. Suppose a particle moves back and forth along a straight line with velocity v(t) , measured in feet per second, and acceleration aft) 120 a. What is the meaning of La muce? v(t) dt? 120 b. What is the meaning of (Odt? 60 120 c. What is the meaning of a(t) dt ? 60

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Answer 1

The meaning of a(t) dt is the change in velocity of the particle over a time interval dt.

(a) La muce: La muce is the displacement of the particle from its initial position. If we integrate the velocity function v(t) over time from t = 0 to t = T, then we get La muce.T is the time elapsed since the particle began to move.

(b) (Odt:We can also write the displacement of the particle as the integral of the velocity function v(t) multiplied by the time differential dt. This is denoted by (Odt.La muce = ∫ v(t) dt

(c) a(t) dt:We know that acceleration a(t) is the rate of change of velocity with respect to time. Therefore, integrating acceleration a(t) over time from t = 0 to t = T gives the change in velocity of the particle over that time period.Taking the limits of the integral as t = 0 and t = T, we get:a(T) - a(0) = ∫ a(t) dt

Therefore, the meaning of a(t) dt is the change in velocity of the particle over a time interval dt.

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Related Questions

Find an equation of the plane The plane that passes through the point (-3, 3, 2) and contains the line of intersection of the planes x+y-22 and 3x + y + 5z = 5

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An equation of the plane that passes through the point (-3, 3, 2) and contains the line of intersection of the planes x+y-22 and 3x + y + 5z = 5 is **x + 10y - 5z = -52**.

To find the equation of the plane that passes through the point (-3, 3, 2) and contains the line of intersection of the planes x+y-22 and 3x + y + 5z = 5, we can follow these steps:

1. Find the line of intersection of the two planes.

2. Find a point on this line.

3. Use this point and the given point (-3, 3, 2) to find a vector that lies in the plane.

4. Use this vector and the given point (-3, 3, 2) to find the equation of the plane.

The line of intersection of the two planes is:

x + y - 22 = 0

3x + y + 5z - 5 = 0

Solving these equations gives:

x = -1

y = 23

z = -8

So a point on this line is (-1, 23, -8).

A vector that lies in the plane is given by:

(-1 - (-3), 23 - 3, -8 - 2) = (2, 20, -10)

Using this vector and the given point (-3, 3, 2), we can write the equation of the plane in vector form as:

(r - (-3, 3, 2)) · (2, 20, -10) = 0

Expanding this equation gives:

2(x + 3) + 20(y - 3) - 10(z - 2) = 0

Simplifying this expression gives:

**x + 10y - 5z = -52**

Therefore, an equation of the plane that passes through the point (-3, 3, 2) and contains the line of intersection of the planes x+y-22 and 3x + y + 5z = 5 is **x + 10y - 5z = -52**.

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Find the Tangent, Normal and Binormal vectors (T, N and B) for the curve r(t) = (5 cos(4t), 5 sin(4t), 2t) at the point t = 0 T(0) = (0, 5 1 26 27 26 N(0) = (-1,0,0) B(O) = 10, B0-27 1 2v 26 V 26

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The tangent vector T(0) is (0, 20, 2). The normal vector N(0) is (0, 10/sqrt(101), 1/sqrt(101)). The binormal vector B(0) is (-20/sqrt(101), -2/sqrt(101), 0).

To find the tangent, normal, and binormal vectors (T, N, and B) for the curve r(t) = (5cos(4t), 5sin(4t), 2t) at the point t = 0, we need to calculate the derivatives of the curve with respect to t and evaluate them at t = 0.

Tangent vector (T): The tangent vector is given by the derivative of r(t) with respect to t:

r'(t) = (-20sin(4t), 20cos(4t), 2)

Evaluating r'(t) at t = 0:

r'(0) = (-20sin(0), 20cos(0), 2)

= (0, 20, 2)

Therefore, the tangent vector T(0) is (0, 20, 2).

Normal vector (N): The normal vector is obtained by normalizing the tangent vector. We divide the tangent vector by its magnitude:

|T(0)| = sqrt(0^2 + 20^2 + 2^2) = sqrt(400 + 4) = sqrt(404) = 2sqrt(101)

N(0) = T(0) / |T(0)|

= (0, 20, 2) / (2sqrt(101))

= (0, 10/sqrt(101), 1/sqrt(101))

Therefore, the normal vector N(0) is (0, 10/sqrt(101), 1/sqrt(101)).

Binormal vector (B): The binormal vector is obtained by taking the cross product of the tangent vector and the normal vector:

B(0) = T(0) x N(0)

Taking the cross product:

B(0) = (20, 0, -2) x (0, 10/sqrt(101), 1/sqrt(101))

= (-20/sqrt(101), -2/sqrt(101), 0)

Therefore, the binormal vector B(0) is (-20/sqrt(101), -2/sqrt(101), 0).

In summary:

T(0) = (0, 20, 2)

N(0) = (0, 10/sqrt(101), 1/sqrt(101))

B(0) = (-20/sqrt(101), -2/sqrt(101), 0).

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A product is introduced to the market. The weekly profit (in dollars) of that product decays exponentially -0.04.x as function of the price that is charged (in dollars) and is given by P(x) = 75000 ·

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The given equation P(x) = 75000 · e^(-0.04x) represents the weekly profit of a product as a function of the price charged. It demonstrates exponential decay, with the coefficient -0.04 determining the rate of decay.

The first paragraph summarizes the main information provided. It states that the weekly profit of the product is modeled by an exponential decay function, where the price is the independent variable. The profit function, P(x), is given as P(x) = 75000 · e^(-0.04x).

In the second paragraph, we can further explain the equation and its components. The function P(x) represents the weekly profit, which depends on the price x. The coefficient -0.04 determines the rate of decay, indicating that as the price increases, the profit decreases exponentially. The exponential term e^(-0.04x) describes the decay factor, where e is the base of the natural logarithm. As x increases, the exponential term decreases, causing the profit to decay. Multiplying this decay factor by 75000 scales the decay function to the appropriate profit range.

In summary, the given equation P(x) = 75000 · e^(-0.04x) represents the weekly profit of a product as a function of the price charged. It demonstrates exponential decay, with the coefficient -0.04 determining the rate of decay.

