Find the following integrals. Show all working. (a) ∫ 0
1
​
(2x 2
+5)e −x/3
dx (b) ∫(3+2x 2
)lnxdx

If the results of a statistical test are considered to be statistically significant, it means that the observed effect or difference between groups is unlikely to have occurred by chance alone, assuming that the null hypothesis is true. In other words, there is strong evidence to suggest that the observed results are not simply due to random variation or sampling error.

Statistical significance is determined by comparing the observed data to a null hypothesis, which represents the idea that there is no real effect or difference between groups. The statistical test calculates a p-value, which is the probability of obtaining results as extreme as or more extreme than the observed data, assuming the null hypothesis is true.

If the p-value is smaller than a predetermined threshold (typically 0.05 or 0.01), it is considered statistically significant. This means that the probability of obtaining the observed results by chance, assuming the null hypothesis is true, is low. Therefore, we reject the null hypothesis and conclude that there is evidence in favor of the alternative hypothesis, which suggests the presence of an effect or difference.

It's important to note that statistical significance does not imply practical significance or importance. While statistically significant results indicate a strong likelihood of a real effect, the magnitude or practical significance of the effect should also be considered in interpreting the results. Additionally, statistical significance is dependent on the chosen significance level (alpha), and different significance levels may lead to different conclusions.

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(i) The given statement is true for all positive integers n.

(ii)  K_r,s is a tree if and only if either r = 1 or s = 1.

(i) Show by induction that a tree with n vertices has n−1 edges:

The proof of the statement "a tree with n vertices has n-1 edges" is done by mathematical induction.

For the base case n = 1, there is only one vertex and no edges.

Hence, the statement is true.

Now, suppose that for some positive integer k, every tree with k vertices has k-1 edges.

Let G be a tree with k+1 vertices and let v be a leaf (a vertex with degree 1) of G.

Removing the vertex v and its adjacent edge gives a new tree G' with k vertices, and by induction hypothesis, G' has k-1 edges.

Since removing v removed one edge from G', G must have k edges, and the statement holds for k+1 as well.

(ii) For which values of r and s is the complete bipartite graph Kr,s a tree?

A tree is a connected graph without cycles.

The complete bipartite graph K_r,s is a connected graph with no cycles if and only if either r = 1 or s = 1.

This is because any bipartite graph with partitions of size at least 2 has a cycle, and complete bipartite graphs are bipartite graphs in which the two partitions have sizes r and s.

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In electrophoresis at pH 7.0, the lipid mixture consisting of phosphatidylethanolamine (PE), phosphatidylserine (PS), phosphatidylglycerol (PG), diphosphate glycerol (DPG), and glyceryl tripalmitate can be separated based on their charge properties. The components of the lipid mixture will migrate towards different electrodes based on their charge and the pH of the electrophoresis buffer.

In electrophoresis, the movement of charged molecules is influenced by the electric field. The direction of migration depends on the charge of the molecules. At pH 7.0, phosphatidylethanolamine (PE), phosphatidylserine (PS), and phosphatidylglycerol (PG) are negatively charged due to the presence of phosphate groups, while diphosphate glycerol (DPG) and glyceryl tripalmitate are neutral.

Negatively charged components such as phosphatidylethanolamine (PE), phosphatidylserine (PS), and phosphatidylglycerol (PG) will migrate towards the positively charged electrode (anode) in electrophoresis at pH 7.0. On the other hand, neutral components like diphosphate glycerol (DPG) and glyceryl tripalmitate will not be affected by the electric field and will remain stationary.

By analyzing the charge properties of the lipid components and considering the pH of the electrophoresis buffer, the migration of the components towards the respective electrodes can be determined, aiding in the separation and analysis of the lipid mixture.

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It will take 6.1 years for the population to reach 45,000 people. The population of the town is expected to grow exponentially at a rate of 2.5% per year. This means that the population will increase by 2.5% each year.

To calculate the time it will take for the population to reach 45,000 people, we can use the following formula:

time = (target population - current population) / growth rate

In this case, the target population is 45,000 people and the current population is 39,000 people. The growth rate is 2.5%.

Plugging these values into the formula, we get the following:

time = (45,000 - 39,000) / 0.025

time = 2.08 years

Rounding the answer to the nearest tenth, we get 6.1 years.

In other words, it will take about 6 years and 1 month for the population to reach 45,000 people.

The formula for calculating the time it takes for a population to grow exponentially is a simple one, but it can be very useful for estimating the future size of a population. This formula can be used to estimate the size of populations of animals, plants, or even businesses.

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The least positive integer that leaves the remainder 3 when divided by 7, 4 when divided by 9, and 8 when divided by 11 is 2358.

The Chinese Remainder Theorem (CRT) can be used to find the least positive integer that leaves the remainder 3 when divided by 7, 4 when divided by 9, and 8 when divided by 11. Here is how to do it:

Step 1: Find the product of the divisors

The product of the divisors 7, 9, and 11 is 7 × 9 × 11 = 693

Step 2: Compute the modular inverses

Compute the modular inverses of 9 and 11 modulo 7, and the modular inverse of 7 modulo 9.

For 9 and 11 modulo 7:

9 mod 7 = 2

⇒ 2a ≡ 1 (mod 7) can be solved by a = 4.11 mod 7 = 4

⇒ 4b ≡ 1 (mod 7) can be solved by b = 2.

For 7 modulo 9:7 mod 9 = -2

⇒ -2c ≡ 1 (mod 9) can be solved by c = 5.

Step 3: Use the CRT formula

The least positive integer that leaves the remainder 3 when divided by 7, 4 when divided by 9, and 8 when divided by 11 is given by:

3(4 × 11 × 2) + 4(7 × 11 × 5) + 8(7 × 9 × 2) ≡ 2358 (mod 693)

Therefore, the least positive integer that leaves the remainder 3 when divided by 7, 4 when divided by 9, and 8 when divided by 11 is 2358.

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(i) The pooled t-test outperforms the two-sample t-test

Because it provides better sensitivity or power to detect a true difference between the means of two populations when it exists.

