Your new flashlight uses a parabolic mirror which can be modeled by the equation (x – 2)² = 5(y + 1), where x and y are measured
in centimeters. You need to place a new light bulb in your flashlight. How far away from the vertex of the parabolic mirror should you place the bulb to ensure a perfect beam of light?
2 answers:
The given equation is
(x - 2)² = 5(y + 1).
The given equation for the parabola is in the standard form
(x - h)² = 4p(y - k)
where
h = 2
4p = 5, so that p = 5/4
k = -1
The vertex is at
(h,k) or (2, -1)
The focus is located at
(h, k + p) or (2, -1 + 5/4) = (2, 1/4)
We should place the bulb at p = 5/4 from the vertex.
Answer: 1 1/4 or 1.25 cm
(x - 2)² = 5(y + 1).
The given equation for the parabola is in the standard form
(x - h)² = 4p(y - k)
where
h = 2
4p = 5, so that p = 5/4
k = -1
The vertex is at
(h,k) or (2, -1)
The focus is located at
(h, k + p) or (2, -1 + 5/4) = (2, 1/4)
We should place the bulb at p = 5/4 from the vertex.
Answer: 1 1/4 or 1.25 cm
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Answer:
a.) y = 3/(2ˣ)
b.) y = 1/(2ˣ)
c.) y = (π^π)ˣ
d.) y = (1/27)(1/√(3))ˣ
e.) y = .002908/(.119025ˣ)
f.) y = .00000004808/(.0413ˣ)
Step-by-step explanation:
Concept need to know is:
- a negative exponent will flip the numerator with the denominator
- a fraction as an exponent is just a root. so if the exponent is x^(1/2) then the root is 2 and x^(1/3) is a cube root
- adding and subtracting exponent is the same thing as multiplying the same base.
- so x^(1+2) = (x^1)(x^2)
- and x^(1-2) = (x^1)(x^-2)
