A vector v has an initial (2,-3) point and terminal point (3,-4)
Write in component form.

Using the arrangements formula, it is found that 1680 arrangements can be made with these numbers.

The number of possible arrangements of n elements is given by the factorial of n, that is:

[tex]A_n = n![/tex]

When there are repeated elements, repeating [tex]n_1, n_2, \cdots, n_n[/tex] times, the number of arrangements is given by:

[tex]A_n^{n_1, n_2, \cdots, n_n} = \frac{n!}{n_1!n_2! \cdots n_n!}[/tex]

For the number 28535852, we have that:

Hence the number of arrangements is:

[tex]A_8^{3,2,2} = \frac{8!}{3!2!2!} = 1680[/tex]

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The area of the garden is 74 square feet

The complete question is added as an attachment

From the attached figure, we have the following shapes and dimensions:

The rectangular area is

A1 = 10 * 5

Evaluate

A1 = 50

The area of the trapezoid is

A2 = 0.5 * Sum of parallel bases * height

This gives

A2 = 0.5 * (10 + 6) * 3

Evaluate

A2 = 24

The total area is

Total = A1 + A2

This gives

Total = 50 + 24

Evaluate

Total = 74

Hence, the area of the garden is 74 square feet

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The domain of the relation shown in the graph is -4 <= x <= 4

The relation on the graph is an ellipse function.

The domain is the set of x values the function can take

From the graph, we have the following x values

This means that the domain of x in the graph is -4 <= x <= 4

Hence, the domain of the relation shown in the graph is -4 <= x <= 4

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a. Given the 2nd order ODE

[tex]y''(x) = 4y(x) + 4[/tex]

if we substitute [tex]z(x)=y'(x)+2y(x)[/tex] and its derivative, [tex]z'(x)=y''(x)+2y'(x)[/tex], we can eliminate [tex]y(x)[/tex] and [tex]y''(x)[/tex] to end up with the ODE

[tex]z'(x) - 2y'(x) = 4\left(\dfrac{z(x)-y'(x)}2\right) + 4[/tex]

[tex]z'(x) - 2y'(x) = 2z(x) - 2y'(x) + 4[/tex]

[tex]\boxed{z'(x) = 2z(x) + 4}[/tex]

and since [tex]y(0)=y'(0)=1[/tex], it follows that [tex]z(0)=y'(0)+2y(0)=3[/tex].

b. I'll solve with an integrating factor.

[tex]z'(x) = 2z(x) + 4[/tex]

[tex]z'(x) - 2z(x) = 4[/tex]

[tex]e^{-2x} z'(x) - 2 e^{-2x} z(x) = 4e^{-2x}[/tex]

[tex]\left(e^{-2x} z(x)\right)' = 4e^{-2x}[/tex]

[tex]e^{-2x} z(x) = -2e^{-2x} + C[/tex]

[tex]z(x) = -2 + Ce^{2x}[/tex]

Since [tex]z(0)=3[/tex], we find

[tex]3 = -2 + Ce^0 \implies C=5[/tex]

so the particular solution for [tex]z(x)[/tex] is

[tex]\boxed{z(x) = 5e^{-2x} - 2}[/tex]

c. Knowing [tex]z(x)[/tex], we recover a 1st order ODE for [tex]y(x)[/tex],

[tex]z(x) = y'(x) + 2y(x) \implies y'(x) + 2y(x) = 5e^{-2x} - 2[/tex]

Using an integrating factor again, we have

[tex]e^{2x} y'(x) + 2e^{2x} y(x) = 5 - 2e^{2x}[/tex]

[tex]\left(e^{2x} y(x)\right)' = 5 - 2e^{2x}[/tex]

[tex]e^{2x} y(x) = 5x - e^{2x} + C[/tex]

[tex]y(x) = 5xe^{-2x} - 1 + Ce^{-2x}[/tex]

Since [tex]y(0)=1[/tex], we find

[tex]1 = 0 - 1 + Ce^0 \implies C=2[/tex]

so that

[tex]\boxed{y(x) = (5x+2)e^{-2x} - 1}[/tex]

6.a) The differential equation for z(x) is z'(x) = 2z(x) + 4, z(0) = 3.

6.b) The value of z(x) is [tex]z(x) = 5e^{2x} - 2[/tex].