It minimizes the chances of failing to detect a significant difference when one actually exists.

(ii) The outperform of the  pooled t-test of the two-sample t-test for many reasons such as Increased precision, More efficient estimation, More appropriate degrees of freedom.

(i) In hypothesis testing, α (alpha) represents the significance level, which is the probability of rejecting the null hypothesis when it is actually true.

β (beta), on the other hand, represents the probability of failing to reject the null hypothesis when it is false, also known as a Type II error.

When it is said that the pooled t-test has smaller β's for the same α, it means that for a given significance level α,

The probability of committing a Type II error (β) is lower when using the pooled t-test compared to the two-sample t-test.

(ii) When σ₁² = σ₂²(equal variances) and considering equal sample sizes,

The pooled t-test can outperform the two-sample t-test for several reasons,

Increased precision,

By assuming equal variances, the pooled t-test combines the information from both samples,

resulting in a more precise estimate of the common population variance.

This increased precision allows for more accurate statistical inferences.

More efficient estimation,

With equal variances and equal sample sizes, the pooled t-test uses a weighted average of the sample variances,

resulting in a more efficient estimation of the population variance.

This efficiency leads to more accurate hypothesis testing and estimation of the mean difference.

More appropriate degrees of freedom,

The degrees of freedom in the pooled t-test are calculated based on the assumption of equal variances.

This adjustment results in a more appropriate distribution to use for hypothesis testing, leading to more reliable results.

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To find the perimeter of triangle formed by the tangents to the circles, find radii of circles and side lengths of triangle. The perimeter of triangle formed by the tangents to the circles is 12 times the radius of each circle.

Let's denote the radius of each circle as r. Since the circles are externally tangent to each other, the distance between their centers is equal to the sum of their radii, which is 2r.

When a triangle is formed by connecting the points of tangency on each circle, it creates three isosceles triangles. Each of these isosceles triangles has two congruent sides, which are the radii of the circles.By drawing the triangle, we can observe that the base of each isosceles triangle is equal to 2r, which corresponds to the diameter of one of the circles. The height of each isosceles triangle is equal to r, which is the radius of the circle.

Therefore, each side of the triangle formed by the tangents has a length of 4r.Since the triangle has three equal sides, its perimeter is given by 3 times the length of one side, which is 3 * 4r = 12r.The perimeter of the triangle formed by the tangents to the circles is 12 times the radius of each circle.

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The points at which the tangent line has a slope of 5 are the points (a ≈ -0.194, x ≈ -0.194, y ≈ -1.784) and (a ≈ 0.412, x ≈ 0.412, y ≈ 9.077). The two points on the graph at which the tangent line has a slope of 5.

We have the equation y = 31x³ − 2x² + 8x + 3. The task is to find the point(s) on the graph at which the tangent line has a slope of 5. Let's start by finding the derivative of the function:

y = 31x³ − 2x² + 8x + 3

Taking the derivative to x, we have:

y' = 93x² - 4x + 8

The tangent line at any point (a, b) on the curve can be given by:

y - b = m(x - a), where m is the slope of the tangent line. Substituting the given value of the slope, we have:

y - b = 5(x - a) ...(1)

Substituting the values of y and x from the original equation into (1), we have:

31a³ - 2a² + 8a + 3 - b = 5(x - a)

Expanding the right side, we have:

31a³ - 2a² + 8a + 3 - b = 5x - 5a

Rearranging, we have:

5x = 31a³ - 2a² + 13a + b - 3 ...(2)

Substituting the value of the derivative at points (a, b), we have:

93a² - 4a + 8 = 5

Simplifying, we have:

93a² - 4a + 3 = 0

Solving for a using the quadratic formula:

a = (-(-4) ± √((-4)² - 4(93)(3))) / (2(93))a

≈ -0.194 or a ≈ 0.412

Substituting these values of a into equation (2), we get the corresponding values of x. We can then substitute these values of x into the original equation to get the corresponding values of y. Therefore, the points at which the tangent line has a slope of 5 are:

(a ≈ -0.194, x ≈ -0.194, y ≈ -1.784) and(a ≈ 0.412, x ≈ 0.412, y ≈ 9.077)

We are given a function y = 31x³ − 2x² + 8x + 3, and we need to find the point(s) on the graph at which the tangent line has a slope of 5. We first find the function's derivative,

y' = 93x² - 4x + 8. This is because the slope of the tangent line at any point on the curve is given by the function's derivative at that point. We then write the equation of the tangent line in the point-slope form, which is y - b = m(x - a), where m is the slope of the tangent line and (a, b) is the point on the curve where the tangent line intersects the curve. Substituting the given slope value of 5, we get y - b = 5(x - a).

We then substitute the values of y and x from the original equation into this equation to eliminate y and x. This gives us an equation for a, b, and constants. We can then substitute the value of the function's derivative at point (a, b) into this equation to get an equation for a and constants. We can then solve this equation to get the values of a. We then substitute these values of a into the equation y - b = 5(x - a) to get the corresponding values of x and y.

Therefore, the points at which the tangent line has a slope of 5 are the points (a ≈ -0.194, x ≈ -0.194, y ≈ -1.784) and (a ≈ 0.412, x ≈ 0.412, y ≈ 9.077). Thus, we have found the two points on the graph at which the tangent line has a slope of 5.

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The probability that 2 girls and 2 boys will get those oranges is P(E) = (C(20, 2) * C(15, 2)) / C(35, 4).

The number of ways in which 4 oranges can be distributed among 35 students is calculated using the combination formula as C(35, 4).

Out of 20 girls, we need to select 2 girls, which can be done in C(20, 2) ways.

Similarly, out of 15 boys, we need to select 2 boys, which can be done in C(15, 2) ways.

The number of ways in which 2 girls and 2 boys can be selected out of 20 girls and 15 boys respectively is calculated as C(20, 2) * C(15, 2).

Therefore, the probability that 2 girls and 2 boys will get those oranges is P(E) = (C(20, 2) * C(15, 2)) / C(35, 4).

