Based on the given inequality of y≥ x - 1, the graph that matches it best is C. Graph A.
<h3>How are inequalities graphed?</h3>
When graphing an inequality with the "greater or equal" and "less or equal" signs, the line will be a steady line as shown by graphs A and B.
If y is the larger or equal variable, the shaded area will be to the left of the line.
The inequality is y≥ x - 1 which means that the line should be steady and the shading should be on the left of the line. Graph A is therefore the correct graph.
Find out more on graphing inequalities at brainly.com/question/11234618.
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Answer:
it is equivalent
Step-by-step explanation:
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:
1,75
Step-by-step explanation:
The Mosteller formula is ![[tex]M=\sqrt{\frac{(172.72)(63.95)}{3600}} = 1.75](https://tex.z-dn.net/?f=%5Btex%5DM%3D%5Csqrt%7B%5Cfrac%7B%28172.72%29%2863.95%29%7D%7B3600%7D%7D%20%3D%201.75)
[/tex]
We know that height is 68 inches and weight 141 pounds
We must remember that 1 inches = 2.54 cm and 1 pound 0.453 kg
This means 68 inches = 172.72 cm and 141 pounds = 63.95
We substitute on the formula
Answer:
See below, please
Step-by-step explanation:
Hence