6.c) The value of y(x) is [tex]y(x) = \frac{5e^{2x}}{4} - \frac{1}{4e^{2x}} -1[/tex].

The given ordinary differential equation is y''(x) = 4y(x) + 4, y(0) = y'(0) = 1 ... (d).

We are also given a substitution function, z(x) = y'(x) + 2y(x) ... (z).

Putting x = 0, we get:

z(0) = y'(0) + 2y(0),

or, z(0) = 1 + 2*1 = 3.

Rearranging (z), we can write it as:

z(x) = y'(x) + 2y(x),

or, y'(x) = z(x) - 2y(x) ... (i).

Differentiating (z) with respect to x, we get:

z'(x) = y''(x) + 2y'(x),

or, y''(x) = z'(x) - 2y'(x) ... (ii).

Substituting the value of y''(x) from (ii) in (d) we get:

y''(x) = 4y(x) + 4,

or, z'(x) - 2y'(x) = 4y(x) + 4.

Substituting the value of y'(x) from (i) we get:

z'(x) - 2y'(x) = 4y(x) + 4,

or, z'(x) - 2(z(x) - 2y(x)) = 4y(x) + 4,

or, z'(x) - 2z(x) + 4y(x) = 4y(x) + 4,

or, z'(x) = 2z(x) + 4y(x) - 4y(x) + 4,

or, z'(x) = 2z(x) + 4.

The initial value of z(0) was calculated to be 3.

6.a) The differential equation for z(x) is z'(x) = 2z(x) + 4, z(0) = 3.

Transforming z(x) = dz/dx, and z = z(x), we get:

dz/dx = 2z + 4,

or, dz/(2z + 4) = dx.

Integrating both sides, we get:

∫dz/(2z + 4) = ∫dx,

or, {ln (z + 2)}/2 = x + C,

or, [tex]\sqrt{z+2} = e^{x + C}[/tex],

or, [tex]z =Ce^{2x}-2[/tex] ... (iii).

Substituting z = 3, and x = 0,  we get:

[tex]3 = Ce^{2*0} - 2\\\Rightarrow C - 2 = 3\\\Rightarrow C = 5.[/tex]

Substituting C = 5, in (iii), we get:

[tex]z = 5e^{2x} - 2[/tex].

6.b) The value of z(x) is [tex]z(x) = 5e^{2x} - 2[/tex].

Substituting the value of z(x) in (z), we get:

z(x) = y'(x) + 2y(x),

or, 5e²ˣ - 2 = y'(x) + 2y(x),

which gives us:

[tex]y(x) = \frac{5e^{2x}}{4} - \frac{1}{4e^{2x}} -1[/tex] for the initial condition y(x) = 0.

6.c) The value of y(x) is [tex]y(x) = \frac{5e^{2x}}{4} - \frac{1}{4e^{2x}} -1[/tex].

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The most reasonable unit to measure the time spent at school on an average school day is in hours.

In anything we are going to measure, we have to pay special attention to the unit. When doing this, we want to keep the absolute value of the measure, that is, the number small, hence the do this conversion of units may be used.

For example, it is easier and more practical to say one day instead of 87,840 seconds. The same logic is applied to this problem, in which the most reasonable unit to measure the time spent at school on an average school day is in hours, as it keeps the numerical measure smaller than it would be in minutes or seconds.

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Answer:

3/10

Step-by-step explanation:

There are totally 5 disks. On the first pick, there are 3 red discs. So

P(red disc on first pick ) = 3/5

Assuming that a red disc has been picked the first time, there will be 2 red discs and 2 blue discs so

(P red disc on second pick given red disc on first pick) = 2/4

P(red disc on both picks) = 3/5 x 2/4 = 3/10

The equation for the reflected function is:

g(x) =  (1/2)*(3)^(-x)

Here we know that function g(x) is a reflection of f(x) across the y-axis.

