C) (7,7), (-5,4)
I hope it helps
Step-by-step explanation:
(a) ∫₋ₒₒ°° f(x) dx
We can split this into three integrals:
= ∫₋ₒₒ⁻¹ f(x) dx + ∫₋₁¹ f(x) dx + ∫₁°° f(x) dx
Since the function is even (symmetrical about the y-axis), we can further simplify this as:
= ∫₋₁¹ f(x) dx + 2 ∫₁°° f(x) dx
The first integral is finite, so it converges.
For the second integral, we can use comparison test.
g(x) = e^(-½ x) is greater than f(x) = e^(-½ x²) for all x greater than 1.
We can show that g(x) converges:
∫₁°° e^(-½ x) dx = -2 e^(-½ x) |₁°° = -2 e^(-∞) − -2 e^(-½) = 0 + 2e^(-½).
Therefore, the smaller function f(x) also converges.
(b) The width of the intervals is:
Δx = (3 − -3) / 6 = 1
Evaluating the function at the beginning and end of each interval:
f(-3) = e^(-9/2)
f(-2) = e^(-2)
f(-1) = e^(-1/2)
f(0) = 1
f(1) = e^(-1/2)
f(2) = e^(-2)
f(3) = e^(-9/2)
Apply Simpson's rule:
S = Δx/3 [f(-3) + 4f(-2) + 2f(-1) + 4f(0) + 2f(1) + 4f(2) + f(3)]
S ≈ 2.5103
Answer:
-4/3, -1, 0, 2/3, 4/5, 6, 9. They were already in the correct order.
Step-by-step explanation:
I hope this helped and if it did I would appreciate it if you marked me Brainliest. Thank you and have a nice day!
Answer:
Yes, (6, -2) is a solution to the given system of equations.
Step-by-step explanation:
Please write y = –1/6 x − 1 y = 1/6 x − 3 as follows, for greater clarity:
y = (–1/6)x − 1
y = (1/6)x − 3
Let's actually solve this system:
y = (–1/6)x − 1
y = (1/6)x − 3
-----------------------
2y = -4, or y = -2
Now find x. Arbitrarily we choose to use the first equation for this purpose:
y = (-1/6)x - 1. We set y = -2 and find x: -2 = (-1/6)x - 1
Combining the constants, we get -1 = (-1/6)x, or 6 = x
Yes, (6, -2) is a solution to the given system of equations.
Its not showing any thing to answer sorry