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Tom is travelling on a train which is moving at a constant speed of 15 m s- on a horizontal track. Tom has placed his mobile phone on a rough horizontal table. The coefficient of friction
between the phone and the table is 0.2. The train moves round a bend of constant radius. The phone does not slide as the train travels round the bend. Model the phone as a particle
moving round part of a circle, with centre O and radius r metres. Find the least possible value of r

Answers

Tom's mobile phone is placed on a rough horizontal table inside a train moving at a constant speed of 15 m/s on a horizontal track. The phone does not slide as the train goes around a bend of constant radius.

When the train moves around the bend, the phone experiences a centripetal force towards the center of the circular path. This force is provided by the friction between the phone and the table. To prevent the phone from sliding, the frictional force must be equal to or greater than the maximum possible frictional force. Considering the forces acting on the phone, the centripetal force is provided by the frictional force: F_centripetal = F_friction = μN.

The centripetal force can also be expressed as F_centripetal = mv²/r, where v is the velocity of the train and r is the radius of the circular path. Equating the two expressions for the centripetal force, we have mv²/r = μN. Substituting the values, we get m(15)²/r = 0.2mg. The mass of the phone cancels out, resulting in 15²/r = 0.2g.

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2
Problem 3 Fill in the blanks: a) If a function fis on the closed interval [a,b], then f is integrable on [a,b]. b) Iffis and on the closed interval [a,b], then the area of the region bounded by the gr

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a) If a function f is continuous on the closed interval [a, b], then f is integrable on [a, b].

b) If f is continuous and non-negative on the closed interval [a, b], then the area of the region bounded by the graph of f, the x-axis, and the vertical lines x = a and x = b can be calculated using definite integration.

a) The statement "If a function f is continuous on the closed interval [a, b], then f is integrable on [a, b]" is known as the Fundamental Theorem of Calculus. It implies that if a function is continuous on a closed interval, it can be integrated over that interval. This means we can find the definite integral of f from a to b, denoted by ∫[a, b] f(x) dx.

b) The second part states that if a function f is continuous and non-negative on the closed interval [a, b], then we can calculate the area of the region bounded by the graph of f, the x-axis, and the vertical lines x = a and x = b using definite integration. The area is given by the definite integral ∫[a, b] f(x) dx, where f(x) represents the height of the function at each x-value within the interval [a, b]. The non-negativity condition ensures that the area is always positive or zero.

In conclusion, the first statement asserts the integrability of a continuous function on a closed interval, while the second statement relates the area calculation of a bounded region to definite integration for a continuous and non-negative function on a closed interval. These concepts form the foundation of integral calculus.

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Evaluate the derivative of the function. y = sec^(-1) (9 In 8x) dy/dx =

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The derivative is equal to -9/(ln(8x) * |8x| * sqrt((8x)^2 - 1)), where |8x| represents the absolute value of 8x.

The derivative of the function y = sec^(-1)(9ln(8x)) with respect to x, denoted as dy/dx, can be calculated using the chain rule and the derivative of the inverse secant function.

To find the derivative of y = sec^(-1)(9ln(8x)) with respect to x, we can use the chain rule. Let's break down the calculation step by step.

First, let's differentiate the inverse secant function, which has the derivative d/dx(sec^(-1)(u)) = -1/(u * |u| * sqrt(u^2 - 1)), where |u| represents the absolute value of u.

Now, we have y = sec^(-1)(9ln(8x)), and we need to apply the chain rule. The chain rule states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x).

In our case, f(u) = sec^(-1)(u), and g(x) = 9ln(8x).

Taking the derivative of g(x) with respect to x, we get g'(x) = 9 * (1/x) = 9/x.

Next, we need to calculate f'(g(x)). Substituting u = 9ln(8x), we have f'(u) = -1/(u * |u| * sqrt(u^2 - 1)).

Combining all the derivatives, we get dy/dx = f'(g(x)) * g'(x) = -1/(9ln(8x) * |9ln(8x)| * sqrt((9ln(8x))^2 - 1)) * 9/x.

Simplifying this expression, we obtain dy/dx = -9/(ln(8x) * |8x| * sqrt((8x)^2 - 1)).

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Consider the following hypothesis test.
H0: 1 − 2 ≤ 0
Ha: 1 − 2 > 0
The following results are for two independent samples taken from the two populations.
Sample 1 Sample 2
n1 = 40
n2 = 50
x1 = 25.3
x2 = 22.8
1 = 5.5
2 = 6
(a)
What is the value of the test statistic? (Round your answer to two decimal places.)
(b)
What is the p-value? (Round your answer to four decimal places.)
(c)
With
= 0.05,
what is your hypothesis testing conclusion?

Answers

the test statistic and p-value, we need to perform a two-sample t-test. The test statistic is calculated as:

t = [(x1 - x2) - (μ1 - μ2)] / sqrt[(s1²/n1) + (s2²/n2)]

where:x1 and x2 are the sample means,

μ1 and μ2 are the population means under the null hypothesis ,s1 and s2 are the sample standard deviations, and

n1 and n2 are the sample sizes.

In this case, the null hypothesis (H0) is 1 - 2 ≤ 0, and the alternative hypothesis (Ha) is 1 - 2 > 0.

Given the following data:Sample 1: n1 = 40, x1 = 25.3, and s1 = 5.5

Sample 2: n2 = 50, x2 = 22.8, and s2 = 6

(a) To find the test statistic:t = [(25.3 - 22.8) - 0] / sqrt[(5.5²/40) + (6²/50)]

(b) To find the p-value:

Using the test statistic, we can calculate the p-value using a t-distribution table or statistical software.

(c) With α = 0.05, we compare the p-value to the significance level.

hypothesis; otherwise, we fail to reject the null hypothesis.

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Find the center and radius of the sphere
x^2−4x−24+y^2+16y+z^2−12z=0
Halle el centro y radio de la esfera x2 – 4x – 24 + y2 + 16y + z2 – 12z = 0 - Seleccione una: O a. C(-2,8,-6),r=832 9 O b. C(2, -8,6), r = 8 O c. C(2, -8,6), r = 872 O d. C(-2,8,-6), r = 8

Answers

The correct answer is option c. C(2, -8, 6), r = 11.3137 (rounded to the nearest decimal place).

To find the center and radius of the sphere represented by the equation x² - 4x - 24 + y² + 16y + z² - 12z = 0, we can rewrite the equation in the standard form:

(x² - 4x) + (y² + 16y) + (z² - 12z) = 24

Completing the square for each variable group, we get:

(x² - 4x + 4) + (y² + 16y + 64) + (z² - 12z + 36) = 24 + 4 + 64 + 36

Simplifying further:

(x - 2)² + (y + 8)² + (z - 6)² = 128

Now we can compare this equation to the standard equation of a sphere:

(x - h)² + (y - k)² + (z - l)² = r²

From the comparison, we can see that the center of the sphere is (h, k, l) = (2, -8, 6), and the radius squared is r² = 128. Taking the square root of 128, we find the radius r ≈ 11.3137.