Remember that the reflection is written as:

g(x) = f(-x)

Here we know that:

f(x) = (1/2)*(3)^x

Then, replacing that in the equation for g(x), we get:

g(x) =  f(-x) = (1/2)*(3)^(-x)

g(x) =  (1/2)*(3)^(-x)

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[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]

The straight angles are angles that form a straight line, and their measure = 180°

In the given figure, the Straight angles is :

[tex] \qquad \large \sf {Conclusion} : [/tex]

The sales last year of the company is 5617

The given parameters are:

Company sales = 6740

Percentage increase = 20%

The sales of the company last year (x) to date is calculated as:

Company sales = (1 + Proportion) * Last year sales

Substitute the known values in the above equation

6740 = (1 + 20%) * x

Evaluate the sum

6740 = 1.2 * x

Divide both sides by 1.2

x = 5617

Hence, the sales last year is 5617

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Answer:

  1∠22.5°, 1∠112.5°, 1∠202.5°, 1∠292.5°

Step-by-step explanation:

A root of a complex number can be found using Euler's identity.

For some z = a·e^(ix), the n-th root is ...

  z = (a^(1/n))·e^(i(x/n))

Here, we have z = i, so a = 1 and z = π/2 +2kπ.

Using r∠θ notation, this is ...

  i = 1∠(90° +k·360°)

and

  i^(1/4) = (1^(1/4))∠((90° +k·360°)/4)

  i^(1/4) = 1∠(22.5° +k·90°)

For k = 0 to 3, we have ...

 for k = 0, first root = 1∠22.5°

  for k = 1, second root = 1∠112.5°

  for k = 2, third root = 1∠202.5°

  for k = 3, fourth root = 1∠292.5°

if any number or variable is outside a parenthesis it will go to all the number rin the parenthesis.

in this case -y is outside the parenthesis so it will got to both the sides :

-y(-6y-3) = -6y x y - 3 x y

= -6y² -3y//

[tex]v(x) = wlh \\ 2x {}^{3} + {x}^{2} - 16 x- 15 =(2x + 5)(x - 3)h \\ h = \frac{2x {}^{3} + x {}^{2} - 16x - 15 }{2x {}^{2} - x - 15 } \\ using \: long \: divison \\ h = x + 1[/tex]

Step-by-step explanation:

An acute angle is an angle less or equal to 90 degrees.

let the other angle be x.

x + 36 = 90

grouping like terms

X = 90 - 36

X = 54 degrees

Answer:

54

Step-by-step explanation:

= ∠A + ∠B + ∠C = 180o … [∵sum of the angles of a triangle is 180]

or 180 = 36 + 90 + ∠C

C=54

[tex]x > \ln(x)[/tex] for all [tex]x[/tex], so

[tex]\displaystyle \lim_{x\to\infty} (\ln(x) - x) = - \lim_{x\to\infty} x = \boxed{-\infty}[/tex]

Similarly, [tex]\displaystyle \lim_{x\to\infty} (x-e^x) = - \lim_{x\to\infty} e^x = \boxed{-\infty}[/tex]

We can of course see the limits are identical by replacing [tex]x\mapsto e^x[/tex], so that

[tex]\displaystyle \lim_{x\to\infty} (\ln(x) - x) = \lim_{x\to\infty} (\ln(e^x) - e^x) = \lim_{x\to\infty} (x - e^x)[/tex]

You can also rewrite the limands to accommodate the application of l'Hôpital's rule. For instance,

[tex]\displaystyle \lim_{x\to\infty} (x - e^x) = \ln\left(\exp\left(\lim_{x\to\infty} (x - e^x)\right)\right) = \ln\left(\lim_{x\to\infty} e^{x-e^x}\right) = \ln\left(\lim_{x\to\infty} \frac{e^x}{e^{e^x}}\right)[/tex]

Using the rule, the limit here is

[tex]\displaystyle \lim_{x\to\infty} \frac{(e^x)'}{\left(e^{e^x}\right)'} = \lim_{x\to\infty} \frac{e^x}{e^x e^{e^x}} = \lim_{x\to\infty} \frac1{e^{e^x}} = 0[/tex]

so the overall limit is

[tex]\displaystyle \lim_{x\to\infty} (x - e^x) = \ln\left(\lim_{x\to\infty} \frac{e^x}{e^{e^x}}\right) = \ln(0) = \lim_{x\to0^+} \ln(x) = -\infty[/tex]

The correct option regarding whether the relation is a function is:

B. The relation is a function. Each input value is paired with one output value.

A relation represent a function if each value of the input is paired with one value of the output.

In this problem, when the input - output mappings are given by:

{(–2, 3), (–1, 3), (0, 2), (1, 4), (5, 5)}.

