Through how many radians does the minute hand of a clock turn in three hours?

3 pi
9 pi
6 pi

Let

[tex]\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots[/tex]

Differentiating twice gives

[tex]\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots[/tex]

[tex]\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n[/tex]

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

[tex]\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0[/tex]

[tex]\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0[/tex]

Then the coefficients in the power series solution are governed by the recurrence relation,

[tex]\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}[/tex]

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

[tex]k=0 \implies n=0 \implies a_0 = a_0[/tex]

[tex]k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}[/tex]

[tex]k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}[/tex]

[tex]k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}[/tex]

It should be easy enough to see that

[tex]a_{n=2k} = \dfrac{a_0}{(2k)!}[/tex]

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

[tex]k = 0 \implies n=1 \implies a_1 = a_1[/tex]

[tex]k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}[/tex]

[tex]k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}[/tex]

[tex]k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}[/tex]

so that

[tex]a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}[/tex]

So, the overall series solution is

[tex]\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)[/tex]

[tex]\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}[/tex]

If the given equation is

[tex]27^{\frac13} = 9^{14} + 3^{x + 1}[/tex]

first simplify the left side using the fact that 27 = 3³:

[tex]27^{\frac13} = (3^3)^{\frac13} = 3^{\frac33} = 3^1 = 3[/tex]

and on the right side, 9 = 3²:

[tex]9^{14} = (3^2)^{14} = 3^{2\times14} = 3^{28}[/tex]

So we have

[tex]3 = 3^{28} + 3^{x + 1}[/tex]

Next,

[tex]3^{x+1} = 3^x \times 3^1 = 3 \times 3^x[/tex]

so that

[tex]3 = 3^{28} + 3 \cdot 3^x[/tex]

and dividing both side by 3 gives

[tex]1 = 3^{27} + 3^x[/tex]

Isolate 3ˣ :

[tex]3^x = 1 - 3^{27}[/tex]

Solve for x by taking the base-3 logarithm of both sides:

[tex]\log_3(3^x) = \log_3(1-3^{27})[/tex]

[tex]x \log_3(3) = \log_3(1-3^{27})[/tex]

[tex]\boxed{x = \log_3(1-3^{27})}[/tex]

Answer:

No

No

Yes

Yes

No

Step-by-step explanation:

In this case we are looking at substitution, so any where you find x put three(3) in replace of it.

And remember the answer should be 36,

I find it difficult to type the answer so I will take a picture instead.

Answer:

LJ< JK<LK

Step-by-step explanation:

for a theorem, the order of the side of the triangle is the same as the order of the opposite angles since for the angles it's K<L<J, you replace each with the opposite side (quick trick, it's the one not containing the letter!)

Answer:

Answer# 1: A. 26

Answer# 2: A. 25

1. What is m + 5n where m = 1 and n = 5?

Substitute 1 for "m" and 5 for "n":

[tex]\sf (1)+5(5)[/tex]

Multiply 5*5 =25

[tex](1)+25[/tex]

Add :

[tex]\sf 26[/tex]

____________________

2. What is 5m + 5n where m = 10 and n = -5?

Substitute 10 for "m" and -5 for "n".

[tex]\sf 5\left(10\right)\:+\:5\left(-5\right)[/tex]

Multiply 5 * 10= 50

[tex]\sf 50+5\left(-5\right)[/tex]

Multiply 5 * -5 = -25

[tex]\sf 50-25[/tex]

Subtract:

[tex]\sf 25[/tex]

___________________________________

Answer:

wa

at you ım prenses

Step-by-step explanation:

you ind soluk

Answer: the answer is in the attachment.

Answer:

Hope this helps :)

Step-by-step explanation:

Answer:

0.00706666666

Step-by-step explanation:

Using the probability concept, it is found that there is a [tex]\mathbf{\frac{1}{k}}[/tex] probability that “Relaxation Through Mathematics” is first to review.

A probability is given by the number of desired outcomes divided by the number of total outcomes.

In this problem:

The probability is:

[tex]p = \frac{D}{T} = \frac{1}{k}[/tex]

A similar problem is given at https://brainly.com/question/24483829

Answer:

about 20 cents

Step-by-step explanation:

#Shop A

Pack of 70pens:-

[tex]\\ \sf\longmapsto 7(2.30)=16.10[/tex]

#2

Pack of 70

[tex]\\ \sf\longmapsto 14(1.60)=20.80[/tex]

But you get half price

Difference:-

[tex]\\ \sf\longmapsto 16.10-10.40=5.70\£[/tex]

The measures of the angles and the lengths of the sides of the triangle can

be found using sine rule.

