Find the product. Assume variables represent positive real numbers.
-$$7 \mathrm { k } ^ { 1 / 8 } \left( 3 \mathrm { k } ^ { 3 / 8 } + 3 \mathrm { k } ^ { 3 / 4 } \right)$$
A) $21 \mathrm { k } ^ { - 1 / 4 } - 21 \mathrm { k } ^ { - 5 / 8 }$

B) $21 \mathrm { k } ^ { 3 / 64 } + 21 \mathrm { k } ^ { 3 / 32 }$

C) $21 \mathrm { k } ^ { 1 / 2 } + 21 \mathrm { k } ^ { 7 / 8 }$

D) $21 \mathrm { k } ^ { 1 / 2 } - 21 \mathrm { k } ^ { 7 / 8 }$

Question

Find the product. Assume variables represent positive real numbers.
-$$7 \mathrm { k } ^ { 1 / 8 } \left( 3 \mathrm { k } ^ { 3 / 8 } + 3 \mathrm { k } ^ { 3 / 4 } \right)$$
A) $21 \mathrm { k } ^ { - 1 / 4 } - 21 \mathrm { k } ^ { - 5 / 8 }$

B) $21 \mathrm { k } ^ { 3 / 64 } + 21 \mathrm { k } ^ { 3 / 32 }$

C) $21 \mathrm { k } ^ { 1 / 2 } + 21 \mathrm { k } ^ { 7 / 8 }$

D) $21 \mathrm { k } ^ { 1 / 2 } - 21 \mathrm { k } ^ { 7 / 8 }$

Quizplus · Accepted Answer

To find the product, distribute $7k^{1/8}$ across the terms inside the parentheses. Use the property of exponents that $a^m \cdot a^n = a^{m+n}$ to combine the exponents. For the first term, $7k^{1/8} \cdot 3k^{3/8} = 21k^{1/2}$ because $1/8 + 3/8 = 4/8 = 1/2$. For the second term, $7k^{1/8} \cdot 3k^{3/4} = 21k^{7/8}$ because $1/8 + 6/8 = 7/8$. Therefore, the product is $21k^{1/2} + 21k^{7/8}$.

Find the Product 7k1/8(3k3/8+3k3/4)7 \mathrm { k } ^ { 1 / 8 } \left( 3 \mathrm { k } ^ { 3 / 8 } + 3 \mathrm { k } ^ { 3 / 4 } \right)7k1/8(3k3/8+3k3/4)

Find the Product $7 \mathrm { k } ^ { 1 / 8 } \left( 3 \mathrm { k } ^ { 3 / 8 } + 3 \mathrm { k } ^ { 3 / 4 } \right)$