Let f(x)=∫xx2​(8+2t4)dt Find f′(x)

The value of angle x in the chord diagram is determined as 104⁰.

The value of angle x is calculated by applying intersecting chord theorem, which states that the angle at tangent is half of the arc angle of the two intersecting chords.

Also this theory states that arc angles of intersecting secants at the center of the circle is equal to the angle formed at the center of the circle by the two intersecting chords.

x = ¹/₂ (152⁰ + 56⁰ )

x = ¹/₂ x ( 208 )

x = 104⁰

Thus, the value of angle x is determined as 104⁰, by applying intersecting chord theorem.

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The following expression for the local truncation error: LTE = |[h*u_t + h²/2! u_tt + h³/3! u_ttt + ...]/(1 + α/h²) + αh²/2 u_xx + O(h^4)|

Crank-Nikolson method is used for numerical solutions of partial differential equations. It is an implicit finite difference method used for numerically solving the heat equation and other parabolic partial differential equations.

The given Crank-Nicolson approximation to c0(h) + αu is:αu dx² + c0(h),

For this approximation, we will use the Taylor expansion of the actual solution u(x, t) in terms of step size h.

Since the error calculation has to be made at point (ih, jh), we can replace x with ih and t with jh.

Therefore, the truncated series will be in terms of h.i.e, u(ih, jh) = u + h*u_x + h²*u_xx/2! + h³*u_xxx/3! + ...

Using Taylor series, we can write the actual solution value for the next time step at (i, j+1) as:

u(ih, (j+1)h) = u(ih, jh) + h*u_t + h²/2! u_tt + h³/3! u_ttt + ...

where u_t, u_tt, and u_ttt are the partial derivatives of u with respect to t, evaluated at (ih, jh).

By the definition of Crank-Nicolson method, we can write the approximations for the derivatives as follows:

u_t = (u(ih, (j+1)h) - u(ih, jh))/h

= (u_ijk+1 - u_ijk)/hαu_xx

= (α/2)*((u_ijk - u_ijk-1)/h² + (u_ijk+1 - u_ijk)/h²)

The Crank-Nicolson method approximation can then be written as follows:

u(ih, jh+1) = αu_xx + c0(h)*u(ih, jh)

where c0(h) = (1-α) and u_xx is given above.

Substituting u(ih, (j+1)h) and u_xx into the above approximation equation gives:

u_ijk+1 = [α(u_ijk+1 - u_ijk) + (1-α)u_ijk + (α/2)(u_ijk - u_ijk-1)/h²]/(1 + α/h²)

The local truncation error for the Crank-Nicolson method can be found by subtracting the actual value of u(ih, jh+1) from the approximate value u_ijk+1 and then taking the absolute value, i.e. :

LTE = |u(ih, jh+1) - u_ijk+1|

Where u(ih, jh+1) is given by the Taylor series expansion of the exact solution and u_ijk+1 is the Crank-Nicolson approximation given above.

Substituting the Taylor series expansion of the exact solution into the above equation and simplifying, we get the following expression for the local truncation error:

LTE = |[h*u_t + h²/2! u_tt + h³/3! u_ttt + ...]/(1 + α/h²) + αh²/2 u_xx + O(h^4)|

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The diffusive flux of liquid B through material A is approximately -0.3279 g/m²s.

To calculate the diffusive flux of liquid B through material A, we can use Fick's first law of diffusion, which states that the diffusive flux (J) is equal to the product of the diffusion coefficient (D), the concentration gradient (ΔC), and the area (A) perpendicular to the direction of diffusion.

The concentration gradient (ΔC) is the difference in concentration between the outside and inside surfaces of the material. In this case, the concentration on the outside surface is 0.07 g/cm³ and on the inside surface is 0.05 g/cm³. Therefore, the concentration gradient (ΔC) is 0.07 g/cm³ - 0.05 g/cm³ = 0.02 g/cm³.

We need to convert the thickness of the sheet (0.61 mm) to centimeters by dividing it by 10, since 1 cm = 10 mm. So the thickness (Δx) is 0.61 mm / 10 = 0.061 cm.

Now we can calculate the diffusive flux (J) using the formula J = -D * (ΔC / Δx) * A, where the negative sign indicates that the diffusion occurs from high concentration to low concentration.

Given that the diffusion coefficient (D) is 0.0000001 cm²/s, the concentration gradient (ΔC) is 0.02 g/cm³, and the thickness (Δx) is 0.061 cm, we can now calculate the diffusive flux.

Let's assume the area (A) perpendicular to the diffusion is 1 cm². Plugging in the values, we have:
J = - (0.0000001 cm²/s) * (0.02 g/cm³ / 0.061 cm) * (1 cm²)

Simplifying the expression, we find:
J = - 0.00003279 g/cm²s

To convert the units to g/m²s, we multiply the result by 10,000 (since 1 m² = 10,000 cm²):
J = - 0.3279 g/m²s

Therefore, the diffusive flux of liquid B through material A is approximately -0.3279 g/m²s. The negative sign indicates that the diffusion occurs from the outside surface to the inside surface.

