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3 years ago

The radius of a semicircle is 1 meter. What is the semicircle's area?

Mathematics

1 answer:

Stells [14]3 years ago

6 0

Answer:

Step-by-step explanation:

$A -of -circle= \pi*r^{2}\\ Note : A -semicircle- is -half -of- a- circle\\Area = \frac{ \pi*r^{2}}{2} \\r = 1\\A =\frac{3.14*1^{2}}{2}\\A = \frac{3.14}{2} \\A = 1.57$

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3.) Simplify the expression by combining like

Evgesh-ka [11]

$45s^2-19-s^2+16\\=45s^2-s^2-19+16\\=s^2(45-1)-19+16\\=44s^2+(-19+16)\\=44s^2-3$

Hope that helps and hope you get better!

3 0

2 years ago

A department store buys 400 shirts at a cost of $7 200 and sells them at a selling

noname [10]

Answer:

down below.

Step-by-step explanation:

20 × 400 = 8000

8000 - 7200 = 800

therefore the % mark up is 10%, because 10% of 8000 is 800.

800 + 7200 = 8000

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3 years ago

Percent of students with brown eys

Over [174]

Answer:

between 55 to 79 percent

6 0

2 years ago

Compare the triangles and determine whether they can be proven congruent by SSS, SAS, ASA, AAS, or HL. If not, type "NONE".

Komok [63]

Answer:

SAS

Step-by-step explanation:

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3 years ago

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:

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