Therefore, the correct answer is option c. C(2, -8, 6), r = 11.3137 (rounded to the nearest decimal place).

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Suppose that the manufacturing cost of a particular item is approximated by M(x, y) = 2x2 – 2x²y3 +35, where x is the cost of materials and y is the cost of labor. Find the y following: Mz(x, y) = = My(x, y) = = Mxx(x, y) = Mry(x, y) = =

Answers

To find the partial derivatives of the function M(x, y) = 2x^2 - 2x^2y^3 + 35, we differentiate the function with respect to all variables (x,y) separately while treating the other variable as a constant.

My(x, y) = -2x^2 * 3y^2 = -6x^2y^2

Mxx(x, y) = d/dx(2x^2 - 2x^2y^3) = 4x - 4xy^3

Mry(x, y) = d/dy(2x^2 - 2x^2y^3) = -6x^2 * 2y^3 = -12x^2y^2

So the partial derivatives are:

Mz(x, y) = 0

My(x, y) = -6x^2y^2

Mxx(x, y) = 4x - 4xy^3

Mry(x, y) = -12x^2y^2

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Consider the function f(x)=√x - 2 on the interval [1,9]. Using the Mean Value Theorem we can conclude that: The Mean Value Theorem does not apply because this function is not continuous on [1,9]. Th

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The Mean Value Theorem(MVT) does not apply to the function f(x) = √x - 2 on the interval [1, 9] because this function is not continuous on [1, 9].

The Mean Value Theorem states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in (a, b) where the derivative of the function is equal to the average rate of change of the function over the interval [a, b].

In this case, the function f(x) = √x - 2 is not continuous on the interval [1, 9]. The square root function √x is not defined for negative values of x, and since the interval [1, 9] includes the point x = 0, the function is not defined at that point. Therefore, the function is not continuous on the interval [1, 9], and as a result, the Mean Value Theorem does not apply.

For the Mean Value Theorem(MVT) to be applicable, it is necessary for the function to satisfy the conditions of continuity and differentiability on the given interval. Since f(x) = √x - 2 is not continuous at x = 0, it fails to meet the conditions required by the Mean Value Theorem. Consequently, we cannot apply the theorem to make any conclusions about the existence of a point where the derivative of the function equals the average rate of change on the interval [1, 9].

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3. The function yı = 2+1 is a solution of the differential equation (1 - 2x - ²)y+ 2(1+)y – 2y = 0 The method of Reduction of order produces the second solution y2 = (correct) (a) (b) (c) (d) (e) m2 + +2 2.2 - 1+1 22 - +3 x²+x+3 x²+2 O - 32°C .

Answers

The method of Reduction of order produces the second solution y2 = y1(x)· ∫ [exp (-∫p(x) dx) / y1²(x)] dx. The given differential equation is (1 - 2x - x²)y' + 2(1+x)y – 2y = 0, which is a second-order linear differential equation.

Let's find the homogeneous equation first as follows: (1 - 2x - x²)y' + 2(1+x)y – 2y = 0     ...(i)

Using the given function y1 = 2 + x, let's assume the second solution y2 as y2 = v(x) y1(x).

Substituting this in equation (i), we have y1(x) [(1 - 2x - x²)v' + (2 - 2x)v] + y1'(x) [2v] = 0 ⇒ (1 - 2x - x²)v' + (2 - 2x)v = 0.

Dividing both sides by v y' /v + (-2x-1) / (x² + x - 2) + 2 / (x + 1) = 0...[∵Integrating factor, I.F = 1 / (y1(x))² = 1 / (2 + x)²].

Integrating the above equation, we get v(x) = C / (2 + x)² + x + 1/2C is the constant of integration.

Substituting this in y2 = v(x) y1(x), we get:y2 = (C / (2 + x)² + x + 1/2)(2 + x) ...[∵ y1 = 2 + x]y2 = C (2 + x) + x(2 + x) + 1/2(2 + x) ...(ii)

Therefore, the required second solution is y2 = C (2 + x) + x(2 + x) + 1/2(2 + x) ...[from (ii)].

Hence, the correct option is (d) C (2 + x) + x(2 + x) + 1/2(2 + x).

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3. (a) Calculate sinh (log(5) - log(4)) exactly, i.e. without using a calculator. (3 marks) (b) Calculate sin(arccos )) exactly, i.e. without using a calculator. V65 (3 marks) (e) Using the hyperbolic identity Coshºp - sinh?t=1, and without using a calculator, find all values of cosh r, if tanh x = (4 marks)

Answers

(a) To calculate sinh(log(5) - log(4)) exactly, we can use the properties of logarithms and the definition of sinh function. First, we simplify the expression inside the sinh function using logarithm rules: log(5) - log(4) = log(5/4).

Now, using the definition of sinh function, sinh(x) = (e^x - e^(-x))/2, we substitute x with log(5/4): sinh(log(5/4)) = (e^(log(5/4)) - e^(-log(5/4)))/2.Using the property e^(log(a)) = a, we simplify the expression further: sinh(log(5/4)) = (5/4 - 4/5)/2 = (25/20 - 16/20)/2 = 9/20. Therefore, sinh(log(5) - log(4)) = 9/20.

(b) To calculate sin(arccos(√(65))), we can use the Pythagorean identity sin²θ + cos²θ = 1. Since cos(θ) = √(65), we can substitute into the identity: sin²(θ) + (√(65))² = 1. Simplifying, we have sin²(θ) + 65 = 1. Rearranging the equation, sin²(θ) = 1 - 65 = -64. Since sin²(θ) cannot be negative, there is no real solution for sin(arccos(√(65))).

(e) Using the hyperbolic identity cosh²(x) - sinh²(x) = 1, and given tanh(x) = √(65), we can find the values of cosh(x). First, square the equation tanh(x) = √(65) to get tanh²(x) = 65. Then, rearrange the identity to get cosh²(x) = 1 + sinh²(x). Substituting tanh²(x) = 65, we have cosh²(x) = 1 + 65 = 66.

Taking the square root of both sides, we get cosh(x) = ±√66. Therefore, the values of cosh(x) are ±√66.

