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2 years ago

A group of students each are given a certain number of green and yellow marbles

Mathematics

1 answer:

allochka39001 [22]2 years ago

6 0

Answer: A,d,f

Step-by-step explanation:

Because you have to simplify the ratios of the colors.

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Please help. 20 points and brainliest.

julia-pushkina [17]

Answer:

minimum -45, maximum 32

Step-by-step explanation:

C=4x-3y

x≥0, y≥4, x+y≤15

Maximum value of C can be achieved at max x and min y

min y=4 => max x=15-4=11

C=4*11-3*4= 32

Minimum value of C can be achieved at min x and max y

min x=0, max y =15

C=4*0-3*15= -45

So answer is minimum -45, maximum 32

7 0

2 years ago

Water lilies are often used to decorate ponds, as shown in the photo. But they are also famous for their unusual growth pattern!

Alex17521 [72]

There really isn't a right or wrong answer since this is a prediction but to solve this look at the graph and try to see the pattern

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2 years ago

Read 2 more answers

If the area of a circle is 576 pie squared what is the circumference

kenny6666 [7]

576^2 = 24 ÷ 3 = 8 × 2r = 8 × 6 = 48

7 0

2 years ago

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-5/32a ÷ 7/20a reduce to the lowest terms

Virty [35]

The answer is -25/56

4 0

2 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

2 years ago

A group of students each are given a certain number of green and yellow marbles​

A group of students each are given a certain number of green and yellow marbles