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makvit [3.9K]
3 years ago
8

The perimeter of a rectangle is 176 m. The length is 4 m less than three times the width. Find the dimensions of the rectangle.

Mathematics
1 answer:
lutik1710 [3]3 years ago
5 0

Answer:

Length = 65 m

Width = 23 m

Step-by-step explanation:

Width = w

Length = 3w - 4

Perimeter of rectangle = 176 m

2*(length + width) = 176

2*( 3w - 4 + w ) = 176

2* ( 4w - 4) = 176

2*4w - 2*4 = 176

      8w - 8 = 176    { add 8 to both sides}

           8w = 176 +8

           8w = 184

        8w/8 = 184/8

              w = 23 m

Length = 3w - 4

            = 3*23 - 4

            = 69 - 4

Length = 65 m

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Answer:

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Step-by-step explanation:

(x-2/3)(x-2/3)

= x^2 -2/3x -2/3x + (2/3)^2

= x^2 -4/3x + 4/9

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6 0
3 years ago
Please help me on step 4
Zigmanuir [339]

The beginning part of the question is missing and it says;

The measures of center can be used to summarize the number of pets that students own. Cameron asked eight of his classmates how many pets they own. The results are listed below.

1, 0, 2, 0, 3, 7, 0, 2, 4

Answer:

Median

Step-by-step explanation:

Let's rearrange the results from least to most.

0, 0, 0, 1, 2, 2, 3, 4, 7

Mode is the most occurring number which in this case is zero, so Cameron can't use that to convince his parents to get another pet.

Median is middle number and in this case is 2.

Mean is; sum of results/ number of results = 19/9 = 2.11

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7 0
3 years ago
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
Marina86 [1]

Answer:

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

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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"

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

3 0
3 years ago
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Answer:

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