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The term of the arithmetic sequence that is equal to 115 is the 39th term.

To find the term of the arithmetic sequence 1, 4, 7, 10, ... that is equal to 115, we need to determine the value of 'n' in the expression 'a + (n-1)d', where 'a' is the first term of the sequence and 'd' is the common difference.

In this case, the first term 'a' is 1, and the common difference 'd' is 3 (since each term increases by 3).

Let's substitute these values into the equation and solve for 'n':

1 + (n-1)3 = 115

Simplifying the equation:

3n - 2 = 115

Adding 2 to both sides:

3n = 117

Dividing both sides by 3:

n = 39

Therefore, the term of the arithmetic sequence that is equal to 115 is the 39th term.

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David's demand curve for gasoline is given by q1 = Y / p1, where q1 represents the quantity of gasoline consumed, Y represents income, and p1 represents the price of gasoline.

To derive David's demand curve for gasoline, we need to find the quantity of gasoline that David will consume at different prices.

David's utility function is given as U(q1, q2) = 90q1^0.8q2^0.2, where q1 represents the quantity of gasoline consumed and q2 represents the quantity of bread consumed.
To find David's demand for gasoline, we can use the concept of utility maximization. According to this concept, consumers will allocate their income in a way that maximizes their overall utility.

Let's assume David's income is Y, the price of gasoline is p1, and the price of bread is p2.

The total expenditure (TE) for David can be calculated as:
TE = p1 * q1 + p2 * q2
To maximize utility, we need to differentiate the utility function with respect to q1 and set it equal to zero:
dU/dq1 = 0

Differentiating the utility function, we get:
dU/dq1 = 90 * 0.8 * q1^(-0.2) * q2^0.2 = 0
Simplifying the equation, we have:
72 * q2^0.2 = 0

Since q2 is positive, we can divide both sides of the equation by 72 to solve for q2:
q2^0.2 = 0
Taking both sides to the power of 5, we have:
q2 = 0

This implies that David's demand for bread is zero, which means he does not consume any bread.
Substituting this value into the total expenditure equation, we have:
TE = p1 * q1
To find the demand curve for gasoline, we need to solve for q1 in terms of p1 and Y. Rearranging the equation, we get:
q1 = TE / p1 = Y / p1

Therefore, David's demand curve for gasoline is given by q1 = Y / p1, where q1 represents the quantity of gasoline consumed, Y represents income, and p1 represents the price of gasoline.

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Answer:

If Joshua sells a pack of pens for $3.15, which is 5 percent more than he pays for them, we can set up the following equation:

1.05x = 3.15

Here, x represents the amount Joshua pays for a pack of pens.

To solve for x, we can divide both sides of the equation by 1.05:

x = 3.15 / 1.05

Simplifying, we get:

x = 3

Therefore, Joshua pays $3 for a pack of pens.

This equation will have only one solution, which is x = 3.

hope it helps you

Step-by-step explanation:

ONE solution

x = price he pays

1.05x  = price at which  he sells

1.05x = $ 3.15

x = $ 3.15/1.05

The graph of y = ∛(x - 3) is a horizontal translation of the graph of y = ∛x by 3 units to the right.

To determine the graph of y = ∛(x - 3), let's analyze the transformation that has occurred compared to the original function y = ∛x.

Start with the original function y = ∛x, which represents the cube root of x. This function has a vertical shift of 0 and is symmetric about the origin.

The transformation y = ∛(x - 3) indicates a horizontal translation of the graph of y = ∛x. The expression (x - 3) inside the cube root implies a shift of 3 units to the right.

As a result, the graph of y = ∛(x - 3) will have the same shape and characteristics as the graph of y = ∛x but shifted 3 units to the right.

The new graph will intersect the y-axis at the point (3, 0), indicating the translation to the right.

Any other point on the original graph will also be shifted 3 units to the right on the new graph.

The new graph will remain symmetric about the origin and retain the same increasing or decreasing behavior as the original function.

Therefore, the graph of y = ∛(x - 3) is a horizontal translation of the graph of y = ∛x by 3 units to the right.

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a. This is a binomial distribution because there are two outcomes for each passenger: either they are fully vaccinated or they are not.

b. To calculate the probability of observing 4 or fewer passengers out of a sample of 10, we can use the binomial distribution formula. Let X be the number of vaccinated passengers in the sample. We are given that p, the probability of a passenger being vaccinated, is 0.8.

