An ODE that describes the price p is given by: p"(t) – (k – 1)p' (t) + kp(t)=k. where k > 0 is a constant describing how people's expectations on the rate of inflation changes depending on the observed inflation rate! (a) Show that p(t) = 1 is an equilibrium solution of the ODE. Recall that an equilibrium solution is just a solution that is constant.

To determine the type of triangle, we need to consider the given equation: bcosC + ccosB = asin4.

In a triangle, the angles A, B, and C are related to their respective sides through trigonometric functions. In this equation, we have the cosine functions of angles B and C.

If the triangle is acute, all angles A, B, and C are less than 90 degrees. In an acute triangle, the cosine values of all angles are positive.

If the triangle is obtuse, one angle is greater than 90 degrees. In an obtuse triangle, the cosine value of one angle is negative.

If the triangle is isosceles, two sides are equal, so the corresponding angles are equal as well. In an isosceles triangle, the cosine values of the base angles are equal.

If the triangle is right, one angle is exactly 90 degrees. In a right triangle, the cosine value of the right angle is 0.

Now let's analyze the given equation: bcosC + ccosB = asin4.

Since the equation involves cosine functions, we can conclude the following:

If both b and c are positive and the right side (asin4) is positive, it indicates an acute triangle.

If one of b or c is negative, it indicates an obtuse triangle.

If b and c are positive and the cosine values are equal (bcosC = ccosB), it indicates an isosceles triangle.

If one of b or c is 0, it indicates a right triangle.

Based on the given equation, we cannot determine the specific type of triangle (acute, obtuse, isosceles, or right) without additional information. Therefore, the answer is indeterminate.

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n = 6, taking the sample points to be left endpoints, for the function f(x) = cos(x) over the interval 0 ≤ x ≤ 3/4, we can calculate the sum using the left endpoint rule and round the answer to six decimal places.

The Riemann sum is an approximation of the definite integral of a function using rectangles. In this case, we are given the function f(x) = cos(x) over the interval 0 ≤ x ≤ 3/4.

To evaluate the Riemann sum with n = 6 and left endpoints, we divide the interval [0, 3/4] into six subintervals of equal width. The width of each subinterval is (b - a) / n, where n is the number of subintervals and (b - a) is the interval length (3/4 - 0 = 3/4).

We calculate the left endpoint of each subinterval by using the formula x = a + (i - 1) * (b - a) / n, where i represents the index of each subinterval.

Next, we evaluate the function f(x) = cos(x) at each left endpoint and multiply it by the width of the corresponding subinterval. Then, we sum up the areas of all the rectangles to get the Riemann sum.

Finally, we round the answer to six decimal places to comply with the given precision requirement.

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The area of the shaded region is 17.27 square centimeter.

Given that, θ=80° and the radius of a circle is 5 cm.

The formula to find the area of a sector = θ/360° ×πr².

Here, area of a sector = 80°/360° ×3.14×5²

= 0.22×3.14×25

= 17.27 square centimeter

Therefore, the area of the shaded region is 17.27 square centimeter.

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The probability that a randomly selected woman is taller than 5 ft 10 in (70 inches) is approximately 0.0143 or 1.43%.

To find the probability that a randomly selected woman is taller than 5 ft 10 in, we need to convert the height to inches and then calculate the probability using the Normal distribution.

5 ft 10 in is equivalent to 5(12) + 10 = 70 inches.

Let's calculate the z-score corresponding to a height of 70 inches using the formula: z = (x - μ) / σ

where x is the observed value, μ is the mean, and σ is the standard deviation. In this case, x = 70 inches, μ = 63.8 inches, and σ = 2.8 inches.

[tex]z=\frac{70-63.8}{2.8} = 2.214[/tex]

Using a standard Normal distribution table or calculator, we can find the probability associated with this z-score.The probability of a randomly selected woman being taller than 5 ft 10 in (70 inches) can be found by calculating the area under the Normal distribution curve to the right of z = 2.214.

P(Z > 2.214) = 1 - P(Z ≤ 2.214)

By looking up the corresponding probability in the standard Normal distribution table or using a calculator, we find that P(Z ≤ 2.214) ≈ 0.9857.

Therefore, P(Z > 2.214) = 1 - 0.9857 =0.0143.

Thus, the probability that a randomly selected woman is taller than 5 ft 10 in (70 inches) is approximately 0.0143 or 1.43%.

