Layla got a new job through the Manchester Temporary Services. The job pays $53. 5K per year and the agency fee is equal to 32% of one month’s pay. How much must Layla pay the agency?

Jhon bought a rectangular door mat that was 1/2 meter long and 3/10 meter wide. The area of the rectangular door mat is 3/20 square meters.

Find the area of John's rectangular door mat that is 1/2 meter long and 3/10 meter wide, you'll need to multiply the length by the width.
Identify the length and width.
Length = 1/2 meter
Width = 3/10 meter
Multiply the length and width to find the area.
Area = Length × Width
Area = (1/2) × (3/10)
Calculate the multiplication.
Area = 3/20 square meters
The area of the rectangular door mat is 3/20 square meters.

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The variance of the actuary's wait time is 36 / 121.

From the information, at the airport, there are three counters for checking the luggage. the employees at each counter work independently with the time for each customer modeled as an exponential distribution. the average time is one minute for one counter, two for the next, and three minutes for the third. an actuary, who is next in line, will take the next available counter,

From the complete question, the variance of the actuary's wait time will be:

= 1 / (11/6)²

= (6/11)²

= 36 / 121

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Therefor, the length of mantel in blueprint is > 30 ft

width of the mantel in the blueprint 8ft×2inc/1ft=16inch

The term "width" refers to the length from side to side of anything. For instance, the shorter side of a rectangle would be the width.

we know that

[scale]=[blueprint]/[actual]-------> [actual]=[blueprint]/[scale]

[scale]=3/5 in/ft

for [wall blueprint]=18 in

[wall actual]=[wall blueprint]/[scale]-------> 18/(3/5)----> 30 ft

Part A)

the actual wall is 30 ft  long

Part B) window has actual width of 2.5 ft

[ window blueprint]=[scale]*[actual window]-----> (3/5)*2.5----> 1.5 in

the width of the window in the blueprint is 1.5 in

Part C) Complete the table

For [blueprint length]=4 in

[actual length]=[blueprint length]/[scale]-------> 4/(3/5)----> 20/3 ft

For [blueprint length]=5 in

[actual length]=[blueprint length]/[scale]-------> 5/(3/5)----> 25/3 ft

For [blueprint length]=6 in

[actual length]=[blueprint length]/[scale]-------> 6/(3/5)----> 30/3=10 ft

For [blueprint length]=7 in

[actual length]=[blueprint length]/[scale]-------> 7/(3/5)----> 35/3 ft

For [actual length]=6 ft

[blueprint length]=[actual length]*[scale]-------> 6*(3/5)----> 18/5 in

For [actual length]=7 ft

[blueprint length]=[actual length]*[scale]-------> 7*(3/5)----> 21/5 in

For [actual length]=8 ft

[blueprint length]=[actual length]*[scale]-------> 8*(3/5)----> 24/5 in

For [actual length]=9 ft

[blueprint length]=[actual length]*[scale]-------> 9*(3/5)----> 27/5 in

B) width of the mantel in the blueprint 8ft×2inc/1ft=16inch

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Answer: 5

2.75+(2.45x5) = 15

The width of the rectangular prism popcorn box is approximately 2.27 inches when rounded to the nearest tenth.

The volume of a right rectangular prism is given by:

V = lwh

where V is the volume, l is the length, w is the width, and h is the height.

We are given that the box can hold 46 cubic inches of popcorn, the length is 2¾ inches, and the height is 7½ inches. Let's use w to represent the width we are trying to find.

So we have:

46 = (2¾)w(7½)

To solve for w, we can divide both sides of the equation by (2¾)(7½):

46 / ((2¾)(7½)) = w

Simplifying the right-hand side, we get:

w ≈ 2.27

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Complete Question:

Guadalupe models the volume of a popcorn box as a right rectangular prism and the box can hold 46 cubic inches of popcorn when it is full. Its length is 2¾ inches and its height is 7½ inches. Find the width of the popcorn box in inches. Round your answer to the nearest tenth if necessary.

Angle between the planes is 0.978 radians

To find the angle between the planes 8x + y = -7 and 4x + 9y + 10z = -17, we need to follow these steps:

Step 1: Find the normal vectors of the planes. The coefficients of the variables in the plane equation (Ax + By + Cz = D) represent the components of the normal vector (A, B, C).