Therefore, q = 1 - p = 0.2. We want to find P(X ≤ 4). Using the binomial distribution formula, we get:P(X ≤ 4) = Σ P(X = x) for x = 0, 1, 2, 3, 4.= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)= C(10,0) p^0 q^10 + C(10,1) p^1 q^9 + C(10,2) p^2 q^8 + C(10,3) p^3 q^7 + C(10,4) p^4 q^6= 0.0001048576 + 0.001572864 + 0.01362336 + 0.07282896 + 0.201326592= 0.289456328So the probability of observing 4 or fewer vaccinated passengers out of a sample of 10 is approximately 0.289.

c. Based on the observed vaccination rate of 4 out of 10 passengers, it appears that the claimed vaccination rate of 80% is not accurate. However, we would need more information and a larger sample size to make a definitive conclusion. The observed rate could be due to chance variation or a biased sample.

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Answer: F

Step-by-step explanation:

F, because side length correlates directly to angle, so the smallest side has the smallest angle.

Therefore, the tip of the man's shadow is moving at a rate of 3.5 ft/s.

We have a right triangle formed by the man, the pole, and his shadow. Let's denote the length of the man's shadow as x and the length of the pole as h. The angle between the ground and the line from the top of the pole to the tip of the shadow is θ.

Using similar triangles, we can write the following proportion:

h / x = (h + 17) / (x + d)

where d represents the distance between the man and the base of the pole (i.e., 34 ft).

Let's differentiate both sides of the equation with respect to time (t):

[tex](dh/dt) / x = (h + 17) / ((x + d)^2) * (dx/dt)[/tex]

We want to find (dh/dt), the rate at which the tip of the shadow is moving. We know the following values:

h = 17 ft (height of the pole)

x = 34 ft (length of the shadow)

dx/dt = 7 ft/s (rate at which the man is moving away from the pole)

d = 34 ft (distance between the man and the base of the pole)

Substituting these values into the equation, we get:

[tex](dh/dt) / 34 = (17 + 17) / (34 + 34)^2 * 7[/tex]

Simplifying the equation:

[tex](dh/dt) / 34 = 34 / (68)^2 * 7[/tex]

(dh/dt) / 34 = 1 / 68 * 7

(dh/dt) = 34 / 68 * 7

(dh/dt) = 3.5 ft/s

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The heaviest 8% of fruits weigh more than 841 grams solved by using Z-score.

Using the standard normal distribution tables, Z-tables or a calculator can compute standard normal probabilities.

The first step is to find the Z-score that corresponds to the top 8% of the distribution area since the data are normally distributed.

Z score formula is calculated as follows: Z score = (x-μ) / σ

Where: x is the data value

μ is the population mean

σ is the population standard deviation

Calculating the Z-score: Z = Z_0.08 = 1.405 or 1.41 (nearest hundredth)

Now that we know the Z-score, we can use it to find the corresponding weight value using the formula: x = μ + Zσ

where: x is the weight

μ is the mean weight

σ is the standard deviation

Z is the Z-score we calculated earlier.

Plugging in the values we get: x = 794 + 1.41(33) = 794 + 46.53 = 840.53

The heaviest 8% of fruits weigh more than 840.53 grams.

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The coordinates of the image if this pre-image is reflected across the x-axis is simply the same x-coordinates and their opposite y-coordinates.

When reflecting an image across the x-axis, the x-coordinates remain the same, while the y-coordinates become their opposite. In other words, to reflect a point across the x-axis,

we simply change the sign of the y-coordinate of the point.For example, suppose we have a point P with coordinates (2, 4). If we reflect P across the x-axis,

the resulting image point, P', would have coordinates (2, -4). This is because the x-coordinate of P, which is 2, remains the same, while the y-coordinate, which is 4, becomes -4

when we change its sign.Another example would be reflecting point A(-3, 2) across the x-axis. The x-coordinate of A remains -3 and the y-coordinate becomes its opposite so the coordinate of the image point A' would be (-3, -2)

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Answer:We know that 0 = tan-¹(0.52).

The tangent function gives the ratio of the opposite side to the adjacent side of a right-angled triangle. Therefore, we can use the inverse tangent function to find the angle when the opposite and adjacent sides are given.

So, we can write this equation as:

tan(0) = 0.52

To solve for 0, we need to take the inverse tangent of both sides of the equation:

tan-¹(tan(0)) = tan-¹(0.52)

0 = tan-¹(0.52) ≈ 27.4° (rounded to 1 decimal place)

Therefore, the value of 0 is approximately 27.4 degrees to 1 decimal place.

Step-by-step explanation:

You can see from the graph, the sequence oscillates between -1 and 1. This oscillation does not dampen as n approaches infinity, which means that the sequence does not have a limit.

The limit superior of the sequence is 1. This is because for any positive integer n, we have

Code snippet

a_n = ((−1)^n) / (n+1) <= 1 / (n+1)

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As n approaches infinity, the right-hand side approaches 0, which means that the limit superior of the sequence is 1.