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the standard deviations for X and Y are:

X: 14.57%

Y: 19.59%

To calculate the average returns for X and Y, we sum up the returns for each year and divide by the total number of years (in this case, 5).

Average return for X:

(22.3 - 17.3 + 10.3 + 20.6 + 5.3) / 5 = 8.64%

Average return for Y:

(27.9 - 4.3 + 29.9 - 15.6 + 33.9) / 5 = 14.36%

Therefore, the average returns for X and Y are:

X: 8.64%

Y: 14.36%

To calculate the variances for X and Y, we need to find the sum of squared differences from the mean for each return, divide by the total number of years, and round the result to 6 decimal places.

Variance for X:

((22.3 - 8.64)^2 + (-17.3 - 8.64)^2 + (10.3 - 8.64)^2 + (20.6 - 8.64)^2 + (5.3 - 8.64)^2) / 5 = 211.934933

Variance for Y:

((27.9 - 14.36)^2 + (-4.3 - 14.36)^2 + (29.9 - 14.36)^2 + (-15.6 - 14.36)^2 + (33.9 - 14.36)^2) / 5 = 383.830933

The variances for X and Y are:

X: 211.934933

Y: 383.830933

To calculate the standard deviations for X and Y, we take the square root of their respective variances and express them as percentages rounded to 2 decimal places.

Standard deviation for X:

√(211.934933) = 14.57%

Standard deviation for Y:

√(383.830933) = 19.59%

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The language p is closed under union, concatenation, and complement, as the union of two languages in p, the concatenation of two languages in p, and the complement of a language in p all remain in p.

To prove that a language p is closed under union, concatenation, and complement, we need to demonstrate that the result of each operation on languages in p remains in p.

1. Union: Let L1 and L2 be two languages in p. We need to show that their union, L1 ∪ L2, is also in p. Since both L1 and L2 are in p, it means that every string in L1 and L2 satisfies the property defined by p.

By taking the union, we combine all the strings from L1 and L2, which still satisfy the same property. Therefore, L1 ∪ L2 is also in p.

2. Concatenation: Let L1 and L2 be two languages in p. We want to prove that their concatenation, L1 · L2, is in p. For every string in L1 · L2, it can be split into two parts, one from L1 and the other from L2.

Since both L1 and L2 satisfy the property defined by p, it follows that the strings in L1 · L2 also satisfy the property. Hence, L1 · L2 is in p.

3. Complement: Let L be a language in p. We need to show that its complement, ¬L (all strings not in L), is in p. Since L satisfies the property defined by p, the complement of L will consist of all strings that do not satisfy that property.

However, p is closed under complement, which means that every language in p also satisfies the property of p. Therefore, ¬L is also in p.

In conclusion, we have shown that p is closed under union, concatenation, and complement.

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given that g(x) = 2x + 1 and h(x) = 4x^2 + 4x + 3.Since fog = h, we can write the equation as f(2x + 1) = 4x^2 + 4x + 3To solve for f, we need to isolate it on one side of the equation.

We have to find f such that fog = h

Let's start by substituting y = 2x + 1 in the equation.

f(y) = 4((y - 1)/2)^2 + 4((y - 1)/2) + 3

Simplifying, we get:

f(y) = 2(y - 1)^2 + 2(y - 1) + 3

Thus,

f(x) = 2(x - 1)^2 + 2(x - 1) + 3.

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The correct answer is b. the sampling distribution of means.

The summary of the answer is that the theoretical distribution of all possible random sample means of the same size n is known as the sampling distribution of means.

In the second paragraph, we explain that the sampling distribution of means is a theoretical distribution that represents the distribution of sample means when repeatedly sampling from a population. It is derived from the central limit theorem, which states that as the sample size increases, the sampling distribution of means approaches a normal distribution, regardless of the shape of the population distribution.

The sampling distribution of means is a key concept in statistics and is widely used in hypothesis testing, confidence intervals, and estimating population parameters. It allows us to make inferences about the population based on the characteristics of the sample means. The properties of the sampling distribution of means, such as its mean and standard deviation, are related to the properties of the population distribution and the sample size. Understanding the sampling distribution of means is fundamental in statistical analysis and plays a crucial role in many statistical techniques.