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(10 points) Suppose a virus spreads so that the number N of people infected grows tially with time t. The table below shows how many days it takes from the initial to have various numbers of cases. t=# of days 36 63 N=# of cases 1 million 8 million How many days since the initial outbreak until the virus infects 40 million people? ( to the nearest whole number of days)

Answers

It would take approximately 59 days since the initial outbreak until the virus infects 40 million people.

The growth rate can be found by dividing the final number of cases by the initial number of cases and then taking the t-th root of that value, where t is the number of days it took to reach the final number of cases from the initial.

In this case, the growth rate is (8 million / 1 million)^(1/27), rounded to three decimal places which is 1.297.

Using this growth rate, we can calculate how many days it would take to reach 40 million cases:

40 million = 1 million * (1.297)^d

Solving for d, we get:

d = log(40)/log(1.297)

d ≈ 58.5

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outside temperature over a day can be modelled as a sinusoidal function. suppose you know the high temperature for the day is 63 degrees and the low temperature of 47 degrees occurs at 4 am. assuming t is the number of hours since midnight, find an equation for the temperature, d, in terms of t. g

Answers

In terms of t (the number of hours since midnight), the temperature, d, can be expressed as follows:

d = 8 * sin((π / 12) * t - (π / 3)) + 55

Explanation:

To model the temperature as a sinusoidal function, we can use the form:

d = A * sin(B * t + C) + D

Where:

- A represents the amplitude, which is half the difference between the high and low temperatures.

- B represents the period of the sinusoidal function. Since we want a full day cycle, B would be 2π divided by 24 (the number of hours in a day).

- C represents the phase shift. Since the low temperature occurs at 4 am, which is 4 hours after midnight, C would be -B * 4.

- D represents the vertical shift. It is the average of the high and low temperatures, which is (high + low) / 2.

Given the information provided:

- High temperature = 63 degrees

- Low temperature = 47 degrees at 4 am

We can calculate the values of A, B, C, and D:

Amplitude (A):

A = (High - Low) / 2

A = (63 - 47) / 2

A = 8

Period (B):

B = 2π / 24

B = π / 12

Phase shift (C):

C = -B * 4

C = -π / 12 * 4

C = -π / 3

Vertical shift (D):

D = (High + Low) / 2

D = (63 + 47) / 2

D = 55

Now we can substitute these values into the equation:

d = 8 * sin((π / 12) * t - (π / 3)) + 55

Therefore, the equation for the temperature, d, in terms of t (the number of hours since midnight), is:

d = 8 * sin((π / 12) * t - (π / 3)) + 55

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Does the sequence {a,} converge or diverge? Find the limit if the sequence is convergent. 2 + 4n4 an 4 n + 3n Select the correct choice below and, if necessary, fill in the answer box to complete the

Answers

The limit of the sequence {aₙ} as n approaches infinity is positive infinity (∞). The limit of the sequence is not a finite value, the sequence diverges.

To determine whether the sequence {aₙ} converges or diverges, we need to examine its behavior as n approaches infinity. The sequence is defined as:

[tex]a_n = (2 + 4n^4) / (4n + 3n)[/tex]

We can simplify this expression by factoring out n from the denominator:

[tex]a_n = (2 + 4n^4) / (7n)[/tex]

Now, let's consider the limit of this expression as n approaches infinity:

lim(n→∞) (2 + [tex]4n^4[/tex]) / (7n)

As n approaches infinity, the dominant term in the numerator will be [tex]4n^4[/tex] and in the denominator will be 7n.

Thus, we can ignore the other terms.

lim(n→∞) [tex]4n^4[/tex] / 7n

Simplifying further:

lim(n→∞) (4/7) * ([tex]n^4[/tex]/n)

lim(n→∞) (4/7) * [tex]n^3[/tex]

As n approaches infinity, the limit of [tex]n^3[/tex] will also approach infinity. Therefore, the limit of the sequence {aₙ} as n approaches infinity is positive infinity (∞).

Since the limit of the sequence is not a finite value, the sequence diverges.

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7-8 Find an equation of the tangent to the curve at the given point by two methods: (a) without eliminating the parameter and (6) by first eliminating the parameter. 7. x = 1 + In t, y = x2 + 2; (1,3) 8. x = 1 + Vi, y = f'; (2, e) 2e

Answers

a.  The equation of the tangent to the curve x = 1 + ln(t), y = x^2 + 2 at the point (1, 3) is y = 2x + 1.

b. The equation of the tangent to the curve x = 1 + ln(t), y = x^2 + 2 at the point (1, 3) is y = 2x + 1.

(a) Without eliminating the parameter:

For the curve defined by x = 1 + ln(t) and y = x^2 + 2, we need to find the equation of the tangent at the given point (1, 3).

To do this, we'll find the derivative dy/dx and substitute the values of x and y at the point (1, 3). The resulting derivative will give us the slope of the tangent line.

x = 1 + ln(t)

Differentiating both sides with respect to t:

dx/dt = d/dt(1 + ln(t))

dx/dt = 1/t

Now, we find dy/dt:

y = x^2 + 2

Differentiating both sides with respect to t:

dy/dt = d/dt(x^2 + 2)

dy/dt = d/dx(x^2 + 2) * dx/dt

dy/dt = (2x)(1/t)

dy/dt = (2x)/t

Next, we find dx/dt at the given point (1, 3):

dx/dt = 1/t

Substituting t = e (since ln(e) = 1), we get:

dx/dt = 1/e

Similarly, we find dy/dt at the given point (1, 3):

dy/dt = (2x)/t

Substituting x = 1 and t = e, we have:

dy/dt = (2(1))/e = 2/e

Now, we can find the slope of the tangent line by evaluating dy/dx at the given point (1, 3):

dy/dx = (dy/dt)/(dx/dt)

dy/dx = (2/e)/(1/e)

dy/dx = 2

So, the slope of the tangent line is 2. Now, we can find the equation of the tangent line using the point-slope form:

y - y1 = m(x - x1)

y - 3 = 2(x - 1)

y - 3 = 2x - 2

y = 2x + 1

Therefore, the equation of the tangent to the curve x = 1 + ln(t), y = x^2 + 2 at the point (1, 3) is y = 2x + 1.

(b) By first eliminating the parameter:

To eliminate the parameter, we'll solve the first equation x = 1 + ln(t) for t and substitute it into the second equation y = x^2 + 2.