For the first plane (8x + y = -7), the normal vector is N1 = (8, 1, 0).
For the second plane (4x + 9y + 10z = -17), the normal vector is N2 = (4, 9, 10).

Step 2: Calculate the dot product of the normal vectors.
N1 · N2 = (8 * 4) + (1 * 9) + (0 * 10) = 32 + 9 + 0 = 41

Step 3: Calculate the magnitudes of the normal vectors.
|N1| = √(8² + 1² + 0²) = √(64 + 1) = √65
|N2| = √(4² + 9² + 10²) = √(16 + 81 + 100) = √197

Step 4: Find the cosine of the angle between the planes.
cos(angle) = (N1 · N2) / (|N1| * |N2|) = 41 / (√65 * √197)

Step 5: Calculate the angle in radians.
angle = arccos(cos(angle)) = arccos(41 / (√65 * √197))

Using a calculator, we find the acute angle between the planes to be approximately 0.978 radians (rounded to the nearest thousandth).

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The number 175 in Annie's equation represents the cost per hour per semester for classwork.

This means that for every additional hour of classwork a college student takes per semester, their fee increases by $175. It is important to note that this cost does not include the cost for housing, which is represented by the constant term of the equation, 3375. Therefore, the equation allows us to calculate the total fee a college student would pay for a semester based on the number of hours of classwork they take and the cost per hour.

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The complete factorization of the expression is  (m+n)(x-y).

The complete factorization of the expression is determined as follows;

To factorize the expression (m+n)(2x-y)-x(m+n), we can first factor out the common factor (m+n):

(m+n)(2x-y)-x(m+n) = (m+n)(2x-y-x)

Next, we will factorize completely as follows;

2x - x - y = x - y

(m+n)(2x-y-x)  = (m+n)(x-y)

Therefore, the fully factorized form of the expression is (m+n)(x-y).

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The discount is 26%.

The discount equals the difference between the price paid for and it's par value. Discount is a kind of reduction or deduction in the cost price of a product.

Given:

[tex]\bold{Marked} \ \text{price} = \$25[/tex]

[tex]\bold{Selling} \ \text{price} = \$18.50[/tex]

So,

[tex]\text{Discount = MP - SP}[/tex]

[tex]\text{Discount} = 25-18.50[/tex]

[tex]\bold{Discount} = 6.50[/tex]

Now,

[tex]\text{D}\% = \dfrac{\text{D}}{\text{MP}} \times100[/tex]

[tex]\text{D}\% = \dfrac{6.5}{25} \times100[/tex]

[tex]\text{D}\% = 26\%[/tex]

Hence, the discount percent is 26%.

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The red tractor is heavier.

To determine which tractor is heavier, Mr. Vega needs to compare the weights of the blue and red tractors. The blue tractor weighs [tex]\frac{3}{5}[/tex] of a ton, and the red tractor weighs [tex]\frac{4}{6}[/tex] of a ton.

First, we need to simplify the fractions if possible. In this case, we can simplify the red tractor's fraction by dividing both the numerator and denominator by 2:
[tex]\frac{4}{6} = \frac{\frac{4}{2} }{\frac{6}{2} } = \frac{2}{3}[/tex]

Now we can compare the simplified fractions:
[tex]Blue tractor: \frac{3}{5}[/tex]
[tex]Red tractor: \frac{2}{3}[/tex]

To compare these fractions, we can find a common denominator. The least common multiple of 5 and 3 is 15. To convert the fractions to the same denominator, we multiply the numerators and denominators by the necessary factors:

[tex]Red tractor: (\frac{2}{3}) (\frac{5}{5}) = \frac{10}{15}[/tex]

[tex]Blue tractor: (\frac{3}{5}) (\frac{3}{3}) = \frac{9}{15}[/tex]

Now we can easily compare the weights:
[tex]Blue tractor: \frac{9}{15}[/tex]

[tex]Red tractor: \frac{10}{15}[/tex]


Since [tex]\frac{10}{15}[/tex] is greater than [tex]\frac{9}{15}[/tex] , the red tractor is heavier.