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The equation of the ellipse with focus at (-2,0) and vertices at (±7, 0) is given as follows:

x²/49 + y²/45 = 1.

The equation of an ellipse of center (h,k) is given by the equation presented as follows:

(x - h)²/a² + (y - k)²/b² = 1.

The center of the ellipse is given by the mean of the coordinates of the vertices, as follows:

Hence:

x²/a² + y²/b² = 1.

The vertices are at x + a and x - a, hence the parameter a is given as follows:

a = 7.

Considering the focus at (-2,0), the parameter c is given as follows:

c = -2.

We need the parameter c to obtain parameter b as follows:

c² = a² - b²

b² = a² - c²

b² = 49 - 4

b² = 45.

Hence the equation is given as follows:

x²/49 + y²/45 = 1.

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The correct choice for the use of the Divergence Theorem is (a) Normal vectors pointing away from the enclosed region.

The Divergence Theorem, also known as Gauss's theorem, relates the flux of a vector field across a closed surface to the divergence of the vector field within the enclosed region. It states that the flux through a closed surface is equal to the volume integral of the divergence over the enclosed region.

By convention, the normal vectors on a closed surface are chosen to point outward from the enclosed region. This choice ensures that the divergence of the vector field is positive when it represents a source or outward flow of the field from the enclosed region. If the normal vectors were chosen to point inward, the divergence would be negative for outward flow, leading to incorrect results when applying the Divergence Theorem.

Therefore, to correctly apply the Divergence Theorem, we choose the orientation with normal vectors pointing away from the enclosed region.

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The survey's reliability and validity depend on the methodology and quality of the sample data.

In the given scenario, a survey aims to investigate whether the proportion of today's high school seniors who own their own cars is higher than it was a decade ago. The survey proposes an alternative hypothesis that suggests a change in the proportion, while the null hypothesis assumes no change. The survey also mentions that the observed result would occur by chance 1.7% of the time if the null hypothesis were true.

To evaluate the reasonability of the survey, we need to consider the concept of statistical significance. Statistical significance is a measure of how likely the observed result would occur due to chance alone, assuming the null hypothesis is true. In hypothesis testing, a common threshold for statistical significance is α (alpha), typically set at 0.05 or 5%.

In this case, the survey suggests that the observed result would occur by chance 1.7% of the time if the null hypothesis were true. This is known as the p-value. The p-value represents the probability of obtaining a result as extreme or more extreme than the observed result, assuming the null hypothesis is true.

If the p-value is less than the chosen significance level (α), we reject the null hypothesis in favor of the alternative hypothesis. In this scenario, since the p-value is 1.7%, which is less than 5%, we can conclude that the observed result is statistically significant.

Therefore, it is reasonable to conduct the survey and investigate whether the proportion of high school seniors who own their own cars has increased compared to a decade ago. The survey provides evidence to support the alternative hypothesis and suggests that the observed result is unlikely to occur by chance alone, assuming the null hypothesis is true.

However, it's important to note that the survey's reasonability is based on the assumption that the survey methodology and sample data are reliable and representative. The survey should ensure that the sample is randomly selected and sufficiently large to provide accurate results. Additionally, the survey should consider potential confounding variables and sources of bias that could affect the findings.

In summary, the survey investigating the proportion of high school seniors who own their own cars and proposing a higher proportion than a decade ago is reasonable based on the evidence provided, which suggests a statistically significant result. However, the survey's reliability and validity depend on the methodology and quality of the sample data.

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To solution of  the differential equation (x^2 + 3xy) dx + e^y dy = 0,

A differential equation is a mathematical equation that relates a function or a set of functions to its derivatives. It involves the derivatives of the unknown function(s) with respect to one or more independent variables.

By integrating the equation with respect to x and y separately, we obtain the expression F(x, y) = x^2/2 + xy - 5e^y = C. This equation represents the implicit solution to the given differential equation.

In summary, the solution to the given differential equation is given by the implicit equation F(x, y) = x^2/2 + xy - 5e^y = C, where C is an arbitrary constant.

This solution equation satisfies the original differential equation, and any point (x, y) that satisfies the equation is a solution to the differential equation.

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After approximately two weeks, there will be approximately 18 weeds in the garden.

Out of the given options, the closest number of weeds to 18 is 20 weeds. Hence, the answer would be 20 weeds.

