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makkiz [27]
3 years ago
14

How do you do 4 divided by 0.64 in standard algorithm

Mathematics
1 answer:
Marina CMI [18]3 years ago
7 0
You take .64 turn it into 64,64÷4=16=0.16 I confusing
You might be interested in
What is the slope-intercept form of the line with the point (0, 3) and a slope = -2?
adelina 88 [10]

Answer:

2nd choice OC.y=-2x+3

Step-by-step explanation:

The equation for the slope-intercept form is y=mx+b, where m is the slope and b is the y-intercept. With these two numbers both given to us, we can just plug them in and get y=-2x+3.

Hope this helps you

4 0
3 years ago
PLEASE HELP ME OUT
liubo4ka [24]

Answer:

18.84 ft

Step-by-step explanation:

The formula for the circumference of a circle is C = πd or C = 2πr

We will use C = 2πr since we have r given in the photo

  • We just need to substitute our values into the formula:
  • C = 2(3.14)(3) = 18.84 ft
8 0
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}

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
I need help please someone help i just joined so please help​
mr Goodwill [35]

Answer:

Step-by-step explanation:

4 0
4 years ago
PLZ HELP ME I NEED HELP AND FAST ⚠️
Liula [17]

Answer:

Yor awnser is C and D.

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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