From x = 1 + ln(t), we can rewrite it as ln(t) = x - 1 and exponentiate both sides:

t = e^(x-1)

Substituting t = e^(x-1) into y = x^2 + 2, we have:

y = (1 + ln(t))^2 + 2

y = (1 + ln(e^(x-1)))^2 + 2

y = (1 + (x-1))^2 + 2

y = x^2 + 2

Now, we differentiate y = x^2 + 2 with respect to x to find the slope of the tangent line:

dy/dx = 2x

Substituting x = 1 (the x-coordinate of the given point), we get:

dy/dx = 2(1) = 2

The slope of the tangent line is 2. Now, we can find the equation of the tangent line using the point-slope form:

y - y1 = m(x - x1)

y - 3 = 2(x - 1)

y - 3 = 2x - 2

y = 2x + 1

Therefore, the equation of the tangent to the curve x = 1 + ln(t), y = x^2 + 2 at the point (1, 3) is y = 2x + 1.

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Show all work please!
Solve the initial value problem dy dt = -5/7, y(1) = 1. (Use symbolic notation and fractions where needed.) y = help (decimals) = = 13 find: (1 point) Given that f"(x) = cos(2), f'(7/2) = 5 and f(1/

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The solution to the initial value problem is y = (-5/7) * t + 12/7 where  y at t = 13 is -53/7 or approximately -7.5714 (in decimal form).

To solve the initial value problem dy/dt = -5/7, y(1) = 1, we can integrate both sides of the equation with respect to t.

∫ dy = ∫ (-5/7) dt

Integrating both sides gives:

y = (-5/7) * t + C

To determine the constant of integration, C, we can substitute the initial condition y(1) = 1 into the equation:

1 = (-5/7) * 1 + C

1 = -5/7 + C

C = 1 + 5/7

C = 12/7

Now we can substitute this value of C back into the equation:

y = (-5/7) * t + 12/7

Therefore, the solution to the initial value problem is y = (-5/7) * t + 12/7.

To find the value of y at a specific t, you can substitute the given value of t into the equation. For example, to find y at t = 13, you would substitute t = 13 into the equation:

y = (-5/7) * 13 + 12/7

y = -65/7 + 12/7

y = -53/7

So, y at t = 13 is -53/7 or approximately -7.5714 (in decimal form).

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A private shipping company will accept a box of domestic shipment only if the sum of its length and girth (distance around) does not exceed 90 in. What dimension will give a box with a square end the largest possible volume?

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The dimension the  a box with a square end the largest possible volume is 10 ×10 × 23.3

How to determine the volume

First, we will need to complete the question.

Let us assume that its dimensions are h by h by w and its girth is 2h + 2w.

Volume = h²w

Where h is the length

w is the girth

From the information given, we have;

Length + girth = 90

w+(2h+2w) = 90

2h + 3w = 90

Make 'w' the subject

w = 90- 2h/3

w = 30 - 2h/3

Substitute the values

Volume = h²(30 - 2h/3)

expand the bracket

Volume = 30h² - 2h³/3

Find the differential value

Volume = 60h - 6h²

h = 10

Substitute the values

w =  30 - 2h/3

w = 30 - 2(10)/3

w = 30 - 20/3

w = 23.3 in

The dimensions are 10 ×10 × 23.3

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Calculate the line integral /w + V1 + a2)dx + 3rdy, where C consists of five line segments: from (1,0) to (2,0), from (2,0) to (2,1), from (2,1) to (-2,1), from (-2,1) to (-2, -2), and from (-2, - 2) to (1, -2). Hint: Use the Green's Theorem.

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By applying Green's Theorem and evaluating the double integral of the curl of F, we can calculate the line integral of (w + v + a^2)dx + 3ydy along the given closed curve C.

Green's Theorem states that for a vector field F = (P, Q) and a closed curve C oriented counterclockwise, the line integral of F along C is equal to the double integral of the curl of F over the region R bounded by C.

In this case, the given vector field is F = (w + v + a^2)dx + 3ydy, where w, v, and a are constants. To apply Green's Theorem, we need to calculate the curl of F. The curl of F is given by ∇ x F, which in this case becomes ∇ x F = (∂/∂x)(3y) - (∂/∂y)(w + v + a^2). Simplifying, we have ∇ x F = 3 - 0 = 3.

The region bounded by C consists of five line segments. By evaluating the double integral of the curl of F over this region, we can find the line integral of F along C. However, without knowing the specific values of w, v, and a, we cannot provide the numerical result of the line integral.

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Given t² - 4 f(x) 1² -dt 1 + cos² (t) At what value of x does the local max of f(x) occur? x =

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The value of x at which the local maximum of the function f(x) occurs is within the interval -√2 < x < √2.

To find the value of x at which the local maximum of the function f(x) occurs, we need to find the critical points of f(x) and then determine which one corresponds to a local maximum.

Let's start by differentiating f(x) with respect to x. Using the chain rule, we have:

f'(x) = d/dx ∫[1 to x] (t² - 4) / (1 + cos²(t)) dt.

To find the critical points, we need to find the values of x for which f'(x) = 0.

Setting f'(x) = 0, we have:

0 = d/dx ∫[1 to x] (t² - 4) / (1 + cos²(t)) dt.

Now, we can apply the Fundamental Theorem of Calculus (Part I) to differentiate the integral:

0 = (x² - 4) / (1 + cos²(x)).

To solve for x, we need to eliminate the denominator. We can do this by multiplying both sides of the equation by (1 + cos²(x)):

0 = (x² - 4) * (1 + cos²(x)).

Expanding the equation, we have:

0 = x² + x²cos²(x) - 4 - 4cos²(x).

Combining like terms, we get:

2x²cos²(x) - 4cos²(x) = 4 - x².

Now, let's factor out the common term cos²(x):

cos²(x)(2x² - 4) = 4 - x².

Dividing both sides by (2x² - 4), we have:

cos²(x) = (4 - x²) / (2x² - 4).

Simplifying further, we get:

cos²(x) = 2 / (x² - 2).

To find the values of x for which this equation holds, we need to consider the range of the cosine function. Since cos²(x) lies between 0 and 1, the right-hand side of the equation must also be between 0 and 1. This gives us the inequality:

0 ≤ (4 - x²) / (2x² - 4) ≤ 1.

Simplifying the inequality, we have:

0 ≤ (4 - x²) / 2(x² - 2) ≤ 1.

To find the values of x that satisfy this inequality, we can consider different cases.

Case 1: (4 - x²) / 2(x² - 2) = 0.

This occurs when the numerator is 0, i.e., 4 - x² = 0. Solving this equation, we find x = ±2.

Case 2: (4 - x²) / 2(x² - 2) > 0.