To determine the approximate number of weeds in the garden after two weeks, we need to calculate the exponential growth based on the given rate of 15% per day.

We can use the formula for exponential growth:

[tex]P(t) = P0 \times(1 + r)^t[/tex]  

Where:

P(t) represents the final population after time t

P0 represents the initial population (4 weeds in this case)

r represents the growth rate per period (15% or 0.15 in decimal form)

t represents the number of time periods (in this case, 14 days, as two weeks consist of 14 days)

Let's substitute the values into the formula:

[tex]P(14) = 4 \times (1 + 0.15)^{14[/tex]

Calculating the exponential growth:

[tex]P(14) = 4 \times (1.15)^{14[/tex]

P(14) ≈ [tex]4 \times 4.441703[/tex]

P(14) ≈ 17.766812

Therefore, after approximately two weeks, there will be approximately 18 weeds in the garden.

Out of the given options, the closest number of weeds to 18 is 20 weeds. Hence, the answer would be 20 weeds.

However, it's important to note that this is an approximation as we rounded the value.

The actual number of weeds may not be exactly 20, but it should be close to that value based on the given growth rate.

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The area of the surface above the triangle formed by the points (0, 0), (2, 0), and (2, 4) in the equation 2y + 4z - x² = 5 can be calculated using surface integration techniques.

To find the area, we first need to parameterize the surface. Let's consider the surface as a function of two variables, u and v. We can rewrite the equation as x = u, y = v, and z = (5 - 2v - u²)/4.

Now, we need to find the bounds for u and v that define the region above the triangle. The triangle is bounded by u = 0, u = 2, and v = 0. We can set up the double integral using these bounds:

∫∫[D] √(1 + (∂z/∂u)² + (∂z/∂v)²) du dv

Where [D] represents the region bounded by the triangle.

Next, we calculate the partial derivatives of z with respect to u and v:

(∂z/∂u) = -u/2

(∂z/∂v) = -1/2

Substituting these values into the integral, we have:

∫∫[D] √(1 + (u/2)² + (1/2)²) du dv

Simplifying the expression under the square root:

√(1 + (u/2)² + (1/2)²) = √(1 + u²/4 + 1/4) = √(u²/4 + 1) = √((u² + 4)/4)

The integral becomes:

∫∫[D] √((u² + 4)/4) du dv

Integrating with respect to u first, from u = 0 to u = 2:

∫[0 to 2] ∫[0 to v] √((u² + 4)/4) du dv

Simplifying further:

∫[0 to 2] [(1/2)√(u² + 4)]|[0 to v] dv

= (1/2) ∫[0 to 2] (√(v² + 4) - 2) dv

Now, integrating with respect to v, from v = 0 to v = 4:

(1/2) ∫[0 to 4] (√(v² + 4) - 2) dv

Evaluating the integral, we find the area of the surface above the triangle.

Please note that due to the complexity of the calculations involved, providing an exact numerical result within the specified word limit is not feasible. I recommend using numerical methods or software to evaluate the integral and obtain the final area value.

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The result of the integral is (2x²) + C, where C represents the constant of integration.

To evaluate the integral ∫4x cos(7x) dx using the integration-by-parts formula, we choose u = 4x and dv = dx. Applying the integration-by-parts formula, we find the result of the integral to be (4x/7) sin(7x) - ∫(4/7) sin(7x) dx.

To apply the integration-by-parts formula, we choose one part of the integral to be u and the remaining part as dv. In this case, we select u = 4x and dv = dx. Taking the derivative of u with respect to x gives du/dx = 4, and integrating dv with respect to x gives v = x.

Now, we can use the integration-by-parts formula, which states that ∫u dv = uv - ∫v du. Applying this formula, we have:

∫4x cos(7x) dx = (4x)(x) - ∫x(4) dx

= 4x^2 - ∫4x dx

= 4x^2 - 2x^2 + C (where C is the constant of integration)

Simplifying further, we have:

∫4x cos(7x) dx = (2x^2) + C

Thus, the result of the integral is (2x^2) + C, where C represents the constant of integration.

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r(t) = (8t, 9 - 2t) for line through (0,9) with direction vector (8,-2).

To write a vector function for the line that passes through the point (0, 9) with direction vector (8, -2), we can use the parametric form of a line equation.

Let's denote the vector function as r(t) = (x(t), y(t)), where t is the parameter.

