/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Find the required value by setti... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the required value by setting up the general equation and then evaluating. Find \(y\) for \(x=6\) and \(z=0.5\) if \(y\) varies directly as \(x\) and inversely as \(z,\) and \(y=60\) when \(x=4\) and \(z=10\)

Short Answer

Expert verified
The value of \( y \) is 1800 when \( x = 6 \) and \( z = 0.5 \).

Step by step solution

01

Understanding Direct and Inverse Variation

The statement "y varies directly as x and inversely as z" means that y is directly proportional to x and inversely proportional to z. We can express this relationship with the equation: \( y = k \frac{x}{z} \), where \( k \) is the constant of variation.
02

Finding the Constant of Variation

We know that \( y = 60 \) when \( x = 4 \) and \( z = 10 \). Substitute these values into the equation from Step 1: \( 60 = k \frac{4}{10} \). Solve for \( k \) by multiplying both sides by 10 and then dividing by 4: \( k = 60 \cdot \frac{10}{4} = 150 \).
03

Setting Up the Equation with the Constant

Now that we have found the constant \( k = 150 \), we can write the specific equation that describes the relationship: \( y = 150 \frac{x}{z} \).
04

Substituting the New Values into the Equation

Substitute \( x = 6 \) and \( z = 0.5 \) into the equation \( y = 150 \frac{x}{z} \): \( y = 150 \frac{6}{0.5} \).
05

Calculating the Required Value of y

Compute the expression \( 150 \frac{6}{0.5} \): First, \( \frac{6}{0.5} = 6 \times 2 = 12 \). Then, \( 150 \times 12 = 1800 \). So, \( y = 1800 \).

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Key Concepts

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

Proportionality
Proportionality is a fundamental concept in mathematics that describes how one quantity changes in relation to another. When we say that a value like \( y \) is directly proportional to another value \( x \), it means that as \( x \) increases, \( y \) will also increase at a constant rate.You can imagine this like two friends running together while maintaining a consistent distance between them. Conversely, when \( y \) is inversely proportional to \( z \), it signifies that \( y \) will decrease as \( z \) increases, similar to pulling apart a rubber band where the middle section becomes thinner as you stretch it.To understand proportionality, you can use the formula \( y = k \frac{x}{z} \).Here, \( k \) is the constant that makes this relationship hold true.Understanding how these different types of proportionality work will allow you to solve a variety of algebra problems more easily.
  • Direct Proportionality: As one value increases, the other increases too.
  • Inverse Proportionality: As one value increases, the other decreases.
Recognizing these relationships can be key to solving many textbook problems with ease.
Constant of Variation
The constant of variation \( k \) is like the secret ingredient that makes the equation work.It is a crucial part of understanding proportional relationships, acting like the multiplier that tunes the equation to fit specific conditions provided in the problem.In the equation \( y = k \frac{x}{z} \), \( k \) is essential; it showcases how strongly \( x \) and \( z \) impact the value of \( y \).Finding \( k \) often involves substituting known values into the formula and solving for \( k \). For instance, if \( y = 60 \) when \( x = 4 \) and \( z = 10 \), plugging these into the equation lets us solve for \( k \):\[60 = k \frac{4}{10}\]From this, you can find \( k \) by rearranging:\[k = 60 \cdot \frac{10}{4} = 150\]This constant \( 150 \) is what allows you to use the equation to find \( y \) with other pairs of \( x \) and \( z \). Always keep \( k \) in mind, as it anchors the relationship and makes all the calculations possible.
Algebraic Equations
Algebraic equations are like math sentences showing relationships between different values.They are constructed by combining numbers and variables with operators (+, -, *, /) to describe how these elements interact.In our context, we have the equation:\[y = k \frac{x}{z}\]Each part of the equation reveals something important:
  • \( y \) is the dependent variable that we are trying to find or analyze.
  • \( k \) is the constant of variation that adjusts the equation for accuracy.
  • \( x \) and \( z \) are independent variables representing input values influencing \( y \).
When solving algebraic equations, it's all about isolating the variable you're solving for.By substituting known values and calculating step by step, you can uncover unknowns with precision.Algebra is powerful because it allows you to predict and understand how changing one part of the equation will affect the outcomes.Be consistent in working through each step to avoid errors, and remember that practice makes solving these equations easier over time.

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Most popular questions from this chapter

Solve the given applied problems involving variation. In a physics experiment, a given force was applied to three objects. The mass \(m\) and the resulting acceleration \(a\) were recorded as follows: $$\begin{array}{l|c|c|c} m(\mathrm{g}) & 2.0 & 3.0 & 4.0 \\ \hline a\left(\mathrm{cm} / \mathrm{s}^{2}\right) & 30 & 20 & 15 \end{array}$$ (a) Is the relationship \(a=f(m)\) one of direct or inverse variation? Explain. (b) Find \(a=f(m)\)

Find the required quantities from the given proportions. For two connected gears, the relation \(\frac{d_{1}}{d_{2}}=\frac{N_{1}}{N_{2}}\) holds, where \(d\) is the diameter of the gear and \(N\) is the number of teeth. Find \(N_{1}\) if \(d_{1}=2.60\) in. \(d_{2}=11.7\) in. and \(N_{2}=45 .\) The ratio \(N_{2} / \overline{N_{1}}\) is called the gear ratio. See Fig. \(18.3 .\)

Solve the given applied problems involving variation. The energy \(E\) available daily from a solar collector varies directly as the percent \(p\) that the sun shines during the day. If a collector provides \(1200 \mathrm{kJ}\) for \(75 \%\) sunshine, how much does it provide for a day during which there is \(35 \%\) sunshine?

Find the required ratios. The Mach number of a moving object is the ratio of its speed to the speed of sound \((1200 \mathrm{km} / \mathrm{h})\). Find the Mach number of a military jet that flew at \(7200 \mathrm{km} / \mathrm{h}\) in 2010.

Solve the given applied problems involving variation. The flow of water \(Q\) through a fire hose is proportional to the cross- sectional area \(A\) of the hose. If 250 gal flows through a hose of diameter 2.00 in. in a given time, how much would flow through a hose 3.00 in. in diameter in the same time?

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