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3 years ago

The base of a parallelogram is 10 inches the height is 2 inches more than half the base find the area

1 answer:

8 0

Given is a parallelogram whose base is b = 10 inches and it says height is 2 more than half-value of base i.e. h = 2 + (b ÷ 2).

So height is h = 2 + (10 ÷ 2) = 2 + 5

⇒ h = 7 inches.

We know the formula for area of parallelogram is given as follows :-

Area of parallelogram = base x height

Area of parallelogram = 10 inches x 7 inches

Area of parallelogram = 70 squared inches.

Hence, the final answer is 70 squared inches.

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The perimeter of a rectangle is 338, if the ratio of the length to the width is 8:5, find the length

STALIN [3.7K]

Let the length be 8x and the width = 5x, then
Perimeter = 2(length + width) = 2(8x + 5x) = 2(13x) = 26x
338 = 26x
x = 338/26 = 13

Therefore, length = 8x = 8 x 13 = 104

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3 years ago

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dimaraw [331]

Answer:

Step-by-step explanation:

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2 years ago

Question down below

Bas_tet [7]

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3 years ago

6²+(-4)<br> How do u get a answer

kiruha [24]

So 6^2 would be 36 because it is 6 times 6 and then -4 so 36-4=32

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3 years ago

Read 2 more answers

Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.

Aleks [24]

Answer:

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

Step-by-step explanation:

The actual Series is::

$\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$

The method we are going to use is comparison method:

According to comparison method, we have:

$\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n$

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

$=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}$

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}$

Now:

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}$