Select The Correct Answer.The Variable $b$ Varies Directly As The Square Root Of $c$. If $ B = 100 B=100 B = 100 [/tex] When $c=4$, Which Equation Can Be Used To Find Other Combinations Of $b$ And
Understanding Direct Variation
Direct variation is a relationship between two variables where one variable is a constant multiple of the other. In mathematical terms, if $y$ varies directly as $x$, then $y = kx$, where $k$ is the constant of variation. In this problem, we are given that the variable $b$ varies directly as the square root of $c$, which can be expressed as $b = k\sqrt{c}$.
The Given Information
We are given that $b = 100$ when $c = 4$. This information can be used to find the value of the constant of variation $k$. To do this, we can substitute the given values into the equation $b = k\sqrt{c}$ and solve for $k$.
Finding the Constant of Variation
Substituting $b = 100$ and $c = 4$ into the equation $b = k\sqrt{c}$, we get:
Simplifying the equation, we get:
Dividing both sides of the equation by 2, we get:
Writing the Equation
Now that we have found the value of the constant of variation $k$, we can write the equation that can be used to find other combinations of $b$ and $c$. Substituting $k = 50$ into the equation $b = k\sqrt{c}$, we get:
Verifying the Equation
To verify that this equation is correct, we can substitute the given values $b = 100$ and $c = 4$ into the equation and check if it holds true.
Since the equation holds true, we can be confident that the equation $b = 50\sqrt{c}$ is the correct equation to use to find other combinations of $b$ and $c$.
Conclusion
In this problem, we were given that the variable $b$ varies directly as the square root of $c$. We used the given information to find the value of the constant of variation $k$ and then wrote the equation that can be used to find other combinations of $b$ and $c$. The equation $b = 50\sqrt{c}$ can be used to find other combinations of $b$ and $c$.
Example Use Case
Suppose we want to find the value of $b$ when $c = 9$. We can substitute $c = 9$ into the equation $b = 50\sqrt{c}$ and solve for $b$.
Therefore, when $c = 9$, the value of $b$ is 150.
Final Answer
The final answer is:
Understanding Direct Variation and Square Root Relationship
In the previous article, we discussed the concept of direct variation and how it relates to the square root of a variable. We also learned how to write an equation to find other combinations of variables. In this article, we will answer some frequently asked questions related to direct variation and square root relationship.
Q: What is direct variation?
A: Direct variation is a relationship between two variables where one variable is a constant multiple of the other. In mathematical terms, if $y$ varies directly as $x$, then $y = kx$, where $k$ is the constant of variation.
Q: What is the square root relationship?
A: The square root relationship is a type of direct variation where one variable varies directly as the square root of the other variable. In mathematical terms, if $y$ varies directly as the square root of $x$, then $y = k\sqrt{x}$, where $k$ is the constant of variation.
Q: How do I find the constant of variation?
A: To find the constant of variation, you need to know the values of two variables that are related by direct variation. You can use the formula $k = \frac{y}{x}$ to find the constant of variation.
Q: What is the equation for direct variation and square root relationship?
A: The equation for direct variation and square root relationship is $y = k\sqrt{x}$, where $k$ is the constant of variation.
Q: How do I use the equation to find other combinations of variables?
A: To use the equation to find other combinations of variables, you need to substitute the values of the variables into the equation and solve for the other variable.
Q: What if I have a negative value for the constant of variation?
A: If you have a negative value for the constant of variation, it means that the relationship between the variables is inverse variation, not direct variation.
Q: Can I use the equation to find the value of a variable if I know the value of the other variable?
A: Yes, you can use the equation to find the value of a variable if you know the value of the other variable. Simply substitute the value of the known variable into the equation and solve for the other variable.
Q: What if I have a variable that is not a perfect square?
A: If you have a variable that is not a perfect square, you can still use the equation to find the value of the other variable. However, you may need to use a calculator or a computer to solve the equation.
Q: Can I use the equation to find the value of a variable if I know the value of the other variable and the constant of variation?
A: Yes, you can use the equation to find the value of a variable if you know the value of the other variable and the constant of variation. Simply substitute the values into the equation and solve for the other variable.
Q: What if I have a system of equations with direct variation and square root relationship?
A: If you have a system of equations with direct variation and square root relationship, you can use the same techniques as before to solve the system of equations. However, you may need to use a calculator or a computer to solve the system of equations.
Conclusion
In this article, we answered some frequently asked questions related to direct variation and square root relationship. We also discussed how to use the equation to find other combinations of variables and how to handle different types of variables. We hope that this article has been helpful in understanding direct variation and square root relationship.
Example Use Case
Suppose we have a system of equations with direct variation and square root relationship:
We can use the same techniques as before to solve the system of equations. First, we can substitute the value of $y$ from the first equation into the second equation:
Squaring both sides of the equation, we get:
Solving for $x$, we get:
Substituting the value of $x$ into the first equation, we get:
Simplifying the equation, we get:
Therefore, the solution to the system of equations is $x = \frac{4}{9}$ and $y = \frac{4}{3}$.
Final Answer
The final answer is: