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Sagot :
When a quantity [tex]\( y \)[/tex] varies directly with another quantity [tex]\( x \)[/tex], it means that [tex]\( y = kx \)[/tex] for some constant [tex]\( k \)[/tex].
Given:
- [tex]\( y = 45 \)[/tex] when [tex]\( x = 5 \)[/tex]
First, we need to find the constant of variation, [tex]\( k \)[/tex].
1. Substitute the known values into the equation [tex]\( y = kx \)[/tex]:
[tex]\[ 45 = k \times 5 \][/tex]
2. Solve for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{45}{5} = 9 \][/tex]
Now, we know that [tex]\( k = 9 \)[/tex]. We can use this value to find [tex]\( y \)[/tex] when [tex]\( x = 10 \)[/tex].
3. Substitute [tex]\( k = 9 \)[/tex] and [tex]\( x = 10 \)[/tex] into the equation [tex]\( y = kx \)[/tex]:
[tex]\[ y = 9 \times 10 \][/tex]
4. Calculate [tex]\( y \)[/tex]:
[tex]\[ y = 90 \][/tex]
Therefore, [tex]\( y \)[/tex] is 90 when [tex]\( x \)[/tex] is 10.
Given:
- [tex]\( y = 45 \)[/tex] when [tex]\( x = 5 \)[/tex]
First, we need to find the constant of variation, [tex]\( k \)[/tex].
1. Substitute the known values into the equation [tex]\( y = kx \)[/tex]:
[tex]\[ 45 = k \times 5 \][/tex]
2. Solve for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{45}{5} = 9 \][/tex]
Now, we know that [tex]\( k = 9 \)[/tex]. We can use this value to find [tex]\( y \)[/tex] when [tex]\( x = 10 \)[/tex].
3. Substitute [tex]\( k = 9 \)[/tex] and [tex]\( x = 10 \)[/tex] into the equation [tex]\( y = kx \)[/tex]:
[tex]\[ y = 9 \times 10 \][/tex]
4. Calculate [tex]\( y \)[/tex]:
[tex]\[ y = 90 \][/tex]
Therefore, [tex]\( y \)[/tex] is 90 when [tex]\( x \)[/tex] is 10.
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