. **y" + xy' + y = 0, y(t) = 3 . y'(1)=4 (12pts) 3. Solve the Cauchy-Euler IVP:

Answer:

mean = 10

Step-by-step explanation:

The mean is calculated as

mean = \(\frac{sum}{count}\)

         = \(\frac{6+7+10+12+12+13}{6}\)

        = \(\frac{60}{6}\)

       = 10

The ionization reaction of H2S(aq) by writing formulas for the products is shown below:H2S(aq) + H2O(l) → H3O+(aq) + HS-(aq).

Hydrogen sulfide reacts with water to form hydrosulfuric acid (H2S). The ionization reaction of hydrosulfuric acid is shown below.H2S(aq) ⇌ H+(aq) + HS-(aq).

Here, the acid donates a proton (H+) to water to form hydronium ion (H3O+), and the conjugate base (HS-) is formed. So, the complete ionization reaction of H2S(aq)  H2S(aq) + H2O(l) → H3O+(aq) + HS-(aq)

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If A is a subset of B, it does not necessarily mean that A is a proper subset of B. The union of any set A and its complement is indeed the universal set.

1. False: If A is a subset of B, it does not necessarily mean that A is a proper subset of B. A proper subset is a subset that contains some, but not all, elements of the superset. In other words, if A is a proper subset of B, it means that there exists at least one element in B that is not in A. However, if A is simply a subset of B, it is possible for A and B to have the same elements, making them equal sets.

For example, let's consider two sets:

A = {1, 2}

B = {1, 2, 3}

In this case, A is a subset of B because every element in A (1 and 2) is also in B. However, A is not a proper subset of B since there are no elements in B that are not in A. Therefore, statement 1 is false.

2. True: The union of any set A and its complement is indeed the universal set. The complement of a set A, denoted as A', consists of all elements that are not in A but are in the universal set U. When we take the union of A and its complement, we are effectively combining all elements from A and all elements not in A.

Let's consider a set A and its complement A' within a universal set U. The union of A and A' is denoted as A ∪ A' and can be represented as U. This is because every element in the universal set U is either in A or not in A (in A').

For example, let's consider the following sets:

U = {1, 2, 3, 4, 5}

A = {1, 2}

The complement of A, A', would be {3, 4, 5}. The union of A and A' (A ∪ A') is {1, 2} ∪ {3, 4, 5}, which is equal to U: {1, 2, 3, 4, 5}. Thus, the union of any set A and its complement is indeed the universal set. Therefore, statement 2 is true.

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The domain of the function shown in the graph is the one in option A:

6 ≤ k ≤ 11

The domain of a function y = f(x) is the set of the inputs of the function. To identify the domain in a graph, we need to look at the horizontal axis (also called the x-axis).

On the graph we can see that it starts at x = 6 with a closed dot, and it ends at x = 11 also with a closed dot.

That means that these values belong to the domain, so we can write the domain as follows:

Domain = 6 ≤ k ≤ 11

(notice that the variable in the horizontal axis is k).

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Answer:

(10,85)

Step-by-step explanation:

x=10

y=85

Answer:  -x^2  

Step-by-step explanation:

-1 times x^2  is the same as -1(x^2)  

-1(x^2) = is the same as -1x^2   or -x^2

Answer:

D

Step-by-step explanation:

I have done this before
Top row units is 8 i then multiplied both sides so 5*2 which = 10 then multiplied 8*10 to get 80

Answer:

80 square units (D)

Step-by-step explanation:

Base is 16 units, height is 5 units. 16 * 5 = 80

Answer: 6

He started with 6 costumers

Let X be an element of coo with the norm || ||p, 1 ≤p≤co. Consider the linear function on X, defined by fr(x) = Σ(j=1 to infinity)x(j)/j^r, x ∈ X When p=1, then fr is continuous and ||fr||1 = 1. For 11-1/p=1/q, and then, ||fr|| p = (Σ(j=1 to infinity) 1/j^rq)^(1/q)

:Let X be an element of coo, with the norm || ||p, 1 ≤p≤co. Consider the linear functional fr on X, defined by fr(x) = Σ(j=1 to infinity)x(j)/j^r, x ∈ X. When p=1, then fr is continuous and ||fr||1 = 1. Also, for 11-1/p=1/q, and then, ||fr||p = (Σ(j=1 to infinity) 1/j^rq)^(1/q)The proof is shown below: Let x be a member of X, and ||x||p≤1, for 1≤p≤coLet r>1-1/p = 1/q We want to prove that fr(x) is absolutely convergent. That is, |fr(x)| < ∞|fr(x)| = |Σ(j=1 to infinity)x(j)/j^r| ≤ Σ(j=1 to infinity)|x(j)/j^r| ≤ Σ(j=1 to infinity)(1/j^r)This is a convergent p-series because r>1-1/p = 1/q by the p-test for convergence. Hence, fr(x) is absolutely convergent, and fr is continuous on X. This implies that ||fr||p = sup { |fr(x)|/||x||p: x ∈ X, ||x||p ≤ 1} = (Σ(j=1 to infinity) 1/j^rq)^(1/q)

It has been shown that fr is continuous on X if and only if r>1-1/p=1/q, and then, ||fr||p = (Σ(j=1 to infinity) 1/j^rq)^(1/q). This means that the value of r is important in determining whether fr is continuous or not. Furthermore, ||fr||p is dependent on the value of r. If r>1-1/p=1/q, then fr is continuous and ||fr||p = (Σ(j=1 to infinity) 1/j^rq)^(1/q).

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Answer:

distance = 7.07 units

Step-by-step explanation:

x difference = 7

y difference = -1

using the Pythagorean theorem:

7² + -1² = d²

d² = 49 + 1 = 50

d = 7.07

Answer:

7.1

Step-by-step explanation:

distance between the x is 7 and y is 1

And to find the hypotenuse, you use the quadratic formula, 7^2 + 1^2 = c^2

c=50

and take the square root, which is 7.071, rounds to 7.1

Part A: To find the intersection points of the two polar curves, we need to equate the expressions for r and solve for the angle θ at which they intersect.

For the first polar curve, r = 3cos(8θ).

For the second polar curve, r = 1 + cos(θ).

Setting these two expressions equal to each other:

3cos(8θ) = 1 + cos(θ).

Simplifying the equation, we have:

2cos(θ) = 1.

Solving for θ, we find:

θ = π/3 + 2πn, π/3 + 2πn + 2π/3, where n is an integer.

These solutions represent the angles at which the two polar curves intersect.

Part B: The question is incomplete and it is not clear what is meant by "Let S be t."

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Answer:  1. у %3 2х

Step-by-step explanation: Find the distance between (0, 0) and (-3, 4) pair of points

Answer:

115 lbs = 52.16 kilograms

Step-by-step explanation:

115 pounds is a little over 52kg

Answer:

X = 26

Step-by-step explanation:

3^3 + 5^2 =2x

27 + 25 = 2x

52 = 2x

Divide by 2

X = 26

Answer:

27

Step-by-step explanation:

Answer:

27 cupcakes

Step-by-step explanation:

36 divided by 4 is 9, 9 times 3 is 27

It is expected that the mean absolute deviation would be higher if the temperature is recorded over the course of a year than if it is recorded over the course of a month.

This term refers to the average difference or variation between the mean of all the data and each individual data set. Due to this, the mean absolute deviation is higher if the data collected varies more.

Based on the above, if the data is collected over a year, it is expected the temperature varies more, and therefore, the mean absolute deviation would be higher.

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The value of the given integral is bold 64/15.

To evaluate the given integral, we need to use Green's theorem which states that the line integral of a vector field around a closed curve C is equal to the double integral of the curl of the vector field over the region D enclosed by C.

Let F = ((2x - y^2), xy) be the given vector field. Then, its curl is given by:

curl(F) = d(xy)/dx - d((2x - y^2))/dy

= y - 0

= y

Now, let C be the curve given by x = 8t, y = ((t)), 0 ≤ t ≤ 6. Then, its boundary is C' which consists of four line segments:

1. The segment from (0,0) to (8,1)

2. The segment from (8,1) to (32,2)

3. The segment from (32,2) to (64,2)

4. The segment from (64,2) to (96,2^(1/4))

Using Green's theorem, we have:

∫∫_D curl(F) dA = ∫_C F · dr

where D is the region enclosed by C and dr is the differential element of the curve C.

