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shusha [124]
3 years ago
13

How many digits are there in the repeating block for the decimal equivalent of 3 over 7?

Mathematics
1 answer:
Sindrei [870]3 years ago
7 0

Step-by-step explanation:

\text{By long division, we know that}

\frac37=0.425871425871425871\ldots

\text{Observing the pattern, we know the repeating block is }\overline{425871}\text{. Therefore, there are 6 digits in the repeating block}

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Find vertex of f(x)=(x+3)(x-5)
astra-53 [7]

The x coordinate of the vertex will be the average of the two zeros, here -3 and 5, so x=(-3+5)/2 = 1, f(1)=(1+3)(1-5) = -16.

Answer: (1, -16)

Let's do it some other ways.  How about completing the square to turn f in to vertex form?

f(x) = (x+3)(x-5) = x² - 2x - 15 = (x² - 2x + 1) - 1 - 15 = (x-1)² - 16

and now we can read off (1, -16) as the vertex.

The other method is the vertex is x= - b/2a =  - (-2)/2(1) = 1.

Three methods, same answer. Good.

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san4es73 [151]

Step-by-step explanation:

Given: f'(x) = x^2e^{2x^3} and f(0) = 0

We can solve for f(x) by writing

\displaystyle f(x) = \int f'(x)dx=\int x^2e^{2x^3}dx

Let u = 2x^3

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Then

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Therefore,

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The answer is 18^8 because u add exponents

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Answer:

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