Which means that yes, each input value is paired with one output value, hence the relation is a function and option B is correct.

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The equation that represents the total cost, c, of the purchases made by Jake is: c = 3(0.89) + 2(2.98) + 4.99.

To write the equation that represents the total cost of a given scenario like the one given, you can use variables to represent each component that makes up the total cost in the given situation.

Thus, let:

a = apple = 3a

s = can of soup = 2s

b = box of cereal = b

c = total cost

We know that unit price of each items are:

Apple = $0.89

Can of soup = $2.98

Box of cereal = $4.99

Equation would be:

c = 3a + 2s + b

Plug in the values

c = 3(0.89) + 2(2.98) + 4.99

Therefore, the equation that represents the total cost, c, of the purchases made by Jake is: c = 3(0.89) + 2(2.98) + 4.99.

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Using translation concepts, it is found that pi is the phase shift of the graph, and since it is negative, the graph is shifted right pi units.

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction either in it’s definition or in it’s domain. Examples are shift left/right or bottom/up, vertical or horizontal stretching or compression, and reflections over the x-axis or the y-axis.

In this problem, the change is given as follows:

x -> x - pi

It means that the change is in the domain, in which pi is the phase shift of the graph, and since it is negative, the graph is shifted right pi units.

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Answer:

-2s-18 is the simplified answer

Step-by-step explanation:

s-3(s+6)

distributive property

s-3s-18

-2s-18

Answer:35443

Step-by-step explanation:

Answer:

d. (-6, 0)

Step-by-step explanation:

no explanation bc u need this quick

Answer: 15

Step-by-step explanation: Using the pythagorean theorem A^2 + B^2 = C^2 where C is the hypotenuse of the triangle and any A and B are any sides of the triangle that are not the hypotenuse; we can rewrite this as A^2 = C^2 - B^2 to find out the missing side. I picked A to be the unknown side but you could also choose B. We know the hypotenuse is 39, and side B is 36. 39^2 - 36^2, or 1521 - 1296, is 225.

Now we have A^2 = 225. To get rid of the square, we have to take the square root of both sides. The A^2 cancels to be A, and sqrt(225) is 15. Thus, side A, or the missing side is 15.

Hope this helped!

The Young Modulus of the brass rod ≈ 8.5 × 10¹⁰ Pa or N/m².

Given: Physical Properties of the Brass rod exist

Length, L = 250 mm

Radius, R = Diameter / 2 = 25/2 mm

Load of 50 kN hanging on the brass rod which results in elongation of the rod to 0.3 mm.

To estimate the Young Modulus of the brass rod we will require Stress on the rod, Ratio of elongation to the total length.

To find the stress on the rod,

Stress(Pressure) = Force / Area

⇒ Stress = 50,000 / (Area of cross-section)

The rod exists shaped like a cylinder then the area of its cross-section will be πR² ( R = radius)

⇒ Stress = 50000 / (π (25/2 mm)² )

The force exists given in newton, then we must convert the unit of the area into m² to obtain the solution in pascal.

1 mm = 0.001 m

⇒ Stress = 50,000 / 22/7 ( 25/2 × 10⁻³ m)²

take, π = 22/7

⇒ Stress = 50000 / ( 22/7 × 625 / 4 × 10⁻⁶ )

⇒ Stress = 50000 / ( 11/14 × 625 × 10⁻6 )

⇒ Stress = 50000 × 14 × 10⁶ / 625×11

⇒ Stress = 80 × 14 × 10⁶ / 11

⇒ Stress = 1120/11 × 10⁶ Pa

Now, that we maintain Stressed, to estimate Strain (ratio of elongation to original length),

⇒ Strain = (Elongation) / (Original length)

⇒ Strain = 0.3 / 250

⇒ Strain = 3 / 2500

Therefore, Young Modulus, Y = Stress / Strain

⇒ Y = 1120/11 × 10⁶ / 3/2500

⇒ Y = 1120/11 × 2500/3 × 10⁶

⇒ Y = 101.8 × 833.3 × 10⁶

⇒ Y = 84829.94 × 10⁶

⇒ Y ≈ 8.5 × 10¹⁰ Pa

Therefore, the Young Modulus of the brass rod ≈ 8.5 × 10¹⁰ Pa or N/m².

