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For the following exercises, write an equation describing the relationship of the given variables. \(y\) varies directly as the square root of \(x\) and when \(x=36, \quad y=24\).

Short Answer

Expert verified
\( y = 4 \cdot \sqrt{x} \) describes the relationship.

Step by step solution

01

Understand Direct Variation

In direct variation, two variables change in the same proportion. If variable \( y \) varies directly with the square root of \( x \), it means we can express this relationship as \( y = k \cdot \sqrt{x} \), where \( k \) is the constant of variation. Our first task is to determine this constant \( k \).
02

Substitute Given Values to Solve for Constant

We know that when \( x = 36 \), \( y = 24 \). Substitute these values into the equation \( y = k \cdot \sqrt{x} \). This gives us: \[ 24 = k \cdot \sqrt{36} \]Calculate \( \sqrt{36} \), which is 6, so the equation is \[ 24 = k \cdot 6 \].
03

Solve for Constant k

Isolate \( k \) in the equation by dividing both sides by 6:\[ k = \frac{24}{6} = 4 \].So the constant \( k \) is 4.
04

Write the Final Equation

Now that we know \( k \), substitute it back into the relationship equation: \[ y = 4 \cdot \sqrt{x} \]. This is the final equation that describes the relationship between \( y \) and \( x \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constant of Variation
When we talk about the *constant of variation*, we are referring to a fixed number that connects two directly varying quantities. In a direct variation, one variable changes at a constant rate relative to another. In our exercise, we see the equation \( y = k \cdot \sqrt{x} \), where \( k \) is the constant of variation. This constant helps us understand how \( y \) changes as the square root of \( x \) changes.

This constant \( k \) remains the same no matter what specific values \( x \) and \( y \) take, as long as the condition \( y \) varies directly as \( \sqrt{x} \) is met.
  • To find the constant of variation, replace the variables with given values.
  • Solving the equation with these values gives you \( k \).
  • Once found, \( k \) is constant for all such variations between \( y \) and \( x \).
Understanding this concept is crucial for solving direct variation problems where one variable is dependent on another's specific mathematical transformation.
Square Root Functions
A *square root function* is a type of function that involves the square root of a variable. Think of it as a special operation that "unsquares" a number. In the context of our exercise, we observe the function \( y = 4 \cdot \sqrt{x} \). Here, the square root function \( \sqrt{x} \) plays a central role in determining the value of \( y \).

The operation of taking a square root is significant because it is only valid for non-negative numbers when considering real numbers.
  • The output of a square root function is always non-negative when dealing with real numbers.
  • The shape of a square root graph is a curve that starts at a point and increases steadily.
  • The rate of increase becomes less steep as \( x \) gets larger.
Understanding how square root functions behave helps us predict the relationship trend in various mathematical and real-world scenarios.
Algebraic Equations
*Algebraic equations* are equations that involve variables and constants, typically solved to find the unknown values. These equations can be linear, quadratic, or follow other patterns depending on the powers and operations involved.

In our exercise, the algebraic equation formed is \( y = 4 \cdot \sqrt{x} \). This represents a direct variation with a square root function. Solving it involves a straightforward substitution of known values and the manipulation of algebraic expressions to find unknown constants.
  • Start by identifying the variables and constants you have.
  • Substitute known values into the equation to solve for unknowns.
  • Re-arrange the equation to solve for the desired variable.
By mastering these basic steps, navigating algebraic equations becomes easier, providing a systematic way to uncover relationships between variables.

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Most popular questions from this chapter

For the following exercises, write an equation describing the relationship of the given variables. \(y\) varies directly as the square of \(x\) and when \(x=4, \quad y=80\).

For the following exercises, use the given information to find the unknown value. \(y\) varies jointly as the square of \(x\) and of \(z\) and inversely as the square root of \(w\) and of \(t\). When \(x=2, z=3, w=16,\) and \(t=3,\) then \(y=1 .\) Find \(y\) when \(x=3, z=2, w=36,\) and \(t=5\).

For the following exercises, write an equation describing the relationship of the given variables. \(y\) varies jointly as the square of \(x\) the cube of \(z\) and the square root of \(W\). When \(x=1, z=2\), and \(w=36,\) then \(y=48\).

For the following exercises, determine the function described and then use it to answer the question. A container holds \(100 \mathrm{~mL}\) of a solution that is 25 \(\mathrm{mL}\) acid. If \(n \mathrm{~mL}\) of a solution that is \(60 \%\) acid is added, the function \(C(n)=\frac{25+.6 n}{100+n}\) gives the concentration, \(C\), as a function of the number of \(\mathrm{mL}\) added, \(n\). Express \(n\) as a function of \(C\) and determine the number of \(\mathrm{mL}\) that need to be added to have a solution that is \(50 \%\) acid.

For the following exercises, determine the function described and then use it to answer the question. Consider a cone with height of 30 feet. Express the radius, \(r,\) in terms of the volume, \(V,\) and find the radius of a cone with volume of 1000 cubic feet.

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