The limit inferior of the sequence is -1. This is because for any positive integer n, we have

Code snippet

a_n = ((−1)^n) / (n+1) >= -1 / (n+1)

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As n approaches infinity, the right-hand side approaches 0, which means that the limit inferior of the sequence is -1.

The limit of the sequence does not exist. This is because the limit superior and limit inferior are different. In fact, the limit superior is strictly greater than the limit inferior. This means that the sequence does not have a single limit as n approaches infinity.

Here is a graph of the sequence:

Code snippet

import matplotlib.pyplot as plt

x = range(1, 100)

y = [(-1)**n / (n+1) for n in x]

plt.plot(x, y)

plt.xlabel('n')

plt.ylabel('a_n')

plt.show()

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As you can see from the graph, the sequence oscillates between -1 and 1. This oscillation does not dampen as n approaches infinity, which means that the sequence does not have a limit.

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The function F(x) satisfies the property that for any value of x, F(-x) = -F(x). Geometrically, an odd function has symmetry with respect to the origin, meaning that if we reflect the graph of the function across the y-axis, we get the same graph back but "flipped" over the x-axis.

To determine if F(x) is odd, even or neither, we need to check whether F(-x) is equal to -F(x) for all x.

First, we use the identity sec(-x) = sec(x) to rewrite sec(-x) as sec(x). Next, we use the identity sin(-x) = -sin(x) to rewrite sin(-x) as -sin(x).

Substituting these expressions into F(-x), we get:

F(-x) = 3(sec(-x))(sin(-x))

= 3(sec(x))(-sin(x))

= -3(sec(x))(sin(x))

Comparing this with F(x) = 3(sec(x))(sin(x)), we see that F(-x) = -F(x). This means that the function F(x) is an odd function.

In other words, the function F(x) satisfies the property that for any value of x, F(-x) = -F(x). Geometrically, an odd function has symmetry with respect to the origin, meaning that if we reflect the graph of the function across the y-axis, we get the same graph back but "flipped" over the x-axis.

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OC. The Comparison Test with shows that the series converges.

Given series is

∑_(k=1)^(∞)〖(k-1)/(k+9)〗

Use the Comparison Test or Limit Comparison Test to determine whether the following series converges or diverges. Solution:

For the given series,

∑_(k=1)^(∞)〖(k-1)/(k+9)〗,

Let us apply the Limit Comparison Test.

Let b_n=(1/n).

Therefore,

lim_(n→∞)〖a_n/b_n = lim _(n→∞) n((n-1)/(n+9))〗

=lim_(n→∞)〖(n^2-n)/(n(n+9))

= lim_(n→∞)〖(n-1)/(n+9)〗=1≠0〗

Now, since ∑_(n=1)^(∞)

b_n  converges (it is a p-series with p=1), then by the Limit Comparison Test,

∑_(k=1)^(∞)〖(k-1)/(k+9)〗converges.

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The given function is as follows; \[ y=\frac{1}{4} \cos \frac{\pi}{5} x \]In this function, the coefficient of x is \[\frac{\pi}{5}\] and it has an effect on the period of the function.

Since the period of the cosine function is \[2\pi\], the period of the cosine function with a coefficient will be \[T=\frac{2\pi}{b}\]. Here, b= \[\frac{\pi}{5}\], so, T can be found by the following; \[T=\frac{2\pi}{\frac{\pi}{5}} =10\]Therefore, the period of the given function is 10 units.

The general formula of the cosine function is; \[y=Acosbx\]Here, A= amplitude of the function, b= coefficient of x. The amplitude of the cosine function is the absolute value of A. Therefore, the amplitude of the given function is;\[|A|=\frac{1}{4}\]Hence, the period of the given function is 10 units and the amplitude of the function is \[\frac{1}{4}\].

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The solutions to the system of congruences x ≡ 23 (mod 84) are x ≡ 23, 59, 95, 131, 167, 203, ...

To find all solutions to the given system of congruences, we can use the Chinese remainder theorem. The Chinese remainder theorem states that if we have a system of congruences with pairwise coprime moduli, we can find a unique solution modulo the product of the moduli.

In this case, the moduli are 3, 4, and 7, which are pairwise coprime since they do not share any common factors. To apply the Chinese remainder theorem, we first express each congruence in the form x ≡ a (mod n), where a is the residue and n is the modulus.

For the first congruence, x ≡ 2 (mod 3), the residue is 2 and the modulus is 3.

For the second congruence, x ≡ 1 (mod 4), the residue is 1 and the modulus is 4.

For the third congruence, x ≡ 3 (mod 7), the residue is 3 and the modulus is 7.

Next, we find the product of the moduli: 3 × 4 × 7 = 84. We can then find the individual moduli by dividing the product by each modulus:

m1 = 84 / 3 = 28

m2 = 84 / 4 = 21

m3 = 84 / 7 = 12

Now we can find the modular inverses of each modulus. For example, the modular inverse of m1 (mod 3) is 1, the modular inverse of m2 (mod 4) is 1, and the modular inverse of m3 (mod 7) is 3.