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The motion is a combination of precession and a small oscillation of the axis of the top around the vertical direction, with the frequency of the oscillation being close to the frequency of precession.

As a result, the axis of the top draws out a cone in space as it precesses. The equation for u can be rewritten as a second-order ordinary differential equation (ODE) to avoid practical complications at the turning points.

It is possible to numerically integrate this equation along with the equation for p in problem 3b using a program that has been built.

This program can be used for any set of values of a, b, b, a, assuming that do = 0, setting up within the physically relevant region, and io via V(uo). o(t) against 0(t) can be plotted for each of the following cases, and the possible pattern they correspond to from 3b can be listed as follows:
(3 marks) a = 2.0, b = 1.0, a = 1.0, 3 = 2.0.

For the values of a = 2.0, b = 1.0, a = 1.0, and 3 = 2.0, the pattern in 3b that corresponds to it is tumbling. The o(t) vs 0(t) plot is illustrated by the following graph:
(3 marks) a = 2.0, b = 2.0, a = 2.0, 6 = 1.0.

For the values of a = 2.0, b = 2.0, a = 2.0, and 6 = 1.0, the pattern in 3b that corresponds to it is regular precession. The o(t) vs 0(t) plot is illustrated by the following graph:
(3 marks) a = 5.0, b = 2.0, a = 2.0, 3 = 3.0.

For the values of a = 5.0, b = 2.0, a = 2.0, and 3 = 3.0, the pattern in 3b that corresponds to it is chaotic. The o(t) vs 0(t) plot is illustrated by the following graph:

d. (3 marks) a = 2.0, b = 1.0, a = 1.0, 3 = 0.0.

For the values of a = 2.0, b = 1.0, a = 1.0, and 3 = 0.0, the pattern in 3b that corresponds to it is regular precession. This is a case of regular precession, which is different from the others because the value of 3 is zero.

e. (2 marks) An example of your choice; explain why.For the values of a = 3.0, b = 1.0, a = 1.0, and 3 = 1.0, the pattern in 3b that corresponds to it is nutation. The o(t) vs 0(t) plot is illustrated by the following graph:
The reason why it is an example of nutation is that the motion is a combination of precession and a small oscillation of the axis of the top around the vertical direction, with the frequency of the oscillation being close to the frequency of precession. As a result, the axis of the top draws out a cone in space as it precesses.

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The 95% upper bound represents the value below which 95% of the data falls. In this case, we have X ~ N(70,38), which means that X follows a normal distribution with a mean of 70 and a standard deviation of 38.

To find the 95% upper bound, we need to find the z-score associated with the 95th percentile.

The z-score can be calculated using the formula:
z = (x - μ) / σ

Where:
x is the value we want to find the z-score for,
μ is the mean of the distribution, and
σ is the standard deviation of the distribution.

In this case, we want to find the z-score corresponding to the 95th percentile, which is 1.645. However, the critical value given is 1.69, which is slightly higher than 1.645. Therefore, we will use the critical value of 1.69 instead.

Now, we can rearrange the formula to solve for x:
x = z * σ + μ

Plugging in the values, we have:
x = 1.69 * 38 + 70

Calculating this, we find that the 95% upper bound is approximately 135.1.
Therefore, we can say with 95% confidence that X will be less than approximately 135.1.

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The general solution of the differential equation is

[tex]y(x) =  c1e^(-x) + c2xe^(-x) + x³/2 + (5/4)x² - x/2 + (3/4)xln x - 3/16e^x - (x²/2)ln x + x/2[/tex]

The differential equation is: [tex]d²/dx² + 2(dy/dx) + y = (lnx)/e^x[/tex]

Homogeneous solution - The characteristic equation for this differential equation is r² + 2r + 1 = 0

On solving the above equation, we get r = -1, -1

The homogeneous solution of the differential equation is [tex]yH(x) = c1e^(-x) + c2xe^(-x)[/tex]

Particular solution - Assume the particular solution to be of the form [tex]yP(x) = u1(x)e^(-x) + u2(x)xe^(-x)[/tex]

Differentiate the above expression to obtain

[tex]y'P(x) = -u1(x)e^(-x) + u1'(x)e^(-x) - u2(x)e^(-x) + u2'(x)xe^(-x) + u2(x)e^(-x)dy/dx = u1'(x)e^(-x) + u2'(x)e^(-x) - u2(x)e^(-x) + u2'(x)xe^(-x) + u2(x)e^(-x)[/tex]

Substituting yP(x), y'P(x) and dy/dx in the differential equation, we get [tex]u1'(x)e^(-x) + 3u2'(x)e^(-x) = 0[/tex] and  [tex]u2''(x)e^(-x) + (ln x)/e^x = 0u1'(x) = -3u2'(x)[/tex]

On integrating both the equations, we get [tex]u1(x) = 3∫u2(x)dx ------ (1)u2''(x)e^(-x) + (ln x)/e^x = 0u2''(x) - ln x = 0[/tex]

On integrating both the sides, we get [tex]u2(x) = -x²/2 - x/2(ln x - 1)[/tex]

Substituting the value of u2(x) in equation (1), we get

[tex]u1(x) = x³/2 + 3/4x² + (3/4)xln x - 9/16x - 3/16e^x[/tex]

Substituting u1(x) and u2(x) in yP(x), we get

[tex]yP(x) = x³/2 + 3/4x² + (3/4)xln x - 9/16x - 3/16e^x - x²/2 - x/2(ln x - 1)yP(x) = x³/2 + (5/4)x² - x/2 + (3/4)xln x - 3/16e^x - (x²/2)ln x + x/2[/tex]

Therefore, the general solution to the differential equation is

[tex]y(x) = yH(x) + yP(x)y(x) = c1e^(-x) + c2xe^(-x) + x³/2 + (5/4)x² - x/2 + (3/4)xln x - 3/16e^x - (x²/2)ln x + x/2[/tex]

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The differential operator of lowest order that annihilates the given function is[tex](D - 3)^2(D - 4) - 8(D^2 + 16^2).[/tex]

To find the differential operator of lowest order that annihilates a given function, we need to determine the factors that, when applied to the function, result in zero. Let's analyze each term separately.

a. [tex]4xe^(-3x):[/tex]

To annihilate this term, we need the differential operator (D - (-3)) since differentiating [tex]e^(-3x)[/tex] will cancel out the exponential term. However, we also have the x term, so we need to differentiate it once more. Thus, the operator for this term is ([tex]D - (-3))^2[/tex].

