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Sagot :
Let's break down the solution into several steps:
1. Height of the spray 2 feet away from the sprinkler head:
We use the given formula [tex]\( h(x) = 160x - 16x^2 \)[/tex] to find the height when [tex]\( x = 2 \)[/tex].
[tex]\[ h(2) = 160 \times 2 - 16 \times 2^2 \][/tex]
Simplifying inside the equation:
[tex]\[ h(2) = 320 - 16 \times 4 \][/tex]
[tex]\[ h(2) = 320 - 64 \][/tex]
[tex]\[ h(2) = 256 \][/tex]
Therefore, after 2 feet, the height of the spray is [tex]\( \boxed{256} \)[/tex] inches.
2. Distance along the ground where the spray reaches maximum height:
The height function [tex]\( h(x) = 160x - 16x^2 \)[/tex] represents a parabola that opens downwards (since the coefficient of [tex]\( x^2 \)[/tex] is negative). The maximum height occurs at the vertex of the parabola.
For a quadratic equation in the form [tex]\( ax^2 + bx + c \)[/tex], the x-coordinate of the vertex is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = -16 \)[/tex] and [tex]\( b = 160 \)[/tex]. Substituting these values in:
[tex]\[ x = -\frac{160}{2 \times -16} \][/tex]
[tex]\[ x = -\frac{160}{-32} \][/tex]
[tex]\[ x = 5 \][/tex]
So, the spray reaches its maximum height at [tex]\( \boxed{5} \)[/tex] feet away from the sprinkler head.
3. Maximum height of the water spray:
We already found the x-coordinate where the maximum height occurs (5 feet away). To find the maximum height:
[tex]\[ h(5) = 160 \times 5 - 16 \times 5^2 \][/tex]
Simplifying inside the equation:
[tex]\[ h(5) = 800 - 16 \times 25 \][/tex]
[tex]\[ h(5) = 800 - 400 \][/tex]
[tex]\[ h(5) = 400 \][/tex]
Therefore, the maximum height of the water spray is [tex]\( \boxed{400} \)[/tex] inches.
4. Distance away from the sprinkler head where the water hits the ground again:
The water hits the ground whenever the height [tex]\( h(x) \)[/tex] is zero. Therefore, we solve the equation:
[tex]\[ 0 = 160x - 16x^2 \][/tex]
Factoring out the common terms:
[tex]\[ 0 = x (160 - 16x) \][/tex]
This gives us two solutions:
[tex]\[ x = 0 \quad \text{or} \quad 160 - 16x = 0 \][/tex]
Solving [tex]\( 160 - 16x = 0 \)[/tex]:
[tex]\[ 160 = 16x \][/tex]
[tex]\[ x = 10 \][/tex]
Therefore, the water hits the ground again at [tex]\( \boxed{10} \)[/tex] feet away from the sprinkler head.
1. Height of the spray 2 feet away from the sprinkler head:
We use the given formula [tex]\( h(x) = 160x - 16x^2 \)[/tex] to find the height when [tex]\( x = 2 \)[/tex].
[tex]\[ h(2) = 160 \times 2 - 16 \times 2^2 \][/tex]
Simplifying inside the equation:
[tex]\[ h(2) = 320 - 16 \times 4 \][/tex]
[tex]\[ h(2) = 320 - 64 \][/tex]
[tex]\[ h(2) = 256 \][/tex]
Therefore, after 2 feet, the height of the spray is [tex]\( \boxed{256} \)[/tex] inches.
2. Distance along the ground where the spray reaches maximum height:
The height function [tex]\( h(x) = 160x - 16x^2 \)[/tex] represents a parabola that opens downwards (since the coefficient of [tex]\( x^2 \)[/tex] is negative). The maximum height occurs at the vertex of the parabola.
For a quadratic equation in the form [tex]\( ax^2 + bx + c \)[/tex], the x-coordinate of the vertex is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Here, [tex]\( a = -16 \)[/tex] and [tex]\( b = 160 \)[/tex]. Substituting these values in:
[tex]\[ x = -\frac{160}{2 \times -16} \][/tex]
[tex]\[ x = -\frac{160}{-32} \][/tex]
[tex]\[ x = 5 \][/tex]
So, the spray reaches its maximum height at [tex]\( \boxed{5} \)[/tex] feet away from the sprinkler head.
3. Maximum height of the water spray:
We already found the x-coordinate where the maximum height occurs (5 feet away). To find the maximum height:
[tex]\[ h(5) = 160 \times 5 - 16 \times 5^2 \][/tex]
Simplifying inside the equation:
[tex]\[ h(5) = 800 - 16 \times 25 \][/tex]
[tex]\[ h(5) = 800 - 400 \][/tex]
[tex]\[ h(5) = 400 \][/tex]
Therefore, the maximum height of the water spray is [tex]\( \boxed{400} \)[/tex] inches.
4. Distance away from the sprinkler head where the water hits the ground again:
The water hits the ground whenever the height [tex]\( h(x) \)[/tex] is zero. Therefore, we solve the equation:
[tex]\[ 0 = 160x - 16x^2 \][/tex]
Factoring out the common terms:
[tex]\[ 0 = x (160 - 16x) \][/tex]
This gives us two solutions:
[tex]\[ x = 0 \quad \text{or} \quad 160 - 16x = 0 \][/tex]
Solving [tex]\( 160 - 16x = 0 \)[/tex]:
[tex]\[ 160 = 16x \][/tex]
[tex]\[ x = 10 \][/tex]
Therefore, the water hits the ground again at [tex]\( \boxed{10} \)[/tex] feet away from the sprinkler head.
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