Question

Is anyone here familiar with Algebra 1B problems?

Answer 1

Answer:

Part 5) $3$

Part 6) $2$

Par 7) $S=5\ ft$

Part 8) $S=7\ m$

Step-by-step explanation:

Part 5) we have

$\sqrt[3]{27}$

we know that

$27=3^{3}$

substitute

$\sqrt[3]{27}=\sqrt[3]{3^{3}}=3^{\frac{3}{3}}=3$

Part 6) we have

$\sqrt[4]{16}$

we know that

$16=2^{4}$

substitute

$\sqrt[4]{16}=\sqrt[4]{2^{4}}=2^{\frac{4}{4}}=2$

Part 7) we know that

The volume of the cube is equal to

$V=S^{3}$

where

S is the length side of the cube

In this problem we have

$V=125\ ft^{3}$

substitute

$125=S^{3}$

$S=\sqrt[3]{125}=\sqrt[3]{5^{3}}=5^{\frac{3}{3}}=5\ ft$

Part 8) we know that

The volume of the cube is equal to

$V=S^{3}$

where

S is the length side of the cube

In this problem we have

$V=343\ m^{3}$

substitute

$343=S^{3}$

$S=\sqrt[3]{343}=\sqrt[3]{7^{3}}=7^{\frac{3}{3}}=7\ m$