In this case, both the numerator and denominator have the same sign. Since the numerator is positive (4 - x² > 0), we need the denominator to be positive as well (x² - 2 > 0). Solving x² - 2 > 0, we get x < -√2 or x > √2.

Case 3: (4 - x²) / 2(x² - 2) < 1.

Here, the numerator and denominator have opposite signs. The numerator is positive (4 - x² > 0), so the denominator must be negative (x² - 2 < 0). Solving x² - 2 < 0, we find -√2 < x < √2.

Putting all the cases together, we have the following intervals:

Case 1: x = -2 and x = 2.

Case 2: x < -√2 or x > √2.

Case 3: -√2 < x < √2.

Now, we need to determine which interval corresponds to a local maximum. To do this, we can analyze the sign of the derivative f'(x) in each interval.

For x < -√2 and x > √2, the derivative f'(x) is negative since (x² - 4) / (1 + cos²(x)) < 0.

For -√2 < x < √2, the derivative f'(x) is positive since (x² - 4) / (1 + cos²(x)) > 0.

Therefore, the local maximum of f(x) occurs in the interval -√2 < x < √2.

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π π 7 Find the volume of the region bounded above by the surface z = 4 cos x cos y and below by the rectangle R: 0≤x≤ 0sy≤ 2. 4 V= (Simplify your answer. Type an exact answer, using radicals a

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Substituting this back into the integral: V = 4 sin 2 sin 2 = 4 sin² 2.

The volume of the region is 4 sin² 2.

To find the volume of the region bounded above by the surface z = 4 cos x cos y and below by the rectangle R: 0 ≤ x ≤ π, 0 ≤ y ≤ 2, we can set up a double integral.

The volume can be calculated using the following integral:

[tex]V = ∬R f(x, y) dA[/tex]

where f(x, y) represents the height function, and dA represents the area element.

In this case, the height function is given by f(x, y) = 4 cos x cos y, and the area element dA is dx dy.

Setting up the integral:

[tex]V = ∫[0, π] ∫[0, 2] 4 cos x cos y dx dy[/tex]

Integrating with respect to x first:

[tex]V = ∫[0, π] [4 cos y ∫[0, 2] cos x dx] dy[/tex]

The inner integral with respect to x is:

[tex]∫[0, 2] cos x dx = [sin x] from 0 to 2 = sin 2 - sin 0 = sin 2 - 0 = sin 2[/tex]

Substituting this back into the integral:

[tex]V = ∫[0, π] [4 cos y (sin 2)] dy[/tex]

Now integrating with respect to y:

[tex]V = 4 sin 2 ∫[0, 2] cos y dy[/tex]

The integral of cos y with respect to y is:

[tex]∫[0, 2] cos y dy = [sin y] from 0 to 2 = sin 2 - sin 0 = sin 2 - 0 = sin 2[/tex]

Substituting this back into the integral:

[tex]V = 4 sin 2 sin 2 = 4 sin² 2[/tex]

Therefore, the volume of the region is 4 sin² 2.

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Consider the curves y = 3x2 +6x and y = -42 +4. a) Determine their points of intersection (1.01) and (22,92)ordering them such that 1

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The problem asks us to find the points of intersection between two curves, y = 3x^2 + 6x and y = -4x^2 + 42. The given points of intersection are (1.01) and (22, 92), and we need to order them such that the x-values are in ascending order.

To find the points of intersection, we set the two equations equal to each other and solve for x: 3x^2 + 6x = -4x^2 + 42. Simplifying the equation, we get 7x^2 + 6x - 42 = 0. Solving this quadratic equation, we find two solutions: x ≈ -3.21 and x ≈ 1.01. Given the points of intersection (1.01) and (22, 92), we order them in ascending order of their x-values: (-3.21, -42) and (1.01, 10.07). Therefore, the ordered points of intersection are (-3.21, -42) and (1.01, 10.07).

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Factor completely. Remember you will first need to expand the brackets, gather like terms
and then factor.
a) (x + 4)^2 - 25
b)(a-5)^2-36

Answers

The completely factored form of (x + 4)^2 - 25 is (x - 1)(x + 9), and the completely factored form of (a - 5)^2 - 36 is (a - 11)(a + 1).

To factor completely the expression (x + 4)^2 - 25, we can first expand the square of the binomial, which gives us x^2 + 8x + 16 - 25. Simplifying further, we have x^2 + 8x - 9. Now, we need to factor this quadratic expression. The factors of -9 that add up to 8 are -1 and 9. So, we can rewrite the expression as (x - 1)(x + 9). Therefore, the completely factored form is (x - 1)(x + 9).

Similarly, for the expression (a - 5)^2 - 36, we expand the square of the binomial to get a^2 - 10a + 25 - 36. Simplifying further, we have a^2 - 10a - 11. To factor this quadratic expression, we need to find two numbers that multiply to give -11 and add up to -10. The factors are -11 and 1. Therefore, the completely factored form is (a - 11)(a + 1).

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Is the function below continuous? If not, determine the x values where it is discontinuous. f(x) = {2²²1²² -2²-2x-1 if 5-4 if -4

Answers

The function is not continuous. In fact, it is discontinuous at x = -4 and x = 5.

A continuous function is one for which infinitesimal modifications in the input cause only minor changes in the output. A function is said to be continuous at some point x0 if it satisfies the following three conditions: lim x→x0 f(x) exists. The limit at x = x0 exists and equals f(x0). f(x) is finite and defined at x = x0. Here is a simple method for testing if a function is continuous at a particular point: check if the limit exists, evaluate the function at that point, and compare the two results. If they are equal, the function is continuous at that point. If they aren't, it's not. The function f(x) = {2²²1²² -2²-2x-1 if 5-4 if -4 is not continuous.

The function has two pieces, each with a different definition. As a result, we need to evaluate the limit of each piece and compare the two to determine if the function is continuous at each endpoint. Let's begin with the left end point: lim x→-4- f(x) = 2²²1²² -2²-2(-4)-1= 2²²1²² -2²+8-1= 2²²1²² -2²+7= 4,611,686,015,756,800 - 4 = 4,611,686,015,756,796.The right-hand limit is given by lim x→5+ f(x) = -4 because f(x) is defined as -4 for all x greater than 5.Since lim x→-4- f(x) and lim x→5+ f(x) exist and are equal to 4,611,686,015,756,796 and -4, respectively, the function is discontinuous at x = -4 and x = 5 because the limit does not equal the function value at those points.

