Answer :
To solve the problem, we start by understanding the relationship between the length, width, and volume of the rectangular prism.
Given:
- Height [tex]\( h = 3 \)[/tex] cm
- Volume [tex]\( V = 252 \)[/tex] cubic cm
- Length ([tex]\( f \)[/tex]) is 5 cm longer than the width ([tex]\( w \)[/tex])
Let [tex]\( x \)[/tex] be the width of the tray. Then the length [tex]\( f \)[/tex] will be [tex]\( x + 5 \)[/tex] cm.
Using the volume formula for a rectangular prism:
[tex]\[ V = f \cdot w \cdot h \][/tex]
Substitute the known values:
[tex]\[ 252 = (x + 5) \cdot x \cdot 3 \][/tex]
Simplifying the equation:
[tex]\[ 252 = 3x(x + 5) \][/tex]
Remove the factor of 3 from both sides:
[tex]\[ 252 / 3 = x(x + 5) \][/tex]
[tex]\[ 84 = x^2 + 5x \][/tex]
Thus, the equation modeling the volume of the tray in terms of its width [tex]\( x \)[/tex] is:
[tex]\[ x^2 + 5x = 84 \][/tex]
To make sure this equation is written in the required format:
[tex]\[ 3x^2 + 15x = 252 \][/tex]
For the second part, we check if it is possible for the width of the tray to be 7.5 cm. Substitute [tex]\( x = 7.5 \)[/tex] into the equation:
[tex]\[ 7.5^2 + 5 \times 7.5 = 84 \][/tex]
[tex]\[ 56.25 + 37.5 = 93.75 \][/tex]
Since [tex]\( 93.75 \neq 84 \)[/tex], the width cannot be 7.5 cm.
Therefore, the answer is:
Complete the equation:
[tex]\[ 3x^2 + 15x = 252 \][/tex]
Is it possible for the width of the tray to be 7.5 cm?
[tex]\[ \text{No} \][/tex]
Given:
- Height [tex]\( h = 3 \)[/tex] cm
- Volume [tex]\( V = 252 \)[/tex] cubic cm
- Length ([tex]\( f \)[/tex]) is 5 cm longer than the width ([tex]\( w \)[/tex])
Let [tex]\( x \)[/tex] be the width of the tray. Then the length [tex]\( f \)[/tex] will be [tex]\( x + 5 \)[/tex] cm.
Using the volume formula for a rectangular prism:
[tex]\[ V = f \cdot w \cdot h \][/tex]
Substitute the known values:
[tex]\[ 252 = (x + 5) \cdot x \cdot 3 \][/tex]
Simplifying the equation:
[tex]\[ 252 = 3x(x + 5) \][/tex]
Remove the factor of 3 from both sides:
[tex]\[ 252 / 3 = x(x + 5) \][/tex]
[tex]\[ 84 = x^2 + 5x \][/tex]
Thus, the equation modeling the volume of the tray in terms of its width [tex]\( x \)[/tex] is:
[tex]\[ x^2 + 5x = 84 \][/tex]
To make sure this equation is written in the required format:
[tex]\[ 3x^2 + 15x = 252 \][/tex]
For the second part, we check if it is possible for the width of the tray to be 7.5 cm. Substitute [tex]\( x = 7.5 \)[/tex] into the equation:
[tex]\[ 7.5^2 + 5 \times 7.5 = 84 \][/tex]
[tex]\[ 56.25 + 37.5 = 93.75 \][/tex]
Since [tex]\( 93.75 \neq 84 \)[/tex], the width cannot be 7.5 cm.
Therefore, the answer is:
Complete the equation:
[tex]\[ 3x^2 + 15x = 252 \][/tex]
Is it possible for the width of the tray to be 7.5 cm?
[tex]\[ \text{No} \][/tex]