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Answer:

Point A:  (2, 10)

Point B:  (-3, 0)

Point C:  (-5, -4)

Point D:  (-5, -32)

Step-by-step explanation:

Points A and B are the points of intersection between the two graphs.

Therefore, to find the x-values of the points of intersection, substitute one equation into the other and solve for x:

[tex]\implies 2x+6=-2x^2+18[/tex]

[tex]\implies 2x^2+2x-12=0[/tex]

[tex]\implies 2(x^2+x-6)=0[/tex]

[tex]\implies x^2+x-6=0[/tex]

[tex]\implies x^2+3x-2x-6=0[/tex]

[tex]\implies x(x+3)-2(x+3)=0[/tex]

[tex]\implies (x-2)(x+3)=0[/tex]

[tex]\implies x=2, -3[/tex]

From inspection of the graph:

To find the y-values, substitute the found x-values into either of the equations:

[tex]\begin{aligned} \textsf{Point A}: \quad 2x+6 & =y\\2(2)+6 & =10\\ \implies & (2, 10)\end{aligned}[/tex]

[tex]\begin{aligned} \textsf{Point B}: \quad -2x^2+18 & =y\\-2(-3)^2+18 & =0\\ \implies & (-3,0)\end{aligned}[/tex]

Therefore, point A is (2, 10) and point B is (-3, 0).

If the distance between points C and D is 28 units, the y-value of point D will be 28 less than the y-value of point C.  The x-values of the two points are the same.

Therefore:

[tex]\textsf{Equation 1}: \quad y=2x+6[/tex]

[tex]\textsf{Equation 2}: \quad y-28=-2x^2+18[/tex]

As the x-values are the same, substitute the first equation into the second equation and solve for x to find the x-value of points C and D:

[tex]\implies 2x+6-28=-2x^2+18[/tex]

[tex]\implies 2x^2+2x-40=0[/tex]

[tex]\implies 2(x^2+x-20)=0[/tex]

[tex]\implies x^2+x-20=0[/tex]

[tex]\implies x^2+5x-4x-20=0[/tex]

[tex]\implies x(x+5)-4(x+5)=0[/tex]

[tex]\implies (x-4)(x+5)=0[/tex]

[tex]\implies x=4,-5[/tex]

From inspection of the given graph, the x-value of points C and D is negative, therefore x = -5.

To find the y-value of points C and D, substitute the found value of x into the two original equations of the lines:

[tex]\begin{aligned} \textsf{Point C}: \quad 2x+6 & =y\\2(-5)+6 & =-4\\ \implies & (-5,-4)\end{aligned}[/tex]

[tex]\begin{aligned} \textsf{Point D}: \quad -2x^2+18 & = y \\ -2(-5)^2+18 & =-32\\ \implies & (-5, -32)\end{aligned}[/tex]

Therefore, point C is (-5, -4) and point D is (-5, -32).

Answer:

a) A = (2, 10) and B = (-3, 0)

b) C = (-5, -4) and D = (-5, -32)

Explanation:

a) To determine the coordinates of A and B, find the intersection points of  the line "y = 2x + 6" and curve "y = -2x² + 18".

Solve the equation's simultaneously:

y = y

⇒ 2x + 6 =  -2x² + 18

⇒ 2x² + 2x + 6 - 18 = 0

⇒ 2x² + 2x - 12= 0

⇒ 2x² + 6x - 4x - 12 = 0

⇒ 2x(x + 3) - 4(x + 3) = 0

⇒ (2x - 4)(x + 3) = 0

⇒ 2x - 4 = 0, x + 3 = 0

⇒ x = 2, x = -3

Then find value of y at this x points,

at x = 2, y = 2(2) + 6 = 10

at x = -3, y = 2(-3) + 6 = 0

Intersection points: A(2, 10) and B(-3, 0)

b) Given that CD = 28 units. Also stated parallel to y axis so x coordinates for both will be same but differ in y coordinate.

