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2 years ago

9

A garden at the local library is in the shape of a right triangle the dimensions of the garden are shown in the diagram. What is

the area in the square feet Area of triangle = 1/2 bh

1 answer:

Sladkaya [172]2 years ago

3 0

37.5 square feet

Area=1/2 b*h=1/2 * 7.5 * 10= 1/2 * 75 = 37.5

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The sum of two numbers is 51. The larger number is 17 more than the smaller number. What are the numbers ?

Serggg [28]

Answer:

The numbers are 17 and 34.

Step-by-step explanation:

I just subtracted 17 from 51, it equals 34, and 34 is 17 more than 17. Add 34 and 17 and you get 51.

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2 years ago

Each saddle has exactly two saddle bags on it

neonofarm [45]

So we have 22 riders....each carrying 2 saddle bags.....thats (2 * 22) = 44 bags.....each saddle bag has 3 waters....(44 * 3) = 132 total water bottles

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2 years ago

5/10 = /6 is 5/10 and 3/6 equivalent fraction. if so, please explain

Sonja [21]

$\frac{5}{10}= \frac{1\cdot5}{2\cdot5} = \frac{1}{2} \ \ \ and\ \ \ \frac{3}{6}= \frac{1\cdot3}{2\cdot3} = \frac{1}{2} \ \ \ \ \Rightarrow\ \ \ \frac{5}{10}=\frac{3}{6}$

8 0

2 years ago

A commercial property purchased for 1.2 million is revalued at 995,000. Calculate the percentage decrease in the property’s valu

soldi70 [24.7K]

Answer:

17%

Step-by-step explanation:

First: work out the difference (decrease) between the two numbers you are comparing. Then: divide the decrease by the original number and multiply the answer by 100.

First:

1,200,000 - 995,000 = 205,000

Then:

205,000 / 1,200,000 = 0.17

0.17 × 100% = 17%

4 0

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:

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

So a_n is finite, so it converges.

Similarly b_n converges according to p-test.

P-test:

General form:

$\sum_{n=1}^{inf}\frac{1}{n^p}$

if p>1 then series converges. In oue case we have:

$\sum_{n=1}^{inf}b_n=\frac{1}{n^4}$

p=4 >1, so b_n also converges.

According to comparison test if both series converges, the final series also converges.

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

5 0

3 years ago