Highest common factors of 64 and 40
1 answer:
You might be interested in
Answer:
49 ft
Step-by-step explanation:
h(t) = 56t - 16t²
This is the equation of a parabola.
We must solve the equation to find the time (t) when the ball reaches its maximum height (h).
The coefficient of t² is negative, so the parabola opens downward, and the vertex is a maximum.
One way to solve this problem is to convert the equation to the vertex form.
We do that by completing the square.
Calculation:
h = -16t² + 56t
Divide both sides by -16 to make the coefficient of t² equal to 1.
(-1/16)h = t² - ⁷/₂t
Square half the coefficient of t
(-7/4)² = 49/16
Add and subtract it on the right-hand side
(-1/16)h = t² - ⁷/₂t + 49/16 - 49/16
Write the first three terms as the square of a binomial
(-1/16)h = (t - ⁷/₄)² - 49/16
Multiply both sides by -16
h = -16(t - ⁷/₄)² + 49
You have converted your equation to the vertex form of a parabola:
y = a(t - h)² + k = 0,
where (h, k) is the vertex.
h = ⁷/₄ and k = 49, so the vertex is at (⁷/₄, 49).
The time to reach maximum height is ⁷/₄ s = 1.75 s.
The graph below shows that the ball reaches a maximum height of 49 ft after 1.75 s.
Answer:
The values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001 is 0 < x < 0.3936.
Step-by-step explanation:
Note: This question is not complete. The complete question is therefore provided before answering the question as follows:
Determine the values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001. f(x) = e^x ≈ 1 + x + x²/2! + x³/3!, x < 0
The explanation of the answer is now provided as follows:
Given:
f(x) = e^x ≈ 1 + x + x²/2! + x³/3!, x < 0 …………….. (1)
= (x) = (e^z /4!)x^4
Since the aim is (x) < 0.001, this implies that:
(e^z /4!)x^4 < 0.0001 ………………………………….. (2)
Multiply both sided of equation (2) by (1), we have:
e^4x^4 < 0.024 ……………………….......……………. (4)
Taking 4th root of both sided of equation (4), we have:
|xe^(z/4) < 0.3936 ……………………..........…………(5)
Dividing both sides of equation (5) by e^(z/4) gives us:
|x| < 0.3936 / e^(z/4) ……………….................…… (6)
In equation (6), when z > 0, e^(z/4) > 1. Therefore, we have:
|x| < 0.3936 -----> 0 < x < 0.3936
Therefore, the values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001 is 0 < x < 0.3936.