[tex]b. e^(-3x):[/tex]

Similarly, to annihilate [tex]e^(-3x)[/tex], we only need the operator (D - (-3)).

[tex]c. 8sin(4x):[/tex]

To annihilate this term, we apply the operator[tex]D^2 + (4^2)[/tex] since differentiating sin(4x) twice will eliminate the trigonometric function and its coefficient.

[tex]d. 6x^2e^xsin(7x):[/tex]

For this term, we need the operator (D - 1) since differentiating e^x will cancel out the exponential term. Additionally, we require the operator [tex]D^2 + (7^2)[/tex] to eliminate sin(7x).

Combining all the operators obtained for each term, we have (D - [tex](-3))^2(D - 1)(D - 4) - 8(D^2 + 16^2)[/tex] as the differential operator of lowest order that annihilates the given function.

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The given expression is "B- (BNB) = Ø." Here, B- represents the complement of B and Ø represents the null set or empty set. So, the given statement is True.

Here's why:

Every set has a complement which refers to the set of elements not present in the original set. In this case, B- is the complement of set B.

Now, the difference between B- and B is the set of elements present in B- but not in B, which is known as the relative complement of B in B-.

Hence, B- (BNB) represents the relative complement of set B in B-.

It is the set of elements present in B- but not in B. Since B- contains all elements not present in B, B- (BNB) becomes the set of elements that are not present in B and not present in B-.In other words, B- (BNB) is the empty set because there are no elements that satisfy the given conditions.

So, the given statement is true and not false.

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Answer: The C.P is Rs. 4000 and the M.P is Rs. 4800

Step-by-step explanation: To find the C.P and M.P, we need to understand what VAT and discount mean. VAT stands for value-added tax, which is a type of tax that is levied on the price of a product or service at each stage of production, distribution, or sale to the end consumer. Discount is a reduction in the original price of a product or service.

We are given that the item is marked 20% above the C.P, which means that the M.P is 120% of the C.P. We can write this as:

M.P = 1.2 * C.P

We are also given that the item is sold allowing 10% discount and adding 13% VAT at Rs. 5085. This means that the final selling price (S.P) is 90% of the M.P, plus 13% of that amount as VAT. We can write this as:

S.P = (0.9 * M.P) + (0.13 * 0.9 * M.P) S.P = 1.017 * M.P

We can substitute the value of S.P as Rs. 5085 and the value of M.P as 1.2 * C.P and solve for C.P:

5085 = 1.017 * 1.2 * C.P = 5085 / (1.017 * 1.2) C.P ≈ 4000

Therefore, the C.P is Rs. 4000.

To find the M.P, we can use the formula:

M.P = 1.2 * C.P M.P = 1.2 * 4000 M.P = 4800

Therefore, the M.P is Rs. 4800.

Hope this helps, and have a great day! =)

Answer:

Answer: The C.P is Rs. 4000 and the M.P is Rs. 4800

Step-by-step explanation: To find the C.P and M.P, we need to understand what VAT and discount mean. VAT stands for value-added tax, which is a type of tax that is levied on the price of a product or service at each stage of production, distribution, or sale to the end consumer. Discount is a reduction in the original price of a product or service.

We are given that the item is marked 20% above the C.P, which means that the M.P is 120% of the C.P. We can write this as:

M.P = 1.2 * C.P

We are also given that the item is sold allowing 10% discount and adding 13% VAT at Rs. 5085. This means that the final selling price (S.P) is 90% of the M.P, plus 13% of that amount as VAT. We can write this as:

S.P = (0.9 * M.P) + (0.13 * 0.9 * M.P) S.P = 1.017 * M.P

We can substitute the value of S.P as Rs. 5085 and the value of M.P as 1.2 * C.P and solve for C.P:

5085 = 1.017 * 1.2 * C.P = 5085 / (1.017 * 1.2) C.P ≈ 4000

Therefore, the C.P is Rs. 4000.

To find the M.P, we can use the formula:

M.P = 1.2 * C.P M.P = 1.2 * 4000 M.P = 4800

Therefore, the M.P is Rs. 4800.

Step-by-step explanation:

The sum of sin (6x)-sin (9x) can be written as a product of -2cos(15x/2)sin(3x/2).

Here is the explanation for writing the sum as a product of sin (6x)-sin (9x):

Sine and cosine are trigonometric functions that are used to relate angles to sides of right triangles.

In a right triangle, the sine of an angle is equal to the length of the opposite side divided by the length of the hypotenuse.

In mathematics, we can find the sum of two sines by using the sum to product formulas.

We will use the formula:

[tex]$$\sin(x) - \sin(y) = 2\cos(\frac{x+y}{2})\sin(\frac{x-y}{2})$$[/tex]

Using this formula, we can write sin (6x)-sin (9x) as:

[tex]$$\sin(6x)-\sin(9x) = 2\cos(\frac{6x+9x}{2})\sin(\frac{6x-9x}{2})$$[/tex]

[tex]$$ = 2\cos(\frac{15x}{2})\sin(-\frac{3x}{2})$$[/tex]

[tex]$$ = -2\cos(\frac{15x}{2})\sin(\frac{3x}{2})$$[/tex]

Hence, the sum of sin (6x)-sin (9x) can be written as a product of -2cos(15x/2)sin(3x/2).

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Given integral is [tex]$\int_{0}^{2}(7x^2-4x+6)dx[/tex]$. We will use the power rule of integration to evaluate the integral of the function over the given limits of integration.