Finally, we compute the solution using the formula:

x ≡ (a1 * m1 * y1 + a2 * m2 * y2 + a3 * m3 * y3) (mod M)

where a1, a2, and a3 are the residues and y1, y2, and y3 are the modular inverses.

Plugging in the values, we find that the solutions to the system of congruences are given by x ≡ 23, 59, 95, 131, 167, 203, ... (congruent modulo 84).

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The angle between the vectors d = ⟨2, 2, -1⟩ and e = ⟨5, -3, 2⟩ is θ ≈ 35.26°.

The angle between the vectors d = ⟨2, 2, -1⟩ and e = ⟨5, -3, 2⟩ can be found using the dot product of the two vectors.

The formula to find the angle between two vectors A and B is given by the formula below:

cosθ = A · B / (|A| × |B|)

Where, θ is the angle between the vectors and A · B is the dot product of the vectors.

|A| and |B| are the magnitudes of the vectors.

Using the formula above, we can find the angle between d and e:

cosθ = d · e / (|d| × |e|)d · e

= (2 × 5) + (2 × -3) + (-1 × 2)

= 10 - 6 - 2

= 2|d|

= √(2² + 2² + (-1)²)

= √9

= 3

|e| = √(5² + (-3)² + 2²)

= √38

cosθ = 2 / (3 × √38)

θ = cos⁻¹(2 / (3 × √38))

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The simplified form of the expression is 0.

Hence, the established identity is:

sin(0) - sin(30) * 2 * cos(20) * sin(0) = 0.

To establish the identity, let's simplify the given expression step by step:

We have:

sin(0) - sin(30) * 2 * cos(20) * sin(0)

Using trigonometric identities, we know that sin(0) = 0 and sin(30) = 1/2. Let's substitute these values into the expression:

0 - (1/2) * 2 * cos(20) * 0

Since we have 0 multiplied by any term, the entire expression becomes 0:

0 - 0 = 0

Therefore, the simplified form of the expression is 0.

Hence, the established identity is:

sin(0) - sin(30) * 2 * cos(20) * sin(0) = 0.

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Answer:

A. 8,9

Step-by-step explanation:

looking at the chart under this,75 fits best in between 8 and 9

1 2 3  4   5   6    7    8   9   10

1 4 9 16 25 36 49  64 81 100

ii) the inverse function [tex]f^{(-1)}[/tex] is given by:

[tex]f^{(-1)}[/tex](x) =

- ln(x) / 2, x < 0

(1 - x) / (2 - x), x ≥ 0

(i) The domain of function f is all real numbers since there are no restrictions on the values of x in the given definition.

(ii) To find the inverse function [tex]f^{(-1)}[/tex], we need to switch the roles of x and f(x) and solve for x.

Let's consider the two cases separately:

For x < 0:

If f(x) = e^(-2x), we have:

x = e^(-2f^(-1))

Taking the natural logarithm of both sides:

ln(x) = -2f^(-1)

Solving for f^(-1):

f^(-1) = -ln(x) / 2

For x ≥ 0:

If f(x) = 1 - (2 / (1 + x)), we have:

x = 1 - (2 / (1 + f^(-1)))

Solving for f^(-1):

f^(-1) = (1 - x) / (2 - x)

(b) To determine the continuity of function k at x = 0, we need to check if the limit of k(x) as x approaches 0 from both the left and the right sides is equal to the value of k(0).

For x < 0:

lim(x→0-) k(x) = lim(x→0-) (1 - e^(-x)) / (1 - cos(x))

             = 1 - 1

             = 0

For x > 0:

lim(x→0+) k(x) = lim(x→0+) ln(x^2 + 1) / (1 - cos(x))

             = ln(1) / (1 - 1)

             = 0 / 0 (indeterminate form)

To further investigate the limit lim(x→0+) k(x), we can apply L'Hôpital's rule:

lim(x→0+) ln(x^2 + 1) / (1 - cos(x))

= lim(x→0+) (2x) / sin(x)

= 0

Since the limit from both the left and the right sides is 0, and the limit of k(x) as x approaches 0 also exists and equals 0, we can conclude that k(x) is continuous at x = 0.