The correct responses are;

Reasons:

The given parameter are;

∠B = 65°20', b = 5.38 inches, c = 4.88 inches

∠C ≥ 90°

By sine rule, we have;

[tex]\dfrac{5.38}{sin(65^{\circ} 20')} = \mathbf{ \dfrac{4.88}{sin(\angle C)}}[/tex]

[tex]sin(\angle C) = \mathbf{ \dfrac{sin(65^{\circ} 20') \times 4.88}{5.38}}[/tex]

[tex]\angle C = arcsin \left(\dfrac{sin(65^{\circ} 20') \times 4.88}{5.38} \right) \approx \mathbf{ 55.517 ^{\circ}}[/tex]

∠C = 55.517° < 90°, therefore, the correct response is; None

[tex]\dfrac{sin(65^{\circ} 20')}{5.38} = \dfrac{4.88}{sin(\angle C)}[/tex]

∠A = 180° - 65°20' - 55.517° = 59.15° = 59°9'

∠A = 59°9'

Therefore;

[tex]\dfrac{5.38}{sin(65^{\circ} 20')} = \mathbf{ \dfrac{a}{sin(59.15^{})}}[/tex]

[tex]a = \dfrac{5.38 \times sin(59.15^{\circ}) }{sin(65^{\circ} 20')} \approx \mathbf{ 5.083}[/tex]

a ≈ 5.083 inches

Learn more here:

https://brainly.com/question/15018190

Answer:

four times a number less 10 is equall to 16 ia not

Answer:

20 three-point scores

61/3=20 R1

Answer:

7

Step-by-step explanation:

61 divided by 8 = 7 remainder 5

you can do 7 3's and 7 5's which would make 7x3 = 21  +  7 x 5 = 35  21 + 35 = 56

61 - 56  = 5 (add 1 to the amount of 5's)

7 3's and 8 5's

Answer:

if she pays the bank back it would be 25 but if she just deposited 85 then she has 85 in the bank

Step-by-step explanation:

Answer:

Step-by-step explanation:

5 / 6 = 0.83333333...

0.833333...•100 = 83.333333...

83.3%

Answer:

You can do this type of questions with inverse proportion.

_

%83,3

Answer:

False

Step-by-step explanation:

The sides are not equal.

Answer:

chu papi

Step-by-step explanation:

Answer:

-3

Step-by-step explanation:

a. Simply put (x+h) wherever you have x.

[tex]f(x+h) = -3(x+h)+7 = -3x +7 -3h[/tex]

b. We have the first part, just subtract

[tex]f(x+h) = -3(x+h)+7 = -3x +7 -3h - (-3x+7) = -3x+7-3h +3x -7 = -3h[/tex]

c. Finally, we divide by h[tex]\frac{-3h}h = -3[/tex]

Answer:

You told me to resolve this question

Answer:

No, sorry about that, buddy, I cannot.

Answer:

(6,-2)

Step-by-step explanation:

Answer:

0.09

Step-by-step explanation:

Answer:

A child psychologist is interested in the number of times a newborn baby’s crying wakes its mother after midnight. For a random sample of 50 mothers, the following information was obtained. Let X = the number of times per week a newborn baby’s crying wakes its mother after midnight. For this example, x = 0, 1, 2, 3, 4, 5.

P(x) = probability that X takes on a value x.

x P(x)0 P(x = 0) = 250

1 P(x = 1) = 1150

2 P(x = 2) = 2350

3 P(x = 3) = 950

4 P(x = 4) = 450

5 P(x = 5) = 150

Step-by-step explanation:

pa follow po pls.

Answer:

1386

Step-by-step explanation:

V=πr²*h

V=22/7*7²*9

V=1386

Answer:

6^3

Step-by-step explanation:

3 × 3 × 3 × 3 × 3 × 3=729

6^3=216

3 × 3 × 3 × 3 × 3 × 3=729

3^6=729

Hope this helps :)

3 × 3 × 3 × 3 × 3 × 3

63

9 × 9 × 9

36

Answer:

66

forty-four times 1.5 equals 66

Answer:

No

Step-by-step explanation:

While these values are close, they are not equal, so I know the original fractions cannot be proportional to each other. So my answer is: The fractions are not