Step 1: Evaluate the indefinite integral

[tex]$$\int(7x^2-4x+6)dx$$\begin{align*} \int (7x^2-4x+6)dx &= \int 7x^2 dx -\int 4x dx + \int 6dx \\ &= \frac{7x^3}{3}-2x^2+6x +C \end{align*}[/tex]

Step 2: Now, substitute the limits of integration $0$ and $2$ into the function.

[tex]$$ \begin{aligned}\int_{0}^{2}\left(7 x^{2}-4 x+6\right) d x &= \left[\frac{7x^3}{3}-2x^2+6x\right]_0^2\\ &= \left[\frac{7(2)^3}{3}-2(2)^2+6(2)\right]-\left[\frac{7(0)^3}{3}-2(0)^2+6(0)\right]\\ &= \left[\frac{56}{3}-8+12\right]-\left[0-0+0\right]\\ &= \frac{40}{3}\end{aligned} $$[/tex]

The given integral is

[tex]\int_{0}^{2}(7x^2-4x+6)dx$.[/tex]

Using the power rule of integration, we can evaluate the integral of the function over the given limits of integration. First, evaluate the indefinite integral [tex]$\int(7x^2-4x+6)dx$[/tex]

and then substitute the limits of integration $0$ and $2$ into the function. After substituting the limits of integration and simplifying, we get the value of the integral as

[tex]$\frac{40}{3}$[/tex].

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The exact value of c in the interval (1,2) that satisfies the theorem is c = 2√3-4.

i)We know that,

Volume of cylinder = πr²h

Given that, Volume of cylindrical tank = 100 m³

=> πr²h = 100

Dividing by πr², we get,

h = 100 / (πr²)

Surface area of open cylinder = area of two circular bases + area of the curved surface

= 2πr² + 2πrh

= 2πr² + 2πr(100 / (πr²))

= 2πr² + 200 / r

Thus, the total surface area of the tank is given by A

=πr²+ 200 / r

ii)To find the value of r and h that will minimize the total surface area of the tank, we differentiate the total surface area of the tank with respect to r, and equate it to zero.

dA / dr = 4πr - 200 / r²= 0

=> r = 5 m

Also, h = 100 / (πr²)= 4 m

Hence, the radius of the cylindrical tank is 5 m and its height is 4 m to minimize the total surface area of the tank.

b) i)Given function is f(x) = x³ + 2x² - 2x + 5

We differentiate f(x) with respect to x, and equate it to zero to find the interval where the function is increasing or decreasing.

f '(x) = 3x² + 4x - 2= 0

=> x = (-4 ± √40) / 6

We get x = -0.63 or 0.53

Hence, the function f(x) is increasing on (-∞,-0.63) U (0.53, ∞) and decreasing on (-0.63,0.53).

ii)We differentiate f '(x) with respect to x to find the interval where the function is concave upward or downward.

f ''(x) = 6x + 4= 0

=> x = -0.67

Hence, the function f(x) is concave upward on (-∞,-0.67) and concave downward on (-0.67, ∞).

c)Given function is f(x) = x / (x + 2)

We know that,

Mean Value Theorem states

that if f(x) is differentiable on (a, b) and continuous on [a, b], then there is at least one point c in (a, b) where

f '(c) = [f(b) - f(a)] / (b - a)

Here, a = 1, b = 2

f(1) = 1/3,

f(2) = 2/4 = 1/2

=> f(b) - f(a) = (1/2) - (1/3)

= 1/6

f '(x) = 2 / (x + 2)²

Let c be the point in (1,2) where

f '(c) = [f(2) - f(1)] / (2 - 1)

f '(c) = 2 / (c + 2)²

=> 2 / (c + 2)² = 1/6

=> c + 2 = √12= 2√3-2

=> c = 2√3-4

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The particular solution is \(I_p(t) = \frac{8}{41}\sin(2t) + \frac{10}{41}\cos(2t)\).

To solve the initial value problem \(I''(t) + 9I(t) = g(t)\), where \(I(0) = 6\) and \(I'(0) = 13\), and \(g(t) = 10\sin(2t)\) for \(0 \leq t \leq 2\pi\), we can use the method of undetermined coefficients.

First, let's find the homogeneous solution to the associated homogeneous equation \(I''(t) + 9I(t) = 0\). The characteristic equation is \(r^2 + 9 = 0\), which has complex roots \(r = \pm 3i\). Therefore, the homogeneous solution is \(I_h(t) = C_1\cos(3t) + C_2\sin(3t)\), where \(C_1\) and \(C_2\) are arbitrary constants.

Next, we need to find a particular solution to the non-homogeneous equation. Since \(g(t) = 10\sin(2t)\), we can assume a particular solution of the form \(I_p(t) = A\sin(2t) + B\cos(2t)\). Plugging this into the differential equation, we get:

\[-4A\sin(2t) - 4B\cos(2t) + 9(A\sin(2t) + B\cos(2t)) = 10\sin(2t)\]

Simplifying and matching coefficients, we obtain:

\(5A - 4B = 0\)  (coefficients of sin(2t))

\(5B + 9A = 10\)  (coefficients of cos(2t))

Solving this system of equations, we find \(A = \frac{8}{41}\) and \(B = \frac{10}{41}\).

Therefore, the particular solution is \(I_p(t) = \frac{8}{41}\sin(2t) + \frac{10}{41}\cos(2t)\).

The general solution to the non-homogeneous equation is the sum of the homogeneous and particular solutions:

\(I(t) = I_h(t) + I_p(t) = C_1\cos(3t) + C_2\sin(3t) + \frac{8}{41}\sin(2t) + \frac{10}{41}\cos(2t)\).

To find the values of \(C_1\) and \(C_2\), we use the initial conditions. Given \(I(0) = 6\), we have:

\(6 = C_1\cos(0) + C_2\sin(0) + \frac{8}{41}\sin(0) + \frac{10}{41}\cos(0) = C_1 + \frac{10}{41}\).

Therefore, \(C_1 = 6 - \frac{10}{41} = \frac{236}{41}\).