We know that the line passes through the point (0, 9), so the initial point of the line is r(0) = (0, 9).

Since the direction vector is (8, -2), we can use it to determine the change in x and y coordinates over a certain value of t.

The change in x coordinate is 8t, and the change in y coordinate is -2t.

Therefore, the vector function for the line passing through (0, 9) with direction vector (8, -2) is:

r(t) = (0 + 8t, 9 - 2t) = (8t, 9 - 2t).

This vector function represents the position of points on the line as t varies.

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The two operations used for converting weight measurements are multiplication and division

The two operations used for converting weight measurements are:

Multiplication is used when converting from a smaller unit to a larger unit. To convert a weight from a smaller unit to a larger unit, you multiply by a conversion factor that represents the relationship between the two units.

Division is used when converting from a larger unit to a smaller unit. To convert a weight from a larger unit to a smaller unit, you divide by the conversion factor that represents the relationship between the two units.

By using multiplication and division with the appropriate conversion factors, you can convert weight measurements between different units of measurement.

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In a k-nearest neighbors (k-NN) algorithm, similarity between records is based on a distance metric.

The choice of distance metric is crucial in determining the similarity between data points and plays a significant role in the k-NN algorithm's performance.

The most commonly used distance metric in k-NN algorithms is the Euclidean distance. The Euclidean distance measures the straight-line distance between two points in a Euclidean space. For example, in a two-dimensional space, the Euclidean distance between two points (x1, y1) and (x2, y2) is calculated as:

d = √((x2 - x1)² + (y2 - y1)²)

This distance metric assumes that all dimensions have equal importance and calculates the distance based on the geometric distance between the points. It is widely used because it provides a meaningful measure of similarity between data points.

However, depending on the nature of the data and the problem at hand, alternative distance metrics may be used. Some common alternatives include:

Manhattan distance (also known as city block distance or L1 distance): This metric calculates the distance by summing the absolute differences between the coordinates of two points. In a two-dimensional space, the Manhattan distance between two points (x1, y1) and (x2, y2) is calculated as:

d = |x2 - x1| + |y2 - y1|

Minkowski distance: This is a generalized distance metric that includes both the Euclidean and Manhattan distances as special cases. It is defined as:

d = (∑(|xi - yi|^p))^(1/p)

where p is a parameter that determines the specific distance metric. When p = 1, it reduces to the Manhattan distance, and when p = 2, it becomes the Euclidean distance.

Cosine similarity: This metric measures the cosine of the angle between two vectors. It is often used when dealing with high-dimensional data or text data, where the magnitude of the vectors is less relevant than the direction.

The choice of distance metric depends on the specific characteristics of the data and the problem being solved. It is important to select a distance metric that captures the relevant aspects of similarity and aligns with the underlying structure of the data.

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The correct values for a 4f orbital are:

n = 4, ℓ = 3, m = -2

The quantum number "n" represents the principal quantum number, which determines the energy level of the electron. In this case, it is 4.

The quantum number "ℓ" represents the azimuthal quantum number, which determines the shape of the orbital. For an f orbital, the value of ℓ is 3.

The quantum number "m" represents the magnetic quantum number, which determines the orientation of the orbital in space. In this case, it is -2.

Therefore, the correct values for a 4f orbital are n = 4, ℓ = 3, and m = -2.

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the critical value corresponding to a sample size of 24 and a confidence level of 95% is 2.064.

the critical value corresponds to the z-score that defines the boundary for the confidence interval. In this case, with a sample size of 24 and a confidence level of 95%, we use a two-tailed z-test. Looking up the z-score for a confidence level of 95%, or alpha of 0.025, we can find the critical value of 2.064.

the critical value for a sample size of 24 and a confidence level of 95% is 2.064. This value is important in calculating the confidence interval for the population parameter.

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To find the area under the curve y = 7x^4 over the interval [0, 3], we need to integrate the function with respect to x using the definite integral formula:

∫[0, 3] 7x^4 dx

After integrating, we get:

(7/5)x^5]0^3

Plugging in the upper and lower limits of integration, we get:

(7/5)(3^5 - 0^5)

Simplifying further, we get:

(7/5)(243)

The exact value of the area under the curve y = 7x^4 over the interval [0, 3] is 1701/5.


