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Sagot :
To solve for [tex]\( y \)[/tex] when [tex]\( x = 7 \)[/tex] given that [tex]\( y \)[/tex] varies directly with [tex]\( x \)[/tex], we can follow these steps:
1. Identify the given information:
- We know that [tex]\( y = 40 \)[/tex] when [tex]\( x = 8 \)[/tex].
- The relationship given by the problem is [tex]\( y = kx \)[/tex], where [tex]\( k \)[/tex] is a constant.
2. Find the constant of variation [tex]\( k \)[/tex] using the initial values:
- From the relationship [tex]\( y = kx \)[/tex], substitute [tex]\( y = 40 \)[/tex] and [tex]\( x = 8 \)[/tex]:
[tex]\[ 40 = k \times 8 \][/tex]
- Solve for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{40}{8} = 5 \][/tex]
3. Use the constant [tex]\( k = 5 \)[/tex] to find [tex]\( y \)[/tex] when [tex]\( x = 7 \)[/tex]:
- Substitute [tex]\( k = 5 \)[/tex] and [tex]\( x = 7 \)[/tex] back into the relationship [tex]\( y = kx \)[/tex]:
[tex]\[ y = 5 \times 7 \][/tex]
4. Calculate the value of [tex]\( y \)[/tex]:
[tex]\[ y = 35 \][/tex]
Therefore, [tex]\( y \)[/tex] equals 35 when [tex]\( x = 7 \)[/tex].
1. Identify the given information:
- We know that [tex]\( y = 40 \)[/tex] when [tex]\( x = 8 \)[/tex].
- The relationship given by the problem is [tex]\( y = kx \)[/tex], where [tex]\( k \)[/tex] is a constant.
2. Find the constant of variation [tex]\( k \)[/tex] using the initial values:
- From the relationship [tex]\( y = kx \)[/tex], substitute [tex]\( y = 40 \)[/tex] and [tex]\( x = 8 \)[/tex]:
[tex]\[ 40 = k \times 8 \][/tex]
- Solve for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{40}{8} = 5 \][/tex]
3. Use the constant [tex]\( k = 5 \)[/tex] to find [tex]\( y \)[/tex] when [tex]\( x = 7 \)[/tex]:
- Substitute [tex]\( k = 5 \)[/tex] and [tex]\( x = 7 \)[/tex] back into the relationship [tex]\( y = kx \)[/tex]:
[tex]\[ y = 5 \times 7 \][/tex]
4. Calculate the value of [tex]\( y \)[/tex]:
[tex]\[ y = 35 \][/tex]
Therefore, [tex]\( y \)[/tex] equals 35 when [tex]\( x = 7 \)[/tex].
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