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Determine whether the series is convergent or divergent by expressing the nth partial sum s, as a telescoping sum. If it is convergent, find its sum. (If the quantity diverges, enter DIVERGES.) 8 n2 n = 4 X

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Thus, the given series is a telescoping series. The sequence of the nth partial sum is as follows:S(n) = 4 [1 + 1/(n(n − 1))]We can see that limn → ∞ S(n) = 4Hence, the given series is convergent and its sum is 4. Hence, the option that correctly identifies whether the series is convergent or divergent and its sum is: The given series is convergent and its sum is 4.

Given series is 8n²/n! = 8n²/(n × (n − 1) × (n − 2) × ....... × 3 × 2 × 1)= (8/n) × (n/n − 1) × (n/n − 2) × ...... × (3/n) × (2/n) × (1/n) × n²= (8/n) × (1 − 1/n) × (1 − 2/n) × ..... × (1 − (n − 3)/n) × (1 − (n − 2)/n) × (1 − (n − 1)/n) × n²= (8/n) × [(n − 1)/n] [(n − 2)/n] ...... [(3/n) × (2/n) × (1/n)] × n²= (8/n) × [(n − 1)/n] [(n − 2)/n] ...... [(3/n) × (2/n) × (1/n)] × n²= [8/(n − 2)] × [(n − 1)/n] [(n − 2)/(n − 3)] ...... [(3/2) × (1/1)] × 4

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Would using the commutative property of addition be a good strategy for simplifying 35+82 +65? Explain why or why not.​

Answers

Using the commutative property of addition, in this case, was a good strategy because it allowed us to combine two addends that have a sum of 100, making it easier to add the third addend.

The commutative property of addition states that changing the order of addends does not change the sum. For example, 2 + 5 is the same as 5 + 2. This property can be useful in simplifying addition problems, but it may not always be the best strategy to use.

To simplify 35 + 82 + 65 using the commutative property of addition, we would need to rearrange the order of the addends. We could add 35 and 65 first since they have a sum of 100. Then, we could add 82 to 100 to get a final sum of 182.

35 + 82 + 65 = (35 + 65) + 82 = 100 + 82 = 182. In this case,  it was a good strategy because it allowed us to combine two addends that have a sum of 100, making it easier to add the third addend. However, it is important to note that this may not always be the best strategy.

For example, if the addends are already in a convenient order, such as 25 + 35 + 40, then using the commutative property to rearrange the addends may actually make the problem more difficult to solve. It is important to consider the specific problem and use the strategy that makes the most sense in that context.

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Let P5 denote the vector space of all one-variable polynomials of degree at most 5. Which of the following are subspaces of P? (Mark all that apply.) All p(x) in P, with p(0) > 0. All p(x) in P5 with degree at most 3. All p(x) in P5 with p'(4) = 0. All p(x) in P, with p'(3) = 2. 5

Answers

To determine which of the given sets are subspaces of P5, we need to check if they satisfy the three conditions for being a subspace:

1. The set is closed under addition.

2. The set is closed under scalar multiplication.

3. The set contains the zero vector.

Let's evaluate each set based on these conditions:

1. All p(x) in P, with p(0) > 0:

This set is not a subspace of P5 because it is not closed under addition. For example, if we take two polynomials p(x) = x^2 and q(x) = -x^2, both p(x) and q(x) satisfy p(0) > 0, but their sum p(x) + q(x) = x^2 + (-x^2) = 0 does not have a positive value at x = 0.

2. All p(x) in P5 with degree at most 3:

This set is a subspace of P5. It satisfies all three conditions: closure under addition, closure under scalar multiplication, and contains the zero vector (the zero polynomial of degree at most 3).

3. All p(x) in P5 with p'(4) = 0:

This set is not a subspace of P5 because it is not closed under addition. If we take two polynomials p(x) = x^2 and q(x) = -x^2, both p(x) and q(x) satisfy p'(4) = 0, but their sum p(x) + q(x) = x^2 + (-x^2) = 0 does not have a derivative of 0 at x = 4.

4. All p(x) in P, with p'(3) = 2:

This set is a subspace of P5. It satisfies all three conditions: closure under addition, closure under scalar multiplication, and contains the zero vector (the zero polynomial).

Based on the above analysis, the sets that are subspaces of P5 are:

- All p(x) in P5 with degree at most 3.

- All p(x) in P, with p'(3) = 2.

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Exercise5 : Find the general solution of the ODE 4y'' – 20y' + 25y = (1 + x + x2) cos (3x). Exercise6 : Find the general solution of the ODE d²y + 49 y = 2x² sin (7x). dr2

Answers

The general solution of the ODE 4y'' - 20y' + 25y = (1 + x + x²) cos(3x) is y = c₁ e²(2.5x) + c₂ x e²(2.5x) + A + Bx + Cx² + D cos(3x) + E sin(3x).The general solution of the ODE d²y + 49y = 2x² sin(7x) is y = c₁ e²(7ix) + c₂ e²(-7ix) + (Ax²+ Bx + C) sin(7x) + (Dx² + Ex + F) cos(7x).

Exercise 5: To find the general solution of the given ordinary differential equation (ODE), 4y'' - 20y' + 25y = (1 + x + x²) cos(3x)

Step 1: Find the complementary solution:

Assume y = e²(rx) and substitute it into the ODE:

4(r² e²(rx)) - 20(r e²(rx)) + 25(e²(rx)) = 0

Simplify the equation by dividing through by e²(rx):

4r² - 20r + 25 = 0

Solve this quadratic equation to find the values of r:

r = (20 ± √(20² - 4 ×4 × 25)) / (2 × 4)

r = (20 ± √(400 - 400)) / 8

r = (20 ± √0) / 8

r = 20 / 8

r = 2.5

y-c = c₁ e²(2.5x) + c₂ x e²(2.5x)

Step 2: Find the particular solution:

To find the particular solution the method of undetermined coefficients the particular solution has the form

y-p = A + Bx + Cx² + D cos(3x) + E sin(3x)

Substitute this into the ODE and solve for the coefficients A, B, C, D, and E by comparing like terms.

