Join the IDNLearn.com community and start finding the answers you need today. Get step-by-step guidance for all your technical questions from our knowledgeable community members.
Sagot :
To solve the equation [tex]\( x^3 y L + 4 L = 9 y_0 \)[/tex] for [tex]\( x \)[/tex], let's go step by step:
1. Isolate the [tex]\( x \)[/tex]-term: Move all terms except the term containing [tex]\( x \)[/tex] to one side of the equation.
[tex]\[ x^3 y L = 9 y_0 - 4 L \][/tex]
2. Solve for [tex]\( x \)[/tex]: We want to isolate [tex]\( x^3 \)[/tex], so divide both sides by [tex]\( y L \)[/tex].
[tex]\[ x^3 = \frac{9 y_0 - 4 L}{y L} \][/tex]
3. Take the cube root: To finally get [tex]\( x \)[/tex], we take the cube root of both sides of the equation.
[tex]\[ x = \sqrt[3]{\frac{9 y_0 - 4 L}{y L}} \][/tex]
So, the solution to the equation [tex]\( x^3 y L + 4 L = 9 y_0 \)[/tex] is:
[tex]\[ x = \sqrt[3]{\frac{9 y_0 - 4 L}{y L}} \][/tex]
The correct multiple-choice answer is:
[tex]\[ \boxed{\text{B}) \sqrt[3]{\frac{9 y 0 - 4 L}{y L}}} \][/tex]
1. Isolate the [tex]\( x \)[/tex]-term: Move all terms except the term containing [tex]\( x \)[/tex] to one side of the equation.
[tex]\[ x^3 y L = 9 y_0 - 4 L \][/tex]
2. Solve for [tex]\( x \)[/tex]: We want to isolate [tex]\( x^3 \)[/tex], so divide both sides by [tex]\( y L \)[/tex].
[tex]\[ x^3 = \frac{9 y_0 - 4 L}{y L} \][/tex]
3. Take the cube root: To finally get [tex]\( x \)[/tex], we take the cube root of both sides of the equation.
[tex]\[ x = \sqrt[3]{\frac{9 y_0 - 4 L}{y L}} \][/tex]
So, the solution to the equation [tex]\( x^3 y L + 4 L = 9 y_0 \)[/tex] is:
[tex]\[ x = \sqrt[3]{\frac{9 y_0 - 4 L}{y L}} \][/tex]
The correct multiple-choice answer is:
[tex]\[ \boxed{\text{B}) \sqrt[3]{\frac{9 y 0 - 4 L}{y L}}} \][/tex]
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! IDNLearn.com is your reliable source for accurate answers. Thank you for visiting, and we hope to assist you again.