IDNLearn.com is the place where your questions are met with thoughtful and precise answers. Discover the information you need from our experienced professionals who provide accurate and reliable answers to all your questions.
Sagot :
To find the product and simplify the given expression [tex]\(\sqrt{y}(\sqrt{2y} + 2)\)[/tex], let's go through it step-by-step.
1. Distribute [tex]\(\sqrt{y}\)[/tex] to both terms inside the parentheses:
[tex]\[ \sqrt{y}(\sqrt{2y}) + \sqrt{y}(2) \][/tex]
2. Simplify each term separately:
- The first term is [tex]\(\sqrt{y} \cdot \sqrt{2y}\)[/tex]:
[tex]\[ \sqrt{y} \cdot \sqrt{2y} = \sqrt{y \cdot 2y} = \sqrt{2y^2} \][/tex]
- Since [tex]\(y\)[/tex] is nonnegative, [tex]\(\sqrt{y^2} = y\)[/tex]:
[tex]\[ \sqrt{2y^2} = \sqrt{2} \cdot \sqrt{y^2} = \sqrt{2} \cdot y = y\sqrt{2} \][/tex]
- The second term is:
[tex]\[ \sqrt{y} \cdot 2 = 2\sqrt{y} \][/tex]
3. Combine the simplified terms:
[tex]\[ \sqrt{y}(\sqrt{2y} + 2) = y\sqrt{2} + 2\sqrt{y} \][/tex]
Therefore, the simplified expression is:
[tex]\[ y\sqrt{2} + 2\sqrt{y} \][/tex]
This is the exact answer, using radicals as needed.
1. Distribute [tex]\(\sqrt{y}\)[/tex] to both terms inside the parentheses:
[tex]\[ \sqrt{y}(\sqrt{2y}) + \sqrt{y}(2) \][/tex]
2. Simplify each term separately:
- The first term is [tex]\(\sqrt{y} \cdot \sqrt{2y}\)[/tex]:
[tex]\[ \sqrt{y} \cdot \sqrt{2y} = \sqrt{y \cdot 2y} = \sqrt{2y^2} \][/tex]
- Since [tex]\(y\)[/tex] is nonnegative, [tex]\(\sqrt{y^2} = y\)[/tex]:
[tex]\[ \sqrt{2y^2} = \sqrt{2} \cdot \sqrt{y^2} = \sqrt{2} \cdot y = y\sqrt{2} \][/tex]
- The second term is:
[tex]\[ \sqrt{y} \cdot 2 = 2\sqrt{y} \][/tex]
3. Combine the simplified terms:
[tex]\[ \sqrt{y}(\sqrt{2y} + 2) = y\sqrt{2} + 2\sqrt{y} \][/tex]
Therefore, the simplified expression is:
[tex]\[ y\sqrt{2} + 2\sqrt{y} \][/tex]
This is the exact answer, using radicals as needed.
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. IDNLearn.com is committed to your satisfaction. Thank you for visiting, and see you next time for more helpful answers.