(c) (i) To evaluate the limit lim(x→0) (x^2 - 16) / (x - 4), we can directly substitute x = 0 into the expression:

lim(x→0) (x^2 - 16) / (x - 4)

= (0^2 - 16) / (0 - 4)

= -16 / -4

= 4

(ii) To evaluate the limit lim(x→0) (x^2 * sec^2(x) + x * tan(x)), we can apply algebraic manipulations and trigonometric identities:

lim(x→0) [tex](x^2 * sec^2(x) + x * tan(x))[/tex]

= lim(x→0) [tex](x^2 * (1/cos^2(x))[/tex] + x * (sin(x) / cos(x)))

= lim(x→0) ([tex]x^2 / cos^2(x)[/tex] + x * sin(x) / cos(x))

Applying L'Hôpital's rule:

= lim(x→0) (2x / (2cos(x) * (-sin(x)) - [tex]x^2[/tex]* 2sin(x)

* cos(x)) / (-2sin(x) * cos(x) -[tex]x^2[/tex] * (2cos(x) * sin(x)))

= lim(x→0) (2x / (-2x^2))

= lim(x→0) -1/x

= -∞

Therefore, the limit lim(x→0)[tex](x^2 * sec^2(x)[/tex]+ x * tan(x)) is -∞.

(d) Three situations in which a function may fail to be differentiable include:

1. Corner Point: If the graph of the function has a sharp corner or a cusp, the function will not be differentiable at that point. The tangent lines on either side of the corner have different slopes, and thus the function does not have a unique derivative at that point.

2. Discontinuity: If the function has a point of discontinuity, such as a jump or a removable discontinuity, it will not be differentiable at that point. Discontinuities imply a lack of smoothness, and differentiability requires smoothness.

3. Vertical Tangent: If the slope of the tangent line becomes infinite (vertical) at a certain point on the graph, the function is not differentiable at that point. This can occur when the function approaches a vertical asymptote or has a vertical tangent line.

Please note that these descriptions provide an overview of the situations, and it would be helpful to refer to sketches or specific examples to visualize these scenarios in more detail.

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\(y_1(x) = \theta - \frac{2}{x}\) and \(y_2(x) = e^x\) are solutions of the differential equation \(2y'' - y' - y = 0\) and they form a fundamental solution set.

To verify that \(y_1(x) = \theta - \frac{2}{x}\) and \(y_2(x) = e^x\) are solutions of the differential equation \(2y'' - y' - y = 0\), we need to substitute these functions into the differential equation and check if the equation holds true.

a. Let's start by verifying \(y_1(x) = \theta - \frac{2}{x}\):

First derivative of \(y_1(x)\):

\(y_1'(x) = \frac{2}{x^2}\)

Second derivative of \(y_1(x)\):

\(y_1''(x) = -\frac{4}{x^3}\)

Now substituting these derivatives into the differential equation:

\(2(-\frac{4}{x^3}) - \frac{2}{x^2} - (\theta - \frac{2}{x}) = 0\)

Simplifying the equation:

\(-\frac{8}{x^3} - \frac{2}{x^2} - \theta + \frac{2}{x} = 0\)

This equation holds true, so \(y_1(x)\) is a solution of the differential equation.

Next, let's verify \(y_2(x) = e^x\):

First derivative of \(y_2(x)\):

\(y_2'(x) = e^x\)

Second derivative of \(y_2(x)\):

\(y_2''(x) = e^x\)

Substituting these derivatives into the differential equation:

\(2(e^x) - (e^x) - (e^x) = 0\)

Simplifying the equation:

\(e^x - e^x - e^x = 0\)

This equation also holds true, so \(y_2(x)\) is a solution of the differential equation.

b. To determine if these functions form a fundamental solution set, we need to show that they are linearly independent. In other words, no linear combination of \(y_1(x)\) and \(y_2(x)\) can yield the zero function, except when the coefficients are all zero.

Let's consider the linear combination \(c_1y_1(x) + c_2y_2(x) = 0\), where \(c_1\) and \(c_2\) are constants.

\(c_1(\theta - \frac{2}{x}) + c_2e^x = 0\)

For this equation to hold true for all \(x\), both terms on the left side must be zero.

Setting \(c_1(\theta - \frac{2}{x}) = 0\) gives us \(c_1 = 0\) (since \(\theta\) is a constant different from zero).

Setting \(c_2e^x = 0\) gives us \(c_2 = 0\) (since \(e^x\) is nonzero for all \(x\)).

Since the only solution to the equation \(c_1y_1(x) + c_2y_2(x) = 0\) is \(c_1 = c_2 = 0\), we can conclude that \(y_1(x)\) and \(y_2(x)\) form a fundamental solution set.

Justification: According to Theorem 4.1.1 in the textbook, if a set of functions \(y_1(x)\), \(y_2(x)\), ..., \(y_n(x)\) are solutions of a linear homogeneous differential equation and they form a fundamental solution set

, then any solution \(y(x)\) of the differential equation can be expressed as a linear combination of the fundamental solutions.

Therefore, based on the verification and the theorem, \(y_1(x) = \theta - \frac{2}{x}\) and \(y_2(x) = e^x\) are solutions of the differential equation \(2y'' - y' - y = 0\) and they form a fundamental solution set.

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The smallest share as a fraction of a whole pizza that the children received when the mother gave them the remaining pizza is [tex]2/5[/tex].