Since curl(F) = y, we have:

∫∫_D y dA = ∫_C (2x - y^2) dx + xy dy

To evaluate the left-hand side, we need to find the limits of integration for x and y. Since x ranges from 0 to 96 and y ranges from 0 to (t)), we have

0 ≤ x ≤ 96

0 ≤ y ≤ (t)

Converting to polar coordinates with x = r cosθ and y = r sinθ, we have:

0 ≤ r ≤ (t)

0 ≤ θ ≤ π/2

Then, the double integral becomes:

∫∫_D y dA = ∫_0^(π/2) ∫_0^((6)) r sinθ r dr dθ

= ∫_0^(π/2) (1/4) sinθ [(t))]^4 dθ

= (1/4) [2 - 2/(3)]

To evaluate the right-hand side, we need to parameterize each segment of C and compute the line integral.

1. The segment from (0,0) to (8,1):

x = 8t, y = (t), 0 ≤ t ≤ 1

∫_0^1 (2x - y^2) dx + xy dy

= ∫_0^1 (16t - t) (8 dt) + (8t)((t))) (1/4) dt

= 32/3 + 2/5

2. The segment from (8,1) to (32,2):

x = 8 + 4t, y = (t)), 1 ≤ t ≤ 16

∫_1^16 (2x - y^2) dx + xy dy

= ∫_1^16 (32t - t) (4 dt) + (4t+32)((t))) (1/4) dt

= 64/3 + 128/15

3. The segment from (32,2) to (64,2):

x = 64 - 4t, y = (t)), 16 ≤ t ≤ 36

∫_16^36 (2x - y^2) dx + xy dy

= ∫_16^36 (128 - t) (-4 dt) + (64-4t)((t))) (1/4) dt

= -64/3 + 128/15

4. The segment from (64,2) to (96,2^(1/4)):

x = 96 - 8t, y = 2^(1/8)t^(1/4), 0 ≤ t ≤ 2^6

∫_0^(2^6) (2x - y^2) dx + xy dy

= ∫_0^(2^6) (192 - 2^(5/4)t) (-8 dt) + (96-8t)(2^(1/8)t^(1/4)) (1/4) dt

= -32/3 + 256/15

Adding up the line integrals, we get:

∫_C F · dr = 64/15

Therefore, by Green's theorem, we have:

∫∫_D y dA = ∫_C F · dr

= 64/15

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Answer:

The answer is C

Step-by-step explanation:

the volume equation is b*h*

the b= 2*8

the h= 3.

6 is the diagonal length of the side face which would be incorrect.

The answer would be 48. or C in the pic

Probability that the sum of the numbers rolled is even = 5/8

An even number is 3 times more likely to appear as an odd number, the probability of an even number appearing is 3/4.

The problem states that the sum of the two dies must be even, the numbers must both be even, or both be odd. We either have EE or OO, so we have:

= 3/4x3/4 + 1/4x1/4

= 1/16 + 9/16

= 10/16

Probability that the sum of the numbers rolled is even = 5/8

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or, s + 16 = 3s

or, 16 = 3s - s

or, 2s = 16

or, s = 8

The value of s is 8.

Hope it helps.

Do comment if you have any query.

Answer + Step-by-step explanation:

10 × 5 × 2

= 10 × (5 × 2)   (associative property)

= 10 × (10)

= 10 × 10

= 100

The simplified expression without absolute value signs is x - 2.

To simplify the expression |x - 2| when x > 2, we can use the fact that if x is greater than 2, then x - 2 will be positive. In this case, |x - 2| simplifies to just x - 2.

This simplification is based on the understanding that the absolute value function, denoted by | |, returns the positive value of a number. When x > 2, x - 2 will be positive, and the absolute value function is not needed to determine its value. In this case, the expression simplifies to x - 2.

However, it's important to note that when x ≤ 2, the expression |x - 2| would simplify differently. When x is less than or equal to 2, x - 2 would be negative or zero, and |x - 2| would simplify to -(x - 2) or 2 - x, respectively.

