To solve this problem, we'll proceed step-by-step and utilize the concept of direct variation.
Direct variation means that [tex]\( y \)[/tex] is directly proportional to [tex]\( x \)[/tex]. This can be expressed with the equation:
[tex]\[ y = kx \][/tex]
where [tex]\( k \)[/tex] is the constant of variation.
Given:
1. [tex]\( y = 400 \)[/tex] when [tex]\( x = r \)[/tex].
2. [tex]\( y = r \)[/tex] when [tex]\( x = 4 \)[/tex].
Let's use these conditions to find the value of [tex]\( r \)[/tex].
1. From the first condition, when [tex]\( y = 400 \)[/tex] and [tex]\( x = r \)[/tex]:
[tex]\[ 400 = kr \][/tex]
2. From the second condition, when [tex]\( y = r \)[/tex] and [tex]\( x = 4 \)[/tex]:
[tex]\[ r = k \cdot 4 \][/tex]
[tex]\[ k = \frac{r}{4} \][/tex]
We now have two equations involving [tex]\( k \)[/tex]:
[tex]\[ 400 = kr \][/tex]
[tex]\[ k = \frac{r}{4} \][/tex]
Substituting the value of [tex]\( k \)[/tex] from the second equation into the first equation:
[tex]\[ 400 = \left(\frac{r}{4}\right) r \][/tex]
[tex]\[ 400 = \frac{r^2}{4} \][/tex]
Multiply both sides by 4 to solve for [tex]\( r^2 \)[/tex]:
[tex]\[ 1600 = r^2 \][/tex]
Taking the square root of both sides:
[tex]\[ r = \sqrt{1600} \][/tex]
[tex]\[ r = 40 \][/tex]
So now we know that [tex]\( r = 40 \)[/tex].
Using this result, we can find the constant [tex]\( k \)[/tex]:
[tex]\[ k = \frac{r}{4} \][/tex]
[tex]\[ k = \frac{40}{4} \][/tex]
[tex]\[ k = 10 \][/tex]
Thus, the values we've found are:
- [tex]\( r = 40 \)[/tex]
- [tex]\( k = 10 \)[/tex]
Now we can summarize the constant of variation and the proportional relationship between [tex]\( y \)[/tex] and [tex]\( x \)[/tex]:
[tex]\[ y = 10x \][/tex]
This directly means [tex]\( y \)[/tex] is 10 times [tex]\( x \)[/tex].
So the correct answer to the number relation is:
[tex]\[ 10 \][/tex]
The answer is 10.