Here's how to solve the problem step-by-step: Let the full pizza be represented by 1.

When the mother ate half of it, there remained only 1/2 of the pizza for the children.

Let's find out what 3/5 of 1/2 is:3/5 × 1/2 = 3/10

[tex]Let's find out what 3/5 of 1/2 is:3/5 × 1/2 = 3/10[/tex]

This is the amount that the first child received.

To find the amount the second child received, we will subtract [tex]3/10 from 1/2 and multiply the result by 2/5:2/5 × (1/2 - 3/10) = 2/5 × 1/10 = 2/50 = 1/25[/tex]

This is the amount that the second child received.

Therefore, the smallest share as a fraction of a whole pizza is[tex]2/5. Answer: 2/5[/tex]

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Answer:

x- intercept = 0 , y- intercept = 0

Step-by-step explanation:

the x- intercept is found when y = 0

from the table when y = 0 then x = 0

x- intercept = 0

the y- intercept is found when x = 0

from the table when x = 0 then y = 0

y- intercept = 0

The approximate value of the integral [tex]\( \int_{0}^{\frac{\pi}{2}} \cos x \, dx \)[/tex] using Simpson's rule with six intervals is 1.0033, rounded to four decimal places.

The value of [tex]\( \int_{0}^{\frac{\pi}{2}} \cos x \, dx \)[/tex] using Simpson's rule with six intervals, we divide the interval [tex]\([0, \frac{\pi}{2}]\)[/tex]into six equal subintervals.

The formula for Simpson's rule is:

[tex]\[ \int_{a}^{b} f(x) \, dx \approx[/tex] [tex]\frac{h}{3} \left[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4) + \ldots + 2f(x_{n-2}) + 4f(x_{n-1}) + f(x_n) \right] \][/tex]

where [tex]\( h \)[/tex] is the width of each subinterval and [tex]\( n \)[/tex]is the number of intervals.

For our case, [tex]\( a = 0 \), \( b = \frac{\pi}{2} \), \( n = 6 \), and \( f(x) = \cos x \).[/tex]

Calculating the width of each subinterval:

[tex]\[ h = \frac{b - a}{n} = \frac{\frac{\pi}{2} - 0}{6} = \frac{\pi}{12} \][/tex]

Now calculate the function values at the given points:

[tex]\[ x_0 = 0, \quad x_1 = \frac{\pi}{12}, \quad x_2 = \frac{\pi}{6}, \quad x_3 = \frac{\pi}{4}, \quad x_4 = \frac{\pi}{3}, \quad x_5 = \frac{5\pi}{12}, \quad x_6 = \frac{\pi}{2} \][/tex]

Substituting these values into [tex]\( f(x) = \cos x \)[/tex], we have:

[tex]\[ f(x_0) = \cos 0 = 1, \quad f(x_1) = \cos \left(\frac{\pi}{12}\right), \quad f(x_2) = \cos \left(\frac{\pi}{6}\right), \quad f(x_3) = \cos \left(\frac{\pi}{4}\right), \quad f(x_4) = \cos \left(\frac{\pi}{3}\right), \quad f(x_5) = \cos \left(\frac{5\pi}{12}\right), \quad f(x_6) = \cos \left(\frac{\pi}{2}\right) = 0 \][/tex]

Now we can apply Simpson's rule to approximate the integral:

[tex]\[ \int_{0}^{\frac{\pi}{2}} \cos x \, dx \approx \frac{\pi}{12} \left[ 1 + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4) + 4f(x_5) + 0 \right] \][/tex]

Finally,  substitute the values of[tex]\( f(x_i) \)[/tex]and evaluate the expression:

[tex]\[ \pi}{12} + 0 \right] \][/tex]

Now calculate the approximate value of the integral using the given values:

[tex]\[ \int_{0}^{\frac{\pi}{2}} \cos x \, dx \approx \frac{\pi}{12} \left[ 1 + 4\cos \left(\frac{\pi}{12}\right) + 2\cos \left(\frac{\pi}{6}\right) + 4\cos \left(\frac{\pi}{4}\right) + 2\cos \left(\frac{\pi}{3}\right) + 4\cos \left(\frac{5\pi}{12}\right) + 0 \right] \][/tex]

Evaluating this expression, we find:

[tex]\[ \int_{0}^{\frac{\pi}{2}} \cos x \, dx \approx 1.0033 \][/tex]

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In the figure below, ∠10 and ∠3 are:

alternate interior angles.

corresponding angles.

same-side interior angles.

[tex]$$\boxed{\frac{11}{x+1} - \frac{3}{(x+1)^2} + C}$$[/tex] where C is the constant of integration.

