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DonovanBaily42
08.10.2019 •
Mathematics
Origami is the japanese art of paper folding. the heart shown on the lesson 6 assessment page was created using origami. a classmate classifies this polygon as a hexagon. is this classmate correct? how would you justify your response? half the heart would be a hexagon, but when it is opened up as in the heart shown in my lesson it has more sides. !
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Ответ:
Ответ:
j(x) = 2x² - 5x - 5
General Formulas and Concepts:
Calculus
Intermediate Value Theorem - If f is continuous on [a, b] and k is between f(a) and f(b), then there exists a c∈[a, b] such that f(c) = kStep-by-step explanation:
Step 1: Define functions
f(x) = 2x² + x - 3
g(x) = -4x² + 3x - 4
h(x) = x³ - 2x² - x + 4
j(x) = 2x² - 5x - 5
Step 2: Check
Check if each function is continuous over the interval [3, 4].
1. f(x) = 2x² + x - 3
(a) f is cont on [3, 4]? Yes
2. g(x) = -4x² + 3x - 4
(a) f is cont on [3, 4]? Yes
3. h(x) = x³ - 2x² - x + 4
(a) f is cont on [3, 4]? Yes
4. j(x) = 2x² - 5x - 5
(a) f is cont on [3, 4]? Yes
Step 3: IVT
Find the zeros on the functions in the interval [3, 4].
1. f(x) = 2x² + x - 3
(b) f(3) = 18 < k = 0 < f(4) = 33? No
2. g(x) = -4x² + 3x - 4
(b) f(3) = -31 < k = 0 < f(4) = -56? No
3. h(x) = x³ - 2x² - x + 4
(b) f(3) = 10 < k = 0 < f(4) = 32? No
4. j(x) = 2x² - 5x - 5
(b) f(3) = -2 < k = 0 < f(4) = 7? Yes
Step 4: Justify IVT Function
j(x) = 2x² - 5x - 5
(a) f is cont on [3, 4]? Yes
(b) f(3) = -2 < k = 0 < f(4) = 7? Yes
∴ by IVT, there exists a c∈[3, 4] such that f(c) = 0.