Answer:
Explanation:
Given that,
A system can detect objects as small as 19.1 cm i.e. 0.191 m. It is the wavelength.
We know that,
Frequency,
So,
So, the frequency of such a system is equal to.
HELP !! Maura is deciding which hose to use to water her outdoor plants. Maura noticed that the water coming out of her garden h
oses follows a parabolic path.
When Maura uses her green garden hose the greatest height the water reaches is 8 feet, and it lands on the plants 10 feet from where she is standing. Both the nozzle of the hose and the top of the flowers are 4 feet above the ground.
Maura has determined that when she has the water on full blast, the water from the red hose follows the path y= -(x-3)^2+7.
a. Which hose will throw the water higher ?
b. Write an equation that models the path of the water from Maura's green hose.
c. What domain and range make sense for the water from Majra's green hose?
When Maura uses her green garden hose the greatest height the water reaches is 8 feet, and it lands on the plants 10 feet from where she is standing. Both the nozzle of the hose and the top of the flowers are 4 feet above the ground.
Maura has determined that when she has the water on full blast, the water from the red hose follows the path y= -(x-3)^2+7.
a. Which hose will throw the water higher ?
b. Write an equation that models the path of the water from Maura's green hose.
c. What domain and range make sense for the water from Majra's green hose?
1 answer:
THE GREEN HOSE:
Define the (x,y) coordinate at a height of 4 feet up from the ground to match where Majra is holding the green hose.
This means that the equation for the green hose is of the form
y = a(x - h)² + 4 (1)
Because water from the green hose lands on the ground 10 feet from where Majra is standing, therefore
y(10) = -4 (2)
Because the curve passes through (0,0), therefore
ah² + 4 = 0
ah² = - 4 (3)
To satisfy (2), obtain
a(10 - h)² + 4 = -4
a(10 - h)² = - 8 (4)
Divide (3) by (4).
h²/(10-h)² = 1/2
2h² = (10 - h)² = 100 - 20h + h²
h² + 20h - 100 = 0
Solve with the quadratic formula.
x = 0.5[-20 +/- √(8400)] = 4.142, - 24.142
Reject the negative solution.
The vertex is at (4.142, 4).
From (3), obtain
a = -4/4.142² = -0.2332
The equation for the green hose is
y = 0.2332(x - 4.142)² + 4
THE RED HOSE
The red hose has a vertex at (3,7), according to the equation y = -(x-3)² + 7.
A graph of y(x) for both hoses is shown in the attached figure.
Answers:
a. The red hose will throw the water higher.
b. The equation for the green hose is
y = -0.2332(x - 4.124)² + 4,
with the origin at a height of 4 feet above ground level.
c. The domain for the green hose that makes sense is 0 ≤ x ≤ 10 feet.
The corresponding range is -4 ≤ y ≤ 4 feet.
Define the (x,y) coordinate at a height of 4 feet up from the ground to match where Majra is holding the green hose.
This means that the equation for the green hose is of the form
y = a(x - h)² + 4 (1)
Because water from the green hose lands on the ground 10 feet from where Majra is standing, therefore
y(10) = -4 (2)
Because the curve passes through (0,0), therefore
ah² + 4 = 0
ah² = - 4 (3)
To satisfy (2), obtain
a(10 - h)² + 4 = -4
a(10 - h)² = - 8 (4)
Divide (3) by (4).
h²/(10-h)² = 1/2
2h² = (10 - h)² = 100 - 20h + h²
h² + 20h - 100 = 0
Solve with the quadratic formula.
x = 0.5[-20 +/- √(8400)] = 4.142, - 24.142
Reject the negative solution.
The vertex is at (4.142, 4).
From (3), obtain
a = -4/4.142² = -0.2332
The equation for the green hose is
y = 0.2332(x - 4.142)² + 4
THE RED HOSE
The red hose has a vertex at (3,7), according to the equation y = -(x-3)² + 7.
A graph of y(x) for both hoses is shown in the attached figure.
Answers:
a. The red hose will throw the water higher.
b. The equation for the green hose is
y = -0.2332(x - 4.124)² + 4,
with the origin at a height of 4 feet above ground level.
c. The domain for the green hose that makes sense is 0 ≤ x ≤ 10 feet.
The corresponding range is -4 ≤ y ≤ 4 feet.
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