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Svetllana [295]
2 years ago
7

Help pleaseeeeeeeeeeeeeeeeeeeeeee

Mathematics
1 answer:
bixtya [17]2 years ago
3 0

Answer:  \bold{(1)\ \dfrac{19,683}{64}\qquad (2)\ 16}

<u>Step-by-step explanation:</u>

(1)           (12, 18, 27, ...)

The common ratio is:

r=\dfrac{a_{n+1}}{a_n}\quad r =\dfrac{18}{12}=\boxed{\dfrac{3}{2}}\quad \rightarrow \quad r=\dfrac{27}{18}=\boxed{\dfrac{3}{2}}

The equation is:

a_n=a_o(r)^{n-1}\\\\Given:a_o=12,\  r=\dfrac{3}{2}\\\\\\Equation:\\a_n =12\bigg(\dfrac{3}{2}\bigg)^{n-1}\\\\\\\\9th\ term:\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{9-1}\\\\\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{8}\\\\\\.\quad =\large\boxed{\dfrac{19643}{64}}

(2)\qquad \bigg(\dfrac{1}{16},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{2}\bigg)\\\\\\\text{The common ratio is}:\\\\r=\dfrac{a_{n+1}}{a_n}\quad  r=\dfrac{\frac{1}{8}}{\frac{1}{16}}=\boxed{2}\quad \rightarrow \quad r=\dfrac{\frac{1}{4}}{\frac{1}{8}}=\boxed{2}

The equation is:

a_n=a_o(r)^{n-1}\\\\Given:a_o=\dfrac{1}{16},\  r=2\\\\\\Equation:\\a_n =\dfrac{1}{16}(2)^{n-1}\\\\\\\\9th\ term:\\a_9=\dfrac{1}{16}(2)^{9-1}\\\\\\a_9=\dfrac{1}{16}(2)^{8}\\\\\\.\quad =\large\boxed{16}

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Answer:

The cubic function is f(x) = (27/32)·x  - 3/32·x³ - -9/32·x² - 9/32

Step-by-step explanation:

The given function is f(x) = a·x³ + b·x² + c·x + d

By differentiation, we have;

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Subtracting equation (1) from equation (2) gives;

28·a - 8·b + 4·c = 3

Therefore, we have;

27·a - 6·b + c = 0

3·a + 2·b + c = 0

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Solving the system of equations using an Wolfram Alpha gives;

a = -3/32, b = -9/32, c = 27/32 from which we have;

a + b + c + d = 0 3 × (-3/32) + 2 × (-9/32) + (27/32) + d = 0

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The cubic function is therefore f(x) = (-3/32)·x³ + (-9/32)·x² + (27/32)·x + (-9/32).

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