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3 years ago

Please help, I'm so close to finishing all these.

Mathematics

1 answer:

Vadim26 [7]3 years ago

6 0

Answer:

d. 43°

Step-by-step explanation:

KM is a bisector of ∠LKN, so ...

∠LKN = 2(∠4)

7q +2 = 2(4q -5)

12 = q . . . . . . . . . . . add 10-7q, simplify

Now, we can put 12 where q is in the expression for ∠4:

∠3 = ∠4 = 4·12 -5 = 43

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You bought a large container of fruit punch. The label on the container says there are 128fluid ounces of fruit punch in the con

Alenkasestr [34]

Answer:

16 cups.

Step-by-step explanation:

We have been given that there are 128 fluid ounces of fruit punch in the container. We are asked to find the cups will be in 128 oz.

We know that 1 cup equals 8 fluid oz. To convert 128 oz into cups, we will divide 128 by 8.

$\text{Cups in 128 oz}=\frac{128\text{ oz}}{\frac{8\text{ oz}}{\text{1 cup}}}$

$\text{Cups in 128 oz}=\frac{128\text{ oz}}{8\text{ oz}}\times \text{1 cup}$

$\text{Cups in 128 oz}=16\times \text{1 cup}$

$\text{Cups in 128 oz}=16\text{ cups}$

Therefore, there are 16 cups in the container.

5 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

3 years ago

What s the length of the arc for the half circle

Andrej [43]

Answer`Step-by-step explanation:

The length of an arc depends on the radius of a circle and the central angle Θ. We know that for the angle equal to 360 degrees (2π), the arc length is equal to circumference. Hence, as the proportion between angle and arc length is constant, we can say that:

Hope that help, at least a bit.

3 0

3 years ago

Can you please help and I need help fast

KonstantinChe [14]

Answer:

Step-by-step explanation:

He borrowed money so he has -5 dollars and but he payed 2 back, so now he needs to pay $3

6 0

2 years ago

Read 2 more answers

The gray area is the sidewalk. the area of the sidewalk is ___________ square units.

Bas_tet [7]

The area of the sidewalk is 72 square units. Solution: First, multiply 10 by 12 which equals 120. Then, subtract 6x8=48. So, therefore, 120-48= 72

5 0

3 years ago