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Bas_tet [7]
3 years ago
15

If you could only plot 5 points to graph y=sinx, which points would you choose? How would you connect these points? Please expla

in
If you could only plot 5 points to graph y = cosx, which points would you choose? How would you connect these points? Please explain this too!
Mathematics
1 answer:
fenix001 [56]3 years ago
8 0

Answer:

For both graph I would choose the following points: 0°, 90°, 180°, 270° and 360°. I would connect the points (in both cases) with an smooth arc.

Step-by-step explanation:

x          sin(x)     cos(x)

0°        0            1

90°      1             0

180°     0           -1

270°    -1            0

360°    0            1

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Write the slope intercept form of the equation of the line described.
Lady bird [3.3K]

Answer:

y= -5

Step-by-step explanation:

x=0 is a vertical graph hence the perpendicular graph would be a horizontal line, a y= ____ graph.

Since it passes through the point (-2, -5),

the equation of the line is y= -5.

This would be in slope- intercept form since the gradient of horizontal lines is zero.

y= mx +c, where m is the gradient and c is the y-intercept.

Given that gradient =0, m=0

y= 0x +c

when x= -2, y= -5,

-5= 0(-2) +c

-5= c

c= -5

Thus the equation is y= -5.

6 0
3 years ago
Can someone please help me
pentagon [3]
The first one is 8 and here's a graph. the second one is 5 again here's a graph

7 0
3 years ago
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Rewrite each number sentence using numerals and symbols. Find the difference between 4 and 3 and then add 7 *
Zina [86]

Answer:

8

Step-by-step explanation:

4-3+7=8

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3 years ago
WILL MARK BRANLIEST! find of the volume of the prism round to the nearest tenth if necessary
vodomira [7]

Answer:

The answer is A. 204.7

Step-by-step explanation:

5 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
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