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The root mean squared error (RMSE) of the wisdom of the crowd forecasts for 1960-2015 is 3.72. This is lower than the RMSE of any of the individual forecasting approaches, which were:

3-period moving average: 4.02

Exponential smoothing: 3.98

Rolling regression: 3.87

The wisdom of the crowd method is a forecasting technique that averages the forecasts of multiple individuals or models. This can often lead to more accurate forecasts than any of the individual forecasts.

In this case, the wisdom of the crowd method reduced the RMSE by 0.30, 0.26, and 0.15, respectively, compared to the 3-period moving average, exponential smoothing, and rolling regression models.

The reason why the wisdom of the crowd method can be more accurate than any of the individual forecasts is because it can help to correct for individual biases.

For example, if one individual is consistently over-forecasting, the wisdom of the crowd method will down-weight that individual's forecast. This can help to improve the overall accuracy of the forecast.

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The root mean squared error (RMSE) of the wisdom of the crowd forecasts for 1960-2015 is 3.72. This is lower than the RMSE of any of the individual forecasting approaches, which were:

3-period moving average: 4.02

Exponential smoothing: 3.98

Rolling regression: 3.87

The wisdom of the crowd method is a forecasting technique that averages the forecasts of multiple individuals or models. This can often lead to more accurate forecasts than any of the individual forecasts.

In this case, the wisdom of the crowd method reduced the RMSE by 0.30, 0.26, and 0.15, respectively, compared to the 3-period moving average, exponential smoothing, and rolling regression models.

The reason why the wisdom of the crowd method can be more accurate than any of the individual forecasts is because it can help to correct for individual biases.

For example, if one individual is consistently over-forecasting, the wisdom of the crowd method will down-weight that individual's forecast. This can help to improve the overall accuracy of the forecast.

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According to the given information, the sample period should be from 2010-2020.

a) Model specification

The model specification based on the scenario above is as follows:

SDF= f(T, TI, NL)

Where: SDF= Sustainable Development Framework

T= Tourism

TI= Technological innovation

NL= National leadership

b) Justification for the selection of the dependent and independent variables model:

Dependent variable: The dependent variable in this model is Sustainable Development Framework (SDF). The model seeks to develop a framework for sustainable development that would facilitate growth, environmental and socio-cultural sustainability.

Independent variables:

The independent variables are tourism sustainability, technological innovation, and quality of leadership. These variables drive economic development. The inclusion of tourism sustainability reflects its importance in the global economy and its potential to drive growth.

The inclusion of technological innovation reflects its potential to enhance productivity and create new industries. The inclusion of national leadership reflects the role of governance in promoting sustainable development and managing negative externalities.

c) Justification for the selection of the sample period:

The sample period should be selected based on the availability of data for the variables of interest. Ideally, the period should be long enough to capture trends and patterns in the data. However, it should not be too long that the data becomes obsolete or no longer relevant.

Additionally, the period should also reflect the context and relevance of the research question. Therefore, the sample period for this study should cover the last decade to capture the trends and patterns in the data and reflect the relevance of the research question.

The sample period should be from 2010-2020.

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Answer: range

Step-by-step explanation:      

Answer:

1 is

for 1 disc is $16.50

for 2 disc is $33.00

for 3 disc is $49.50

and

2 is

for 1 book is $6.95

for 2 books is $13.90

for 3 books is $20.85

Step-by-step explanation:

brainliest pls

Answer:

The answer is 24%.

Step-by-step explanation:

If you subtract 798 from 1050 you will get 252. Then, divide 252 by 1050. What you get from this is 0.24, which in percentage form is 24%.

Answer:

-5

Step-by-step explanation:

25 – 6 [15 ÷ (7 – 4)]

25 – 6 (15 ÷ 3)

25 – 6 x 5

25 – 30

-5

Answer:

- 5

Step-by-step explanation:

25 - 6 [ 15 ÷(7 - 4) ]

Following the order as stated in PEMDAS

Evaluate parenthesis

= 25 - 6 [ 15 ÷ 3 ]

= 25 - 6 (5) ← evaluate multiplication

= 25 - 30 ← evaluate subtraction

= - 5

Answer:

i dont see what the top part says??

Step-by-step explanation: