Question

I need the last two

Answer 1

Answer:

Part 5) Option b $2x\sqrt{3}\ ft$

Part 6) Option d. $4y\sqrt[3]{2}\ mm$

Step-by-step explanation:

Part 5) we know that

The area of a square is equal to

$A=b^2$

where

b is the length side of the square

we have

$A=12x^2\ ft^2$

substitute

$12x^2=b^2$

Solve for b

take square root both sides

$b=\sqrt{12x^{2}}$

Remember that

$12=(2^2)(3)$

substitute

$b=\sqrt{(2^2)(3)x^{2}}$

Applying property of exponents

$b=\sqrt{(2^2)(3)x^{2}}=[(2^2)(3)x^{2}]^{\frac{1}{2}}=[2^2x^2]^{\frac{1}{2}}3^{\frac{1}{2}}=2x\sqrt{3}\ ft$

Part 6) we know that

The volume of a cube is equal to

$V=b^3$

where

b is the length side of the cube

we have

$V=128y^3\ mm^3$

substitute

$128y^3=b^3$

Solve for b

take cubic root both sides

$b=\sqrt[3]{128y^3}$

Remember that

$128=(2^7)=(2^6)(2)=(2^2)^3(2)$

substitute

$b=\sqrt[3]{(2^2)^3(2)y^3}$

Applying property of exponents

$b=\sqrt[3]{(2^2)^3(2)y^3}=[(2^2)^3(2)y^3]^{\frac{1}{3}}=[(2^2)^3y^3]^{\frac{1}{3}}2^{\frac{1}{3}}=2^2y\sqrt[3]{2}=4y\sqrt[3]{2}\ mm$