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jonny [76]
2 years ago
4

How does the area of triangle ABC compare to the area of parallelogram GHJK? The area of △ABC is 2 square units greater than the

area of parallelogram GHJK. The area of △ABC is 1 square unit greater than the area of parallelogram GHJK. The area of △ABC is equal to the area of parallelogram GHJK. The area of △ABC is 1 square unit less than the area of parallelogram GHJK.
Mathematics
2 answers:
kotegsom [21]2 years ago
8 0

Answer:

A

Step-by-step explanation:

Snowcat [4.5K]2 years ago
4 0
Area ( GHJK ) = 2  x ( 4 x 2 ) / 2 = 8 square units
Area ( Δ ABC ) =  ( 4 x 6 ) - ( 4 x 4 / 2 ) - ( 1 x 6 / 2 ) - ( 3 x 2 / 2 ) =
= 24 - 8 - 3 - 3 = 34 - 14 = 10 square units
10 - 8 = 2 square units
Answer:
A ) The area of Δ ABC is 2 units greater than area of parallelogram GHJK.

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And when we apply the limit we got that:

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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"

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

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

We can cancel n^2 and we got

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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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