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Sagot :
To find the dimensions of the horse stalls, we need to use the volume formula for a rectangular prism, which is [tex]\( V = l \cdot w \cdot h \)[/tex], where [tex]\( l \)[/tex] is the length, [tex]\( w \)[/tex] is the width, and [tex]\( h \)[/tex] is the height.
Given:
- The height [tex]\( h = 9 \)[/tex] feet
- The volume [tex]\( V = 1080 \)[/tex] cubic feet
- The length [tex]\( l \)[/tex] is 2 feet longer than the width [tex]\( w \)[/tex].
Let's denote the width [tex]\( w \)[/tex] by [tex]\( x \)[/tex]. Therefore, the length [tex]\( l \)[/tex] can be written as [tex]\( x + 2 \)[/tex].
The volume equation in terms of [tex]\( x \)[/tex] becomes:
[tex]\[ 1080 = x \cdot (x + 2) \cdot 9 \][/tex]
Now, let's simplify this equation step by step.
1. Distribute and multiply:
[tex]\[ 1080 = 9x(x + 2) \][/tex]
2. To make it easier to handle, divide both sides of the equation by 9:
[tex]\[ \frac{1080}{9} = x(x + 2) \][/tex]
[tex]\[ 120 = x^2 + 2x \][/tex]
After simplifying, we get:
[tex]\[ x^2 + 2x = 120 \][/tex]
So, the correct formulation of the equation is:
[tex]\[ 1 \cdot x^2 + 2 \cdot x = 120 \][/tex]
To solve for [tex]\( x \)[/tex], we need to factorize or use the quadratic formula:
[tex]\[ x^2 + 2x - 120 = 0 \][/tex]
By solving this quadratic equation, we find the solutions are [tex]\( x = 10 \)[/tex] feet and [tex]\( x = -12 \)[/tex] feet. Since a width cannot be negative, we discard [tex]\( x = -12 \)[/tex].
Hence, the possible correct width is:
[tex]\[ x = 10 \][/tex]
Lastly, confirming:
Is it possible for the width of a stall to be 10 feet?
Yes, it is possible for the width to be 10 feet.
In summary:
- Complete the equation: [tex]\( 1 \cdot x^2 + 2 \cdot x = 120 \)[/tex]
- Is it possible for the width of a stall to be 10 feet? Yes
Given:
- The height [tex]\( h = 9 \)[/tex] feet
- The volume [tex]\( V = 1080 \)[/tex] cubic feet
- The length [tex]\( l \)[/tex] is 2 feet longer than the width [tex]\( w \)[/tex].
Let's denote the width [tex]\( w \)[/tex] by [tex]\( x \)[/tex]. Therefore, the length [tex]\( l \)[/tex] can be written as [tex]\( x + 2 \)[/tex].
The volume equation in terms of [tex]\( x \)[/tex] becomes:
[tex]\[ 1080 = x \cdot (x + 2) \cdot 9 \][/tex]
Now, let's simplify this equation step by step.
1. Distribute and multiply:
[tex]\[ 1080 = 9x(x + 2) \][/tex]
2. To make it easier to handle, divide both sides of the equation by 9:
[tex]\[ \frac{1080}{9} = x(x + 2) \][/tex]
[tex]\[ 120 = x^2 + 2x \][/tex]
After simplifying, we get:
[tex]\[ x^2 + 2x = 120 \][/tex]
So, the correct formulation of the equation is:
[tex]\[ 1 \cdot x^2 + 2 \cdot x = 120 \][/tex]
To solve for [tex]\( x \)[/tex], we need to factorize or use the quadratic formula:
[tex]\[ x^2 + 2x - 120 = 0 \][/tex]
By solving this quadratic equation, we find the solutions are [tex]\( x = 10 \)[/tex] feet and [tex]\( x = -12 \)[/tex] feet. Since a width cannot be negative, we discard [tex]\( x = -12 \)[/tex].
Hence, the possible correct width is:
[tex]\[ x = 10 \][/tex]
Lastly, confirming:
Is it possible for the width of a stall to be 10 feet?
Yes, it is possible for the width to be 10 feet.
In summary:
- Complete the equation: [tex]\( 1 \cdot x^2 + 2 \cdot x = 120 \)[/tex]
- Is it possible for the width of a stall to be 10 feet? Yes
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