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Sagot :
Certainly! Let's solve the problem step-by-step.
Given the function [tex]\( h(x) = 160x - 16x^2 \)[/tex], we need to find several key values.
Step 1: Height of the spray 2 feet away from the sprinkler head
To find the height at 2 feet away from the sprinkler head, we substitute [tex]\( x = 2 \)[/tex] into the height function:
[tex]\[ h(2) = 160(2) - 16(2)^2 \][/tex]
Calculating this:
[tex]\[ h(2) = 320 - 64 = 256 \][/tex]
Therefore, the height of the spray 2 feet away from the sprinkler head is [tex]\( 256 \)[/tex] inches.
Step 2: Time is 2 seconds
After 2 seconds, the height of the spray is also [tex]\( 256 \)[/tex] inches.
Step 3: Finding the number of feet along the ground where the spray will reach maximum height
The height function [tex]\( h(x) = 160x - 16x^2 \)[/tex] is a quadratic equation in standard form [tex]\( ax^2 + bx + c \)[/tex] where [tex]\( a = -16 \)[/tex] and [tex]\( b = 160 \)[/tex]. The maximum height of a parabola given by a quadratic function [tex]\( ax^2 + bx + c \)[/tex] can be determined using the vertex formula [tex]\( x = -\frac{b}{2a} \)[/tex].
Substituting the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ x = -\frac{160}{2(-16)} = -\frac{160}{-32} = 5 \][/tex]
Thus, the spray reaches its maximum height at [tex]\( x = 5 \)[/tex] feet away from the sprinkler head.
Step 4: Calculating the maximum height
To find the maximum height, substitute [tex]\( x = 5 \)[/tex] back into the height function:
[tex]\[ h(5) = 160(5) - 16(5)^2 \][/tex]
[tex]\[ h(5) = 800 - 400 = 400 \][/tex]
Therefore, the maximum height of the water spray is [tex]\( 400 \)[/tex] inches.
Step 5: Finding where the water will hit the ground again
The water will hit the ground when the height is zero, i.e., [tex]\( h(x) = 0 \)[/tex]. We solve for [tex]\( x \)[/tex] in the equation [tex]\( 160x - 16x^2 = 0 \)[/tex].
Factoring the quadratic equation, we get:
[tex]\[ 160x - 16x^2 = 0 \][/tex]
[tex]\[ 16x(10 - x) = 0 \][/tex]
Setting each factor equal to zero gives us:
[tex]\[ 16x = 0 \quad \text{or} \quad 10 - x = 0 \][/tex]
This results in:
[tex]\[ x = 0 \quad \text{or} \quad x = 10 \][/tex]
Thus, the water will also hit the ground again at [tex]\( x = 10 \)[/tex] feet away from the sprinkler head.
So, to summarize:
- [tex]\( h(2) = 256 \)[/tex]
- After 2 seconds, the height of the spray is [tex]\( 256 \)[/tex] inches.
- The spray reaches its maximum height 5 feet away from the sprinkler head.
- The maximum height of the water spray is [tex]\( 400 \)[/tex] inches.
- The water will hit the ground again at 10 feet away from the sprinkler head.
Given the function [tex]\( h(x) = 160x - 16x^2 \)[/tex], we need to find several key values.
Step 1: Height of the spray 2 feet away from the sprinkler head
To find the height at 2 feet away from the sprinkler head, we substitute [tex]\( x = 2 \)[/tex] into the height function:
[tex]\[ h(2) = 160(2) - 16(2)^2 \][/tex]
Calculating this:
[tex]\[ h(2) = 320 - 64 = 256 \][/tex]
Therefore, the height of the spray 2 feet away from the sprinkler head is [tex]\( 256 \)[/tex] inches.
Step 2: Time is 2 seconds
After 2 seconds, the height of the spray is also [tex]\( 256 \)[/tex] inches.
Step 3: Finding the number of feet along the ground where the spray will reach maximum height
The height function [tex]\( h(x) = 160x - 16x^2 \)[/tex] is a quadratic equation in standard form [tex]\( ax^2 + bx + c \)[/tex] where [tex]\( a = -16 \)[/tex] and [tex]\( b = 160 \)[/tex]. The maximum height of a parabola given by a quadratic function [tex]\( ax^2 + bx + c \)[/tex] can be determined using the vertex formula [tex]\( x = -\frac{b}{2a} \)[/tex].
Substituting the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ x = -\frac{160}{2(-16)} = -\frac{160}{-32} = 5 \][/tex]
Thus, the spray reaches its maximum height at [tex]\( x = 5 \)[/tex] feet away from the sprinkler head.
Step 4: Calculating the maximum height
To find the maximum height, substitute [tex]\( x = 5 \)[/tex] back into the height function:
[tex]\[ h(5) = 160(5) - 16(5)^2 \][/tex]
[tex]\[ h(5) = 800 - 400 = 400 \][/tex]
Therefore, the maximum height of the water spray is [tex]\( 400 \)[/tex] inches.
Step 5: Finding where the water will hit the ground again
The water will hit the ground when the height is zero, i.e., [tex]\( h(x) = 0 \)[/tex]. We solve for [tex]\( x \)[/tex] in the equation [tex]\( 160x - 16x^2 = 0 \)[/tex].
Factoring the quadratic equation, we get:
[tex]\[ 160x - 16x^2 = 0 \][/tex]
[tex]\[ 16x(10 - x) = 0 \][/tex]
Setting each factor equal to zero gives us:
[tex]\[ 16x = 0 \quad \text{or} \quad 10 - x = 0 \][/tex]
This results in:
[tex]\[ x = 0 \quad \text{or} \quad x = 10 \][/tex]
Thus, the water will also hit the ground again at [tex]\( x = 10 \)[/tex] feet away from the sprinkler head.
So, to summarize:
- [tex]\( h(2) = 256 \)[/tex]
- After 2 seconds, the height of the spray is [tex]\( 256 \)[/tex] inches.
- The spray reaches its maximum height 5 feet away from the sprinkler head.
- The maximum height of the water spray is [tex]\( 400 \)[/tex] inches.
- The water will hit the ground again at 10 feet away from the sprinkler head.
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