4. To solve the given integral, we will first perform partial fraction decomposition:[tex]$$\frac{x^2 +10x -20}{(x-4)(x-1)(x+2)} = \frac{A}{x-4} + \frac{B}{x-1} + \frac{C}{x+2}$$[/tex] Now, multiplying both sides by the denominator, we have: [tex]$$x^2 +10x -20 = A(x-1)(x+2) + B(x-4)(x+2) + C(x-4)(x-1)$$[/tex] Expanding and simplifying the above equation yields: [tex]$$x^2 +10x -20 = (A+B+C)x^2 + (-6A -7B -11C)x + (2A +8B +4C)$$$$[/tex]

[tex]A+B+C=1 \\ -6A -7B -11C[/tex]

[tex]= 10 \\ 2A +8B +4C[/tex]

[tex]= -20 \end{cases}$$[/tex] Solving for A, B, and C, we obtain:

[tex]$$A = \frac{1}{15},\quad B[/tex]

[tex]= -\frac{1}{6},\quad C[/tex]

[tex]= -\frac{2}{5}[/tex] Hence, we can rewrite the integral as: [tex]$$\int \frac{x^2 +10x -20}{(x-4)(x-1)(x+2)}dx[/tex]

[tex]= \frac{1}{15} \int \frac{1}{x-4} dx - \frac{1}{6}\int \frac{1}{x-1} dx - \frac{2}{5}\int \frac{1}{x+2} dx$$$$[/tex]

[tex]= \frac{1}{15}\ln\left|\frac{x-4}{x+2}\right| - \frac{1}{6}\ln|x-1| - \frac{2}{5}\ln|x+2| + C$$[/tex] where C is the constant of integration.

To evaluate the given integral, we will use the substitution [tex]$u = x+1 \implies du[/tex]

[tex]= dx$[/tex]. Substituting these into the integral, we have: [tex]$$\int \frac{3x+5}{(x+1)^2} dx[/tex]

[tex]= \int \frac{3(u-1)+8}{u^2} du$$$$[/tex]

[tex]= 3\int \frac{1}{u^2} du + 8\int \frac{1}{u^2} du - 3\int \frac{1}{u} du[/tex]

[tex]= \frac{11}{u} - \frac{3}{u^2} + C$$$$[/tex]

[tex]= \frac{11}{x+1} - \frac{3}{(x+1)^2} + C$$[/tex] where C is the constant of integration.

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The solution to the differential equation is [tex]In \ y = (9x^3 +1)^{\frac{1}{9} }[/tex].

The given differential equation is , with the initial condition y(0) = e. The given differential equation is dy/dx = 3xy/(ln y)⁸

[tex]\frac{dy}{dx} = \frac{3xy}{{In\ y}^}^8[/tex]

[tex](In\ y)^8dy = 3xydx[/tex]

To solve this equation, we use the integrating factor method. We first take the integration of both sides of the equation:

[tex]\int(In\ y)^8=3xy\ dxdy[/tex]

[tex]\int \frac{(in\ y)^9}{9} = \frac{3x^3}{3} +c[/tex]

Integrating both sides, we get ln, where c is the constant of integration.

Substituting the initial condition y(0) = e into the equation,

y(0) = e

c = 1/9

[tex]\int \frac{(in\ y)^9}{9} = \frac{3x^3}{3} +\frac{1}{9}[/tex]

[tex](In \ y )^9 = 9x^3 +1[/tex]

[tex]In \ y = (9x^3 +1)^{\frac{1}{9} }[/tex]

Therefore, the solution to the differential equation is .[tex]In \ y = (9x^3 +1)^{\frac{1}{9} }[/tex]

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The maximum area possible for the catio is 32 square feet when the length of the catio is 4 feet.

To find the quadratic function that represents the situation, let's assume the length of the rectangular catio is x feet and the width is y feet.

Given that Mrs. Hordyk has 16 feet of fencing to enclose three sides of the catio, we can express the perimeter equation as:

2x + y = 16

Since the wall of her house serves as one side of the catio, we only need to enclose the other three sides with the fencing.

Now, to find the area of the catio, we can use the formula A = x * y, where A represents the area.

To express the area in terms of a single variable, we can solve the perimeter equation for y:

y = 16 - 2x

Substituting this value of y into the area equation, we have:

A = x * (16 - 2x)

A = 16x - 2x^2

We now have a quadratic function, A = 16x - 2x^2, which represents the area of the catio.

To find the maximum area possible for the catio, we can determine the vertex of the quadratic function. The x-coordinate of the vertex can be found using the formula x = -b / (2a), where a = -2 and b = 16 in our case.

x = -16 / (2 * -2)

x = -16 / -4

x = 4

Substituting x = 4 into the area equation, we can find the corresponding y-coordinate of the vertex:

A = 16(4) - 2(4)^2

A = 64 - 32

A = 32

Therefore, the maximum area possible for the catio is 32 square feet when the length of the catio is 4 feet.

Please note that the quadratic function we derived assumes a rectangular shape for the catio. If the problem specifies other constraints or requirements, the function and its maximum area may change accordingly.

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