7. In this problem, we are given that \( y \) varies directly as \( x \), and we are given a specific value for \( y \) when \( x \) is known. We need to find the constant of variation (\( k \)) and the equation of variation.
When two variables vary directly, it means that they are related by a constant ratio. Mathematically, this can be represented as \( y = kx \), where \( k \) is the constant of variation.
Given that \( y = 30 \) when \( x = 5 \), we can substitute these values into the equation to find the value of \( k \):
\( 30 = k \cdot 5 \)
Simplifying the equation, we have:
\( k = \frac{30}{5} = 6 \)
So, the constant of variation (\( k \)) is 6.
To find the equation of variation, we substitute the value of \( k \) into the equation:
\( y = 6x \)
Therefore, the equation of variation is \( y = 6x \).
8. In this problem, we are given a table showing the number of hours traveled (\( t \)) and the distance traveled (\( d \)) by a car. We need to find the constant of variation and the equation that describes the relation.
To determine if the variables vary directly, we can check if the ratio of \( d \) to \( t \) is constant. Let's calculate the ratios for each row:
\( \frac{30}{1} = 30 \)
\( \frac{80}{2} = 40 \)
\( \frac{130}{3} \approx 43.33 \)
\( \frac{180}{4} = 45 \)
\( \frac{230}{5} = 46 \)
Since the ratios are not constant, it means that the variables do not vary directly. Therefore, we cannot find a constant of variation or an equation of variation for this data.
9. In this problem, we are given that \( x \) varies directly as \( y \), and we are given a specific value for \( x \) when \( y \) is known. We need to find the value of \( y \) when \( x \) is given.
When two variables vary directly, it means that they are related by a constant ratio. Mathematically, this can be represented as \( x = ky \), where \( k \) is the constant of variation.
Given that \( x = 40 \) when \( y = 8 \), we can substitute these values into the equation to find the value of \( k \):
\( 40 = k \cdot 8 \)
Simplifying the equation, we have:
\( k = \frac{40}{8} = 5 \)
So, the constant of variation (\( k \)) is 5.
To find the value of \( y \) when \( x = 32 \), we can rearrange the equation:
\( x = ky \)
\( 32 = 5y \)
Solving for \( y \), we have:
\( y = \frac{32}{5} = 6.4 \)
Therefore, when \( x = 32 \), the value of \( y \) is 6.4.
10. In this problem, we are given that a worker's pay check (\( P \)) varies directly as the number of hours worked (\( h \)). We are given the payment for working 10 hours (\( P = 1500 \)), and we need to find the payment for 35 hours of work.
When two variables vary directly, it means that they are related by a constant ratio. Mathematically, this can be represented as \( P = kh \), where \( k \) is the constant of variation.
Given that \( P = 1500 \) when \( h = 10 \), we can substitute these values into the equation to find the value of \( k \):
\( 1500 = k \cdot 10 \)
Simplifying the equation, we have:
\( k = \frac{1500}{10} = 150 \)
So, the constant of variation (\( k \)) is 150.
To find the payment for 35 hours of work, we can substitute the value of \( k \) and \( h \) into the equation:
\( P = 150 \cdot 35 = 5250 \)
Therefore, the payment for 35 hours of work is Php 5250.
Answers & Comments
Answer:
7. In this problem, we are given that \( y \) varies directly as \( x \), and we are given a specific value for \( y \) when \( x \) is known. We need to find the constant of variation (\( k \)) and the equation of variation.
When two variables vary directly, it means that they are related by a constant ratio. Mathematically, this can be represented as \( y = kx \), where \( k \) is the constant of variation.
Given that \( y = 30 \) when \( x = 5 \), we can substitute these values into the equation to find the value of \( k \):
\( 30 = k \cdot 5 \)
Simplifying the equation, we have:
\( k = \frac{30}{5} = 6 \)
So, the constant of variation (\( k \)) is 6.
To find the equation of variation, we substitute the value of \( k \) into the equation:
\( y = 6x \)
Therefore, the equation of variation is \( y = 6x \).
8. In this problem, we are given a table showing the number of hours traveled (\( t \)) and the distance traveled (\( d \)) by a car. We need to find the constant of variation and the equation that describes the relation.
To determine if the variables vary directly, we can check if the ratio of \( d \) to \( t \) is constant. Let's calculate the ratios for each row:
\( \frac{30}{1} = 30 \)
\( \frac{80}{2} = 40 \)
\( \frac{130}{3} \approx 43.33 \)
\( \frac{180}{4} = 45 \)
\( \frac{230}{5} = 46 \)
Since the ratios are not constant, it means that the variables do not vary directly. Therefore, we cannot find a constant of variation or an equation of variation for this data.
9. In this problem, we are given that \( x \) varies directly as \( y \), and we are given a specific value for \( x \) when \( y \) is known. We need to find the value of \( y \) when \( x \) is given.
When two variables vary directly, it means that they are related by a constant ratio. Mathematically, this can be represented as \( x = ky \), where \( k \) is the constant of variation.
Given that \( x = 40 \) when \( y = 8 \), we can substitute these values into the equation to find the value of \( k \):
\( 40 = k \cdot 8 \)
Simplifying the equation, we have:
\( k = \frac{40}{8} = 5 \)
So, the constant of variation (\( k \)) is 5.
To find the value of \( y \) when \( x = 32 \), we can rearrange the equation:
\( x = ky \)
\( 32 = 5y \)
Solving for \( y \), we have:
\( y = \frac{32}{5} = 6.4 \)
Therefore, when \( x = 32 \), the value of \( y \) is 6.4.
10. In this problem, we are given that a worker's pay check (\( P \)) varies directly as the number of hours worked (\( h \)). We are given the payment for working 10 hours (\( P = 1500 \)), and we need to find the payment for 35 hours of work.
When two variables vary directly, it means that they are related by a constant ratio. Mathematically, this can be represented as \( P = kh \), where \( k \) is the constant of variation.
Given that \( P = 1500 \) when \( h = 10 \), we can substitute these values into the equation to find the value of \( k \):
\( 1500 = k \cdot 10 \)
Simplifying the equation, we have:
\( k = \frac{1500}{10} = 150 \)
So, the constant of variation (\( k \)) is 150.
To find the payment for 35 hours of work, we can substitute the value of \( k \) and \( h \) into the equation:
\( P = 150 \cdot 35 = 5250 \)
Therefore, the payment for 35 hours of work is Php 5250.