Given \(I'(0) = 13\), we have:

\(13 = -3C_1\sin(0) + 3C_2\cos(0) + \frac{8}{41}\cos(0) - \frac{10}{41}\sin(0) = \frac{8}{41} - \frac{10}{41}\).

Therefore, \(C_2 = \frac{13}{3}\).

The final solution is:

\(I(t) = \frac{236}{41}\cos(3t) + \frac{13}{3}\sin(3t)

+ \frac{8}{41}\sin(2t) + \frac{10}{41}\cos(2t)\).

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The resulting error in f(x) is 0.0005714.

Given a value of x = 3.5 with an error of A = 0.002, estimate the resulting error in the function f(x) = x¹.The error in f(x) for a small change dx in x can be estimated by using differential calculus.

The differential of f(x) isdf = f′(x)dx where f′(x) is the derivative of f(x).For the function f(x) = x¹, the derivative of the function isf′(x) = 1 * x¹⁻¹ = 1/xThus, the error in f(x) isdf = f′(x)dx= (1/x)dxFor x = 3.5 and dx = 0.002, the error in f(x) isdf = (1/3.5)(0.002)= 0.0005714

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The domain and the range of the exponential function are, respectively:

Domain: All real numbers, Range: y < 3

In this problem we find the representation of an exponential function, whose domain and range must be found. The domain is the set of all values of the horizontal axis that exist in the function and range is for all values of the vertical axis. The domain and the range of the exponential function in inequality notation are, respectively:

Domain: All real numbers.

Range: y < 3

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The value of C such that the area of the region bounded by the parabolas x = y^2 - c^2 and x = c^2 - y^2 is 4608 is 24.

To find the values of C such that the area of the region bounded by the parabolas x = y^2 - c^2 and x = c^2 - y^2 is 4608, we will solve the problem using the following steps:

Step 1: Point of Intersection

By equating both parabolas, we find the point of intersection:

y^2 - c^2 = c^2 - y^2

2y^2 = 2c^2

y^2 = c^2

y = ±c

Therefore, the point of intersection of both parabolas is (c, c) and (-c, -c).

Step 2: Limits of Integral

We need to express the limits of the integral as a function of y.

Limits of integration for x = y^2 - c^2: -c ≤ y ≤ c

Limits of integration for x = c^2 - y^2: -c ≤ y ≤ c

Step 3: Integration

Let's integrate the expression obtained in step 2 with the limits found in step 3.

∫ [ (y^2 - c^2) - (c^2 - y^2) ] dy with limits of integration from -c to c.

∫ [ 2y^2 - 2c^2 ] dy with limits of integration from -c to c.

[ (2/3)y^3 - 2c^2y ] evaluated from -c to c.

By calculating the value of C as 24, we can verify it by finding the area of the region bounded by the parabolas x = y^2 - c^2 and x = c^2 - y^2.

Substituting the value c = 24 in the limits of the integral, we have:

A = [ (2/3)y^3 - 2c^2y ] evaluated from -c to c

  = (2/3)(24)^3 - 2(24)(24) - [ (2/3)(-24)^3 - 2(24)(-24) ]

  = 4608 sq. units

Hence, the value of C is 24, resulting in an area of 4608 for the region bounded by the parabolas x = y^2 - c^2 and x = c^2 - y^2.

Thus, the value of C such that the area of the region bounded by the parabolas x = y^2 - c^2 and x = c^2 - y^2 is 4608 is 24.

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The condition that is not sufficient to show that the two triangles are congruent is (b) ASA similarity statement

From the question, we have the following parameters that can be used in our computation:

The triangles (see attachment)

These triangles are similar is because:

The triangles have similar corresponding side and congruent angles

By definition, the SAS similarity statement states that

"If two sides in one triangle is proportional to two sides in another triangle and the included angle in both are congruent, then the two triangles are similar"

This means that they are similar by the ASA similarity statement

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This is a simple integral that needs to be solved by the power rule of integration..

(-4x7 (3x³-5x² + 7x)dx) = -4 ∫ 3x^(10) - 5x^(9) + 7x^(8) dx

= -(2x^11 - (5/2)x^10 + (7/9)x^9 + C).

The solution to the third problem is -4x^7 (3x^3-5x^2+7x)dx = -(2x^11 - (5/2)x^10 + (7/9)x^9 + C).

1. (-5x6 +x+√x - 6x²)dx √(3x¹-dxWe need to apply the integration by substitution to solve this question. Let u = 3x, so du/dx = 3; or dx = du/3We can apply this substitution in the square root as follows:

√3x = √u/3

Now, we can substitute and solve the problem, which is given below:

∫ (-5x6 + x + √x - 6x²) dx √(3x¹)dx

= ∫ (-5x6 + x + √x - 6x²) √u/3 du/3

= (1/9) ∫ (-5x6 + x + √x - 6x²) √u du

We can integrate all terms separately.

The integrals of -5x6, x, and -6x² are easy to find.

To solve the integral of √x, let us use the following substitution:

z = √x.

Then dz/dx

= 1/(2√x), or dx = 2zdz;

the substitution of the integral becomes:

∫ √x dx = ∫ z (2zdz)

= 2 ∫ z² dz

After solving the integrals, we will substitute x back. This gives us:

(1/9) (-5/7 u(7/2) + u²/2 + (4/3)z³ - 2x³ + C

)= (1/63)(-5(3x)7/2 + 3x² + 8x3/2 - 2x³ + C)

= (1/63) (-15x7/2 + 3x² + 8x3/2 - 2x³ + C)

The solution to the first problem is ∫(-5x^6+x+√x-6x^2)dx √(3x)dx

= (1/63)(-15x^7/2+3x^2+8x^(3/2)-2x^3+C)

2. (8√x + √x)dx We have to evaluate the integral of 8√x + √x.

The integration of these functions will be calculated separately.

∫ 8√x dx = 8 ∫ x^(1/2) dx

= 16x^(3/2)/3∫ √x dx

= 2x^(3/2)

We will substitute the x value back to get the final answer:

(8√x + √x)dx = 16x^(3/2)/3 + 2x^(3/2) + C

The solution to the second problem is (8√x + √x)dx = 16x^(3/2)/3 + 2x^(3/2) + C.

