{3}\ ft[/tex]
Part 6) we know that
The volume of a cube is equal to
[tex]V=b^3[/tex]
where
b is the length side of the cube
we have
[tex]V=128y^3\ mm^3[/tex]
substitute
[tex]128y^3=b^3[/tex]
Solve for b
take cubic root both sides
[tex]b=\sqrt[3]{128y^3}[/tex]
Remember that
[tex]128=(2^7)=(2^6)(2)=(2^2)^3(2)[/tex]
substitute
[tex]b=\sqrt[3]{(2^2)^3(2)y^3}[/tex]
Applying property of exponents
[tex]b=\sqrt[3]{(2^2)^3(2)y^3}=[(2^2)^3(2)y^3]^{\frac{1}{3}}=[(2^2)^3y^3]^{\frac{1}{3}}2^{\frac{1}{3}}=2^2y\sqrt[3]{2}=4y\sqrt[3]{2}\ mm[/tex]
