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If y varies directly as x, and y is 180 when x is n and y is n when x is 5, what is the value of n?
O 6
O 18
O 30
O 36

Sagot :

GinnyAnswer
To find the value of [tex]\( n \)[/tex] where [tex]\( y \)[/tex] varies directly as [tex]\( x \)[/tex], we can use the concept of direct variation. In a direct variation, the relationship between [tex]\( y \)[/tex] and [tex]\( x \)[/tex] can be expressed as:

[tex]\[ y = kx \][/tex]

where [tex]\( k \)[/tex] is the constant of proportionality.

Given two sets of values:

1. [tex]\( y = 180 \)[/tex] when [tex]\( x = n \)[/tex]
2. [tex]\( y = n \)[/tex] when [tex]\( x = 5 \)[/tex]

We can use this information to determine the value of [tex]\( n \)[/tex].

### Step-by-Step Solution

1. Establish the first direct variation relationship:

When [tex]\( y = 180 \)[/tex] and [tex]\( x = n \)[/tex]:

[tex]\[ 180 = kn \][/tex]

Therefore, the constant [tex]\( k \)[/tex] can be expressed as:

[tex]\[ k = \frac{180}{n} \][/tex]

2. Establish the second direct variation relationship:

When [tex]\( y = n \)[/tex] and [tex]\( x = 5 \)[/tex]:

[tex]\[ n = k \cdot 5 \][/tex]

3. Substitute the value of [tex]\( k \)[/tex] from the first relationship into the second relationship:

[tex]\[ n = \left(\frac{180}{n}\right) \cdot 5 \][/tex]

4. Simplify the equation:

[tex]\[ n^2 = 180 \cdot 5 \][/tex]

[tex]\[ n^2 = 900 \][/tex]

5. Solve for [tex]\( n \)[/tex]:

[tex]\[ n = \sqrt{900} \][/tex]

[tex]\[ n = 30 \][/tex]

### Conclusion
The value of [tex]\( n \)[/tex] is [tex]\( 30 \)[/tex].

Thus, the correct answer is:

- [tex]\( 30 \)[/tex]

Therefore, the value of [tex]\( n \)[/tex] is 30.
semsee45

Answer:

C) 30

Step-by-step explanation:

Given that y varies directly as x, we can express this relationship with the equation:

[tex]\Large\boxed{y=kx}[/tex]

where k is the constant of proportionality.

Given conditions:

  • y = 180 when x = n
  • y = n when x = 5

Substitute these conditions into y = kx to create two equations:

[tex]180=kn \\\\n=5k[/tex]

Rearrange the first equation to isolate k by dividing both sides by n:

[tex]\dfrac{180}{n}=\dfrac{kn}{n} \\\\\\k=\dfrac{180}{n}[/tex]

Now, substitute the expression for k into the second equation and solve for n:

[tex]n=5 \cdot \dfrac{180}{n} \\\\\\ n= \dfrac{900}{n} \\\\\\ n^2=900 \\\\\\ \sqrt{n^2}=\sqrt{900} \\\\\\n=30[/tex]

Therefore, the value of n is:

[tex]\LARGE\boxed{\boxed{n=30}}[/tex]

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