Step 3: Combine the complementary and particular solutions

The general solution is obtained by adding the complementary and particular solutions

y = y-c + y-p

Exercise 6: To find the general solution of the given ODE d²y + 49y = 2x² sin(7x),

Step 1: Find the complementary solution

Assume y = e²(rx) and substitute it into the ODE

(r² e²(rx)) + 49(e²(rx)) = 0

Simplify the equation by dividing through by e²(rx)

r² + 49 = 0

Solve this quadratic equation to find the values of r:

r = ±√(-49)

r = ±7i

The complementary solution is given by:

y-c = c₁ e²(7ix) + c₂ e²(-7ix)

Step 2: Find the particular solution:

To find the particular solution the method of undetermined coefficients  the particular solution has the form:

y-p = (Ax² + Bx + C) sin(7x) + (Dx² + Ex + F) cos(7x)

Substitute this into the ODE and solve for the coefficients A, B, C, D, E, and F

Step 3: Combine the complementary and particular solutions:

The general solution is obtained by adding the complementary and particular solutions:

y = y-c + y-p

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1. DETAILS MY NOTES ASK YOUR TEACHER Suppose that f(4) = 2 and f'(4) = -3. Find h'(4). Round your answer to two decimal places. (a) h(x) = = (3x? + - 5ln (f(x)) ? h'(4) = (b) 60f(x) h(x) = 2x + 3 h'(4

Answers

By using differentiation we find the value of h'(4) = 48.5.

To find h'(4), we need to differentiate the function h(x) with respect to x and then evaluate the derivative at x = 4.

(a) [tex]h(x) = 3x² - 5ln(f(x))[/tex]

To find h'(x), we'll differentiate each term separately using the power rule and chain rule:

[tex]h'(x) = 6x - 5 * (1/f(x)) * f'(x)[/tex]

Since we are given that f(4) = 2 and f'(4) = -3, we can substitute these values into the derivative expression:

[tex]h'(4) = 6(4) - 5 * (1/f(4)) * f'(4)[/tex]

= 24 - 5 * (1/2) * (-3)

= 24 + 15/2

= 48.5

Therefore, h'(4) = 48.5.

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if A(x) = x^2+4 and Q(x) = x^2+8x evaluate the following:

A(8)

Answers

Answer:

A(x)=68

Step-by-step explanation:

Q(x) is unnecessary in finding any value of A(x) in this instance

Plug is 8 for all x values in the function A(x)

A(x)=8^2+4

A(x)=64+4

A(x)=68

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which of the following describe the channels and data transfer rates used for isdn bri? (select two.) answer 30 b channels operating at 64 kbps each. two b channels operating at 64 kbps each. one d channel operating at 64 kbps. one d channel operating at 16 kbps. 23 b channels operating at 64 kbps each. which of the following tends to be the most difficult step to achieve in the process of analogical problem solving? which step in the new-product development process involves seeking feedback from consumers in order to know what response they would give to a proposed product? What is meant by the channels of monetary transmission? Explain the asset price channels ii. Compare uncovered interest rate parity (UIRP) with covered interest rate parity (CIRP). Explain the relatio Which of the following statements is INCORRECT regarding the disadvantages of simulation?a. The summary of the simulation data only provides estimates about the real system.b. The process of developing a simulation model of a complex system can be time-consuming.c. The larger the number of probabilistic inputs a system has, the less likely a simulation will provide the best approach for studying the system.d. Each simulation run only provides a sample of how the real system will operate. the beginning and ending balances of retained earnings for the year were $66,000 and $77,000, respectively. if dividend payments made during the year were $6,600, determine the net income or net loss for the year. where does strategy formulation fit within the polc framework Calibration of dental x-ray equipment can be preformed by a dentist, dental hygienist, or dental assistant. 6. Set up, but do not evaluate an integral representing the arc length of the curve r(t) = (cost, e". Int), where 2 A company manufactures and sells * television sets per month. The monthly cost and revenue equations are C(x) = 72,000+60X R(x)=200x r? 30 OS XS6,000 (1) Find the maximum revenue. [5] (i" known to exert strong anti-inflammatory properties in the human body 25 Cost (dollars per unit) 15 LRAC 10 5 0 5 10 15 20 25 30 Quantity (units per hour) 1. In the above figure, the long run average cost LRAC between 0 and 10 units per hour what does the firm exhibit ( Simplify each expression. When simplifying these algebraic expressions,here are some ideas to think about: Did you cancel out common factors? Did vou combine like terms? Did you clear complex fractions Are all your exponents positive? Did you use any trigonometric identities(see Appendix A at the end)? 4x+1-4x.2x+1.2x x+1 sin x 21+x- i+x (c) 1+x x+4-2x(x+4 x+4 2x-+2y-1 (f) x+y sin2x (e) cosx your customer, rodrigo, is 40 years old. he is married and has four sons. five years ago, he purchased a nonqualified variable annuity for $20,000. it has grown to $25,000 by the time he withdraws $10,000 to pay for his oldest son's college tuition. how will the withdrawal be taxed? Help solveConsider the following cost' function. a. Find the average cost and marginal cost functions. b. Determine the average and marginal cost when x = a. c. Interpret the values obtained in part (b). C(x)= Keyser Corporation, which has only one product, has provided the following data concerning its most recent month of operations Selling price: $1,500Units in beginning inventory: 1,500Units produced: 9,400Units sold: 9,500Units in ending inventory: 1,400Variable costs per unit:Direct materials: $40Direct labor: $57Variable manufacturing overhead: $21Variable selling and administrative expense: $31Fixed costs:Fixed manufacturing overhead: $75,200Fixed selling and administrative expense: $167,500The company produces the same number of units every month, although the sales in units vary from month to month. The company's variable costs per unit and total fixed costs have been constant from month to monthWhat is the net operating income for the month under absorption costing?a. $5,200b. $35,100c. $5,200d. $22,500 Consider the following vector field F(x, y) = Mi + Nj. F(x, y) = x?i + yj (a) Show that F is conservative. = ON ax = = ay (b) Verify that the value of lo F.dr is the same for each parametric representation of C. (1) C: 1/(t) = ti + t2j, ostsi Sa F. dr = = (ii) Cz: r2(0) = sin(o)i + sin(e)j, o SOS T/2 Ja F. dr = C2 complete a service blueprint for each of the following banking transactions. use the service actions listed in the tables below each name of a transaction. a. make a savings deposit using a teller. i teller prints out receipt and hands to customer. iii lunch and rest breaks are managed based on waiting lines. ii teller checks cash or check and enters amount. iv cash is counted and reconciled with transactions. .The variables x and y vary inversely. Use the given values to write an equation relating x and y. Then find y when x = 3. x = 1, y = 9 " I couldn't hardly hear the teacher because the other students were making to much noise" which aspect of the sentence shows evidenceA. slangB. double negative C. sentence structureD. intensive pronoun Steam Workshop Downloader