We used the definite integral formula to find the area under the curve y = 7x^4 over the interval [0, 3]. The integral involves multiplying the function by dx and integrating with respect to x. After performing the integration and plugging in the limits of integration, we simplified the expression to get the exact value of the area.

The exact value of the area under the curve y = 7x^4 over the interval [0, 3] is 1701/5.

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The area of the shapes are ;

1. 155cm²

2. 236.3 cm²

The area of a shape is the space occupied by the boundary of a plane figures like circles, rectangles, and triangles.

1. The shape is divided into parallelogram and trapezium.

area of trapezoid = 1/2(a+b) h

= 1/2( 3+13)8

= 1/2 × 16 × 8

= 64cm²

area of parallelogram

= b× h

= 13 × 7

= 91 cm²

The area of the shape = 91 +64

= 155cm²

2. area of 2 semi circle = area of circle

Therefore the surface area of the shape = πr² + πrh

= πr(r+h)

= 3.14 × 3.5( 3.5 + 18)

= 10.99 × 21.5

= 236.3 cm²

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A normality test is run as part of every experiment, to find out if a sample data comes from a normally distributed population. It is essential to determine whether a sample data comes from a normal distribution before performing any statistical analysis on it.

Normality tests are important because many statistical tests, including the t-test and the analysis of variance (ANOVA), depend on the assumption of normality. If the data are not normally distributed, the results of the analysis may be incorrect, leading to wrong conclusions. Normality tests are used to determine whether the data is normally distributed or not. The most commonly used normality tests are the Shapiro-Wilk test, the Anderson-Darling test, the Kolmogorov-Smirnov test, and the Lilliefors test.

If the p-value is less than or equal to the level of significance, then the null hypothesis is rejected, which means that the data is not normally distributed. In conclusion, a normality test should be run as part of every experiment to check the normality of the data. If the data are not normally distributed, then the results of the analysis may be incorrect, leading to wrong conclusions. Therefore, normality tests are essential for ensuring the validity of the statistical analysis.

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The number of students who take only lunch or dinner but not both is 50

Since Venn diagram is used to visually represent the differences and the similarities between two concepts.

Given that all 70 students take lunch or dinner or both meals at the hostel.

students take lunch or dinner or both meals at the hostel = 70

lunch = 30

dinner= 50

Students who both or any one of the drinks= 900-125=725

Now number of students who take only lunch or dinner but not both

= 50 - 30

= 20

Then 20 + 30 = 50

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The perimeter of the composite shape is 29.4 units.

A perimeter is a closed path that encompasses, surrounds, or outlines either a two dimensional shape or a one-dimensional length.

The given graph has a rectangle and right triangles.

Perimeter of rectangle=2(length + width)

=2(4+3)

=14 units.

Perimeter of triangle=5+4+√25+16

=5+4+6.4

=15.4

Total perimeter of the composite figure is 14+15.4

29.4 units

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Answer:

12

Based on the given conditions, formulate:: 72/90/15

Cross out the common factor: 72/6

Cross out the common factor: 12

1. To find the expected value, we integrate the product of x and the density function over the given interval [0, ln 2]:

E(X) = ∫₀^ln2 x e^x dx

Using integration by parts with u = x and dv = e^x dx, we get:

E(X) = [x e^x]₀^ln2 - ∫₀^ln2 e^x dx

E(X) = ln 2 - 1

2. To find the variance, we use the formula:

Var(X) = ∫₀^ln2 (x - E(X))^2 e^x dx

Expanding the square and simplifying, we get:

Var(X) = ∫₀^ln2 x^2 e^x dx - 2E(X) ∫₀^ln2 x e^x dx + E(X)^2 ∫₀^ln2 e^x dx

Var(X) = ∫₀^ln2 x^2 e^x dx - (ln 2 - 1)^2

Using integration by parts twice with u = x^2 and dv = e^x dx, we get:

Var(X) = [x^2 e^x]₀^ln2 - 2∫₀^ln2 x e^x dx + ∫₀^ln2 e^x dx - (ln 2 - 1)^2

Var(X) = ln 2 - (3/2) + (ln 2 - 1)^2

3. Finally, the standard deviation is the square root of the variance:

σ(X) = √Var(X) = √[ln 2 - (3/2) + (ln 2 - 1)^2] ≈ 0.5218

To know more about expected value refer here:

https://brainly.com/question/13749480#

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