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jonny [76]
2 years ago
7

On a coordinate plane, a curved line with a minimum value of (1.5, negative 1) and a maximum value of (negative 1.5, 13), crosse

s the x-axis at (negative 3, 0), (1, 0), and (2, 0), and crosses the y-axis at (0, 6).
Which lists all of the x-intercepts of the graphed function?
(0,6)
(1,0)(2,0)
(1,0)(2,0) and (-3,0)
(1,0)(2,0)(-3,0) and (0,6)
Mathematics
2 answers:
Elan Coil [88]2 years ago
8 0
Its not a or B because it has to be a negative ima go with C and if its not C it should be D
Olenka [21]2 years ago
8 0

Answer:

c

Step-by-step explanation:

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The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit in
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\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})

And when we apply the limit we got that:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1

Step-by-step explanation:

Assuming this complete problem: "The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit . 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2"

We have the following formula in order to find the sum of cubes:

\lim_{n\to\infty} \sum_{n=1}^{\infty} i^3

We can express this formula like this:

\lim_{n\to\infty} \sum_{n=1}^{\infty}i^3 =\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2

And using this property we need to proof that: 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2

\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2

If we operate and we take out the 1/4 as a factor we got this:

\lim_{n\to\infty} \frac{n^2(n+1)^2}{n^4}

We can cancel n^2 and we got

\lim_{n\to\infty} \frac{(n+1)^2}{n^2}

We can reorder the terms like this:

\lim_{n\to\infty} (\frac{n+1}{n})^2

We can do some algebra and we got:

\lim_{n\to\infty} (1+\frac{1}{n})^2

We can solve the square and we got:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})

And when we apply the limit we got that:

\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1

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