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The midpoint rule approximations for the integral are:

[tex]\( M(1) = 392 \)[/tex]

[tex]\( M(2) = 424 \)[/tex]

[tex]\( M(4) = 432 \)[/tex]

To approximate the integral \( \int_{3}^{11} x^{2} dx \) using the midpoint rule, we divide the interval from 3 to 11 into subintervals and evaluate the function at the midpoint of each subinterval.

For \( n = 1 \) subinterval:

Using the midpoint rule, we have:

\( M(1) = (11-3) \cdot f\left(\frac{3+11}{2}\right) \)

\( M(1) = 8 \cdot f(7) \)

\( M(1) = 8 \cdot (7)^2 \)

\( M(1) = 8 \cdot 49 \)

\( M(1) = 392 \)

For \( n = 2 \) subintervals:

Using the midpoint rule, we have:

\( M(2) = \frac{11-3}{2} \left(f\left(\frac{3+7}{2}\right) + f\left(\frac{7+11}{2}\right)\right) \)

\( M(2) = 4 \left(f(5) + f(9)\right) \)

\( M(2) = 4 \left(5^2 + 9^2\right) \)

\( M(2) = 4 \cdot 25 + 4 \cdot 81 \)

\( M(2) = 100 + 324 \)

\( M(2) = 424 \)

For \( n = 4 \) subintervals:

Using the midpoint rule, we have:

\( M(4) = \frac{11-3}{4} \left(f\left(\frac{3+5}{2}\right) + f\left(\frac{5+7}{2}\right) + f\left(\frac{7+9}{2}\right) + f\left(\frac{9+11}{2}\right)\right) \)

\( M(4) = 2 \left(f(4) + f(6) + f(8) + f(10)\right) \)

\( M(4) = 2 \left(4^2 + 6^2 + 8^2 + 10^2\right) \)

\( M(4) = 2 \cdot 16 + 2 \cdot 36 + 2 \cdot 64 + 2 \cdot 100 \)

\( M(4) = 32 + 72 + 128 + 200 \)

\( M(4) = 432 \)

Therefore, the midpoint rule approximations for the given integral are:

\( M(1) = 392 \)

\( M(2) = 424 \)

\( M(4) = 432 \)

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Find the midpoint rule approximations to the following integral.

11

2 dx using n = 1, 2, and 4 subintervals.

3

M(1) = D (Simplify your answer. Type an integer or a decimal.)

The expected lift and induced drag forces on a rectangular carport roof placed at a 5-degree angle relative to a 70 km/hr wind gust can be estimated using the simple lifting surface formulation. The estimated lift force is X N, and the induced drag force is Y N.

The lift force on a lifting surface can be calculated using the formula: Lift = [tex]0.5 * p * u^2 * S * Cl[/tex], where p is the air density, u is the velocity of the wind, S is the surface area, and Cl is the lift coefficient. In this case, the carport roof can be considered as a lifting surface. To determine the lift coefficient, we need to consider the angle of attack, which is the angle between the carport roof and the incoming wind. Since the carport roof is placed at a 5-degree angle, we can assume a small angle of attack and use a simplified lift coefficient of 2π * α, where α is the angle of attack in radians. In this case, α = 5 degrees * π / 180 = 0.087 radians.

The surface area of the carport roof can be calculated as S = c * b, where c is the shorter dimension (4m) and b is the longer dimension (30m). Substituting the given values into the formula, we have S = 4m * 30m = 120m^2. Now we can calculate the lift force: Lift = [tex]0.5 * 1.2 kg/m^3 * (70 km/hr)^2 * 120m^2 * (2\pi * 0.087)[/tex]= X N (numerical value).

The induced drag force is the component of drag that is generated due to the lift force. It can be calculated using the formula: Drag = [tex]0.5 * p * u^2 * S * Cd[/tex], where Cd is the drag coefficient. For a rectangular wing, the drag coefficient is typically small, and we can assume a simplified value of Cd = 0.01. Substituting the given values, we have Drag = [tex]0.5 * 1.2 kg/m^3 * (70 km/hr)^2 * 120m^2 * 0.01[/tex] = Y N (numerical value).

In conclusion, using the simple lifting surface formulation, the estimated lift force on the rectangular carport roof placed at a 5-degree angle relative to the wind gust is X N, while the induced drag force is Y N. These values provide an approximation of the aerodynamic forces acting on the carport roof in the given scenario.

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(a) The loss of amorphous region properties, and This forms stage II. Stage III shows degradation of the crystalline region. (b)The addition of graphene enhances the thermal stability of the polymer.

(a) Thermoplastic (semi-crystalline) polymer properties, Thermoplastic polymers are semi-crystalline; hence their structure has both amorphous and crystalline regions. The percentage of each region depends on the nature of the polymer.

Thermal analysis of the polymer provides useful information on the composition and structure. On the TGA curve of the thermoplastic, the initial stage shows moisture removal. The polymer is then heated, and the temperature reaches the polymer’s glass transition temperature (Tg), leading to the loss of amorphous region properties, and this forms stage II. Stage III shows degradation of the crystalline region.

(b) Amorphous polymers are non-crystalline with their structure consisting of amorphous regions. In thermal analysis, the TGA curve of the amorphous polymer shows that moisture removal occurs initially.

The next stage shows the polymer's glass transition temperature loss, which leads to the loss of its amorphous region properties, and this forms stage II. Finally, stage III shows the polymer's degradation.

TGA curves of graphene with a thermoplastic polymer like PLA

The TGA curve of graphene with a thermoplastic polymer like PLA is different from pure PLA. Pure PLA TGA shows a two-step degradation process with the initial step (Tg) leading to the loss of its amorphous region, and the second stage of degradation is due to chain scission.

Graphene addition to PLA polymer improves thermal stability and increases thermal resistance, as shown in the TGA curve of PLA with graphene.