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The expression 6\cdot6\cdot6\cdot6\cdot66\cdot6\cdot6\cdot6\cdot66 is equivalent to 60534416.

The given expression is:

6\cdot6\cdot6\cdot6\cdot66\cdot6\cdot6\cdot6\cdot66

To simplify this expression we can first simplify the factors that are multiples of 6:

6\cdot6\cdot6\cdot6\cdot6\cdot6\cdot6\cdot6\cdot11\cdot11

Next, we can use the commutative property of multiplication to group the factors of 6 together:

(6\cdot6\cdot6\cdot6\cdot6\cdot6)\cdot(6\cdot6\cdot6\cdot6)

Simplifying each of these groups of factors separately, we get:

46656\cdot1296

Multiplying these two numbers together, we get the final result:

60534416

Let's break down the given expression and simplify it step by step.

The expression is:

6\cdot6\cdot6\cdot6\cdot66\cdot6\cdot6\cdot6\cdot66

We can start by simplifying the factors that are multiples of 6:

6\cdot6\cdot6\cdot6\cdot6\cdot6\cdot6\cdot6\cdot11\cdot11

Next, we can use the commutative property of multiplication to group the factors of 6 together:

(6\cdot6\cdot6\cdot6\cdot6\cdot6)\cdot(6\cdot6\cdot6\cdot6)

Simplifying each of these groups of factors separately, we get:

6\cdot6\cdot6\cdot6\cdot6\cdot6 = 46656

6\cdot6\cdot6\cdot6 = 1296

Now we can substitute these values back into the expression:

46656\cdot1296

We can multiply these two numbers together to get the final result:

60534416

The given expression is:

6\cdot6\cdot6\cdot6\cdot66\cdot6\cdot6\cdot6\cdot66

To simplify this expression, we can first simplify the factors that are multiples of 6:

6\cdot6\cdot6\cdot6\cdot6\cdot6\cdot6\cdot6\cdot11\cdot11

Next, we can use the commutative property of multiplication to group the factors of 6 together:

(6\cdot6\cdot6\cdot6\cdot6\cdot6)\cdot(6\cdot6\cdot6\cdot6)

Simplifying each of these groups of factors separately, we get:

46656\cdot1296

Multiplying these two numbers together, we get the final result:

60534416

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The third side that we can not see in the image that is shown has a size of 15.

The hypotenuse of a right triangle can be found using the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

If c^2 = a^2 + b^2

c = √a^2 + b^2

c = √12^2 + 9^2

c = 15

Thus the missing side is 15 from the calculation.

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The results suggest that the mean tar content of filtered 100 mm cigarettes is significantly lower than 21.1 mg, which is the mean for unfiltered king size cigarettes. This indicates that the filters are effective in reducing the tar content of cigarettes.

Null hypothesis: The mean tar content of filtered 100 mm cigarettes is greater than or equal to 21.1 mg.

Alternative hypothesis: The mean tar content of filtered 100 mm cigarettes is less than 21.1 mg.

The test statistic to use is the t-statistic, since the population standard deviation is not known.

t = (19.8 - 21.1) / (3.21 / sqrt(25)) = -2.03

Using a t-table with degrees of freedom of 24 and a significance level of 0.05, the critical t-value is -1.711. Since our test statistic is less than the critical t-value, we reject the null hypothesis.

The p-value can also be calculated using the t-distribution with degrees of freedom of 24 and the t-statistic of -2.03. The p-value is 0.029, which is less than the significance level of 0.05. Therefore, we reject the null hypothesis.

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The absolute maximum value of the function f(x) is 222.

To find the absolute minimum value of the function f(x), we need to first find the critical points within the given interval -6 ≤ x ≤ 5. To do this, we take the derivative of f(x) and set it equal to zero:

f'(x) = 6x² + 12x - 144
0 = 6(x² + 2x - 24)
0 = 6(x+6)(x-4)

The critical points are x=-6, x=-4, and x=4. To determine which of these points correspond to a minimum value, we evaluate f(x) at each of these points and at the endpoints of the interval:

f(-6) = -880, f(-4) = -184, f(4) = -136, f(