[tex]2x + 6 = -2x^2 + 18 + 28[/tex]

[tex]-2x^2 + 18 + 28-2x - 6 = 0[/tex]

[tex]-2x^2-2x+40=0[/tex]

[tex]-2x^2-10x+8x+40=0[/tex]

[tex]-2(x+5)+8(x+5)=0[/tex]

[tex](-2x+8)(x+5)=0[/tex]

[tex]x = -5, 4[/tex]

[tex]\leftrightarrow \sf C(-5, y_2), \ D(-5, y_2)[/tex]

Find y value for Point C : 2x + 6 = 2(-5) + 6 = -4

Find y value for Point D :  -2x² + 18 =  -2(-5)² + 18 = -32

[tex]\sf \rightarrow Point \ C = (-5, -4)\\ \\\rightarrow Point \ D = (-5, -32)[/tex]

The expression that represents the perimeter of the rectangle is 30h - 2

The dimensions of the rectangle are given as:

Length = 7h + 3

Width = 8h - 4

The perimeter of the rectangle is given as:

P = 2 * (Length + Width)

Substitute the known values in the above equation

P =2 * (7h + 3 + 8h - 4)

Evaluate the like terms

P = 2 * (15h - 1)

Evaluate the product

P = 30h - 2

Hence, the expression that represents the perimeter of the rectangle is 30h - 2

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Answer: 45%

Step-by-step explanation:

Given:

Find:

The number of groups which contains 9 is 9 and the total number of groups are 20.

So,

P = 9 / 20 = 0.45 = 45%

Therefore, the experimental probability is 45%

Answer: 13

Step-by-step explanation:

4x+3y=-5 solve x :    

4x = -3y + -5           | -3y

1x = -0.75y + -1.25  | : 4

-3x + 7y = 13 solve x :

-3x + 7y = 13              | -7y

-3x = -7y = 13             | : (-3)

Equalization Method Solution: -0.75y+-1.25=2.333y+-4.333

-0, 75y - 1,25 = 2,333y - 4, 333 solve y:

-0, 75y - 1,25 = 2,333y -4, 333    | -2,333y

-3, 083y - 1,25 = -4,333               | + 1, 25

-3. 083y = -3,088                         | : (-3, 083)

y = 1

Plug y = 1 into the equation 4x + 3y = -5 :

4x + 3 · 1           | Multiply 3 with 1

4x + 3 = -5        | -3

4x = -8              | : 4

x = -2

So the solution is:

y = 1, x = -2

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Based on the mean of the sample and the proportion who read below grade level, the probability that a second sample would have a proportion greater than 0.54 is 0.9772.

The probability that the second sample would be selected with a proportion greater than 0.54 can be found as:

P (x > 0.54) = P ( z > (0.54 - 0.30) / 0.12))

Solving gives:

P (x > 0.54) = P (z > 2)

P (x > 0.54) = 0.9772

In conclusion, the probability that the second sample would be selected with a proportion greater than 0.54 is 0.9772.

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Answer: 250:1

Step-by-step explanation:

Divide the measurement of appearance by the actual measurement.

[tex]100,000/400=250[/tex]

The magnification scale is 250:1. The scale is used by a 250X magnifying lense.

(a) We want to find a scalar function [tex]f(x,y,z)[/tex] such that [tex]\mathbf F = \nabla f[/tex]. This means

[tex]\dfrac{\partial f}{\partial x} = 2xy + 24[/tex]

[tex]\dfrac{\partial f}{\partial y} = x^2 + 16[/tex]

Looking at the first equation, integrating both sides with respect to [tex]x[/tex] gives

[tex]f(x,y) = x^2y + 24x + g(y)[/tex]

Differentiating both sides of this with respect to [tex]y[/tex] gives

[tex]\dfrac{\partial f}{\partial y} = x^2 + 16 = x^2 + \dfrac{dg}{dy} \implies \dfrac{dg}{dy} = 16 \implies g(y) = 16y + C[/tex]

Then the potential function is

[tex]f(x,y) = \boxed{x^2y + 24x + 16y + C}[/tex]

(b) By the FTCoLI, we have

[tex]\displaystyle \int_{(1,1)}^{(-1,2)} \mathbf F \cdot d\mathbf r = f(-1,2) - f(1,1) = 10-41 = \boxed{-31}[/tex]

[tex]\displaystyle \int_{(-1,2)}^{(0,4)} \mathbf F \cdot d\mathbf r = f(0,4) - f(-1,2) = 64 - 41 = \boxed{23}[/tex]

[tex]\displaystyle \int_{(0,4)}^{(2,3)} \mathbf F \cdot d\mathbf r = f(2,3) - f(0,4) = 108 - 64 = \boxed{44}[/tex]