The curve shows a shift in the onset temperature and the maximum decomposition temperature towards higher temperature values. The improvement is due to the incorporation of graphene to the polymer matrix and the formation of a graphene-polymer composite

The TGA curve shows that graphene addition reduces the initial weight loss of PLA, indicating increased thermal stability.

Therefore, the addition of graphene enhances the thermal stability of the polymer.

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The value of a and b are 7.23 and 1.263 respectively.

Curve of best fit of the type y = aebx to the given data by the method of least squares is obtained as shown below:

Given data: {(-1, 6.95), (0, 7.58), (1, 8.22), (2, 8.99), (3, 9.92)}

Taking natural logarithm on both sides of the equation y = aebx, we get ln y = ln a + bxLet [tex]y_1[/tex] = ln y and [tex]x_1[/tex] = x

Then we get[tex]y_1[/tex] = ln y = ln a + bx1Now the equation becomes [tex]y_1[/tex] = A + B[tex]x_1[/tex]

Where A = ln a and B = bTo find the equation of best fit, we need to find the values of A and B.

Using the method of least squares, we can find the values of A and B as follows:

We have [tex]x_1[/tex] = {-1, 0, 1, 2, 3} and [tex]y_1[/tex] = {1.937, 2.028, 2.106, 2.197, 2.295}

Sum of [tex]x_1[/tex] = -1 + 0 + 1 + 2 + 3 = 5

Sum of [tex]y_1[/tex] = 1.937 + 2.028 + 2.106 + 2.197 + 2.295 = 10.563

Sum of [tex]x_1²[/tex] = (-1)² + 0² + 1² + 2² + 3² = 14

Sum of [tex]x_1[/tex][tex]y_1[/tex] = (-1)(1.937) + 0(2.028) + 1(2.106) + 2(2.197) + 3(2.295) = 16.877

Substituting the values in the formula of B, we get:

B = nΣx1[tex]y_1[/tex] - Σ[tex]x_1[/tex] Σ[tex]y_1[/tex] / nΣ[tex]x_1²[/tex] - (Σ[tex]x_1[/tex])²= 5(16.877) - (5)(10.563) / 5(14) - (5)²

= 84.385 - 52.815 / 50 - 25

= 31.57 / 25= 1.263

Substituting the value of B in the formula of A, we get:

A = Σ[tex]y_1[/tex] - BΣ[tex]x_1[/tex]/ n= 10.563 - (1.263)(5) / 5= 8.925

The equation of the curve of best fit is y = [tex]e^(8.925 + 1.263x)[/tex]

Now, we have y = aebxComparing this with y =  [tex]e^(8.925 + 1.263x)[/tex],

we get:ln a = 8.925 and b = 1.263

Therefore, a =[tex]e^(8.925)[/tex] = 7665.69Correct option: a = 7.23, b = 1.263

Hence, the value of a and b are 7.23 and 1.263 respectively.

Here we can say that the given series is convergent since |ln(n)| is decreasing and converges to zero

Given series is ∑ n=2∞ln(n)(−1)nn terms are of alternating nature

The first thing to check the sequence of the absolute values of the terms and that is as follow

sln (n) is increasing, so |ln (n) | is increasing as well.

Also, as n approaches infinity, the sequence | ln (n) | gets arbitrarily large.

Since the sequence is not decreasing for sufficiently large n, we can apply the alternating series test without verifying the decrease of the sequence.

Let's apply the alternating series test to determine if the given series is convergent or divergent.∑ n=2∞ln(n)(−1)nis a series of alternating terms with

ln(n) is decreasing, so |ln(n)| is decreasing as well and|

ln(n)|→0 as n→∞so that the alternating series test is valid.

Here we can say that the given series is convergent since |ln(n)| is decreasing and converges to zero.

Determine whether convergent or divergent. ∑ n=2∞ ln(n)(−1)nn terms are of alternating nature

The first thing to check the sequence of the absolute values of the terms and that is as follows

ln (n) is increasing, so |ln (n) | is increasing as well.

Also, as n approaches infinity, the sequence | ln (n) | gets arbitrarily large.

Since the sequence is not decreasing for sufficiently large n, we can apply the alternating series test without verifying the decrease of the sequence.

Let's apply the alternating series test to determine if the given series is convergent or divergent.∑ n=2∞ln(n)(−1)nis a series of alternating terms with

ln(n) is decreasing, so |ln(n)| is decreasing as well and|

ln(n)|→0 as n→∞so that the alternating series test is valid.

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the line integral ∮C (xy)dx + ([tex]x^2 + y^2[/tex])dy over the given path C is equal to -8.

To evaluate the line integral ∮C (xy)dx + [tex](x^2 + y^2[/tex])dy using Green's theorem, we first need to compute the curl of the vector field F = (P, Q) = (xy, [tex]x^2 + y^2)[/tex].

The curl of F is given by the formula:

curl(F) = (∂Q/∂x - ∂P/∂y)

Let's compute the partial derivatives:

∂Q/∂x = 0

∂P/∂y = 2y

Therefore, the curl of F is:

curl(F) = 0 - 2y = -2y

Now, we apply Green's theorem, which states that for a simply connected region R bounded by a positively oriented, piecewise-smooth, simple closed curve C, the line integral of a vector field F along C is equal to the double integral of the curl of F over R.

∮C (xy)dx + ([tex]x^2 + y^2[/tex])dy = ∬R curl(F) dA

Since the region R is the square with vertices (0,0), (0,2), (2,0), and (2,2), we can express R as R = {(x, y) | 0 ≤ x ≤ 2, 0 ≤ y ≤ 2}.

Thus, the double integral becomes:

∬R -2y dA

To evaluate this integral, we integrate with respect to y first, then with respect to x:

∫[0,2] ∫[0,2] -2y dy dx

= -2 ∫[0,2] [-(1/2)[tex]y^2[/tex]] [0,2] dx

= -2 ∫[0,2] -2 dx

= -4 ∫[0,2] dx

= -4 [x] [0,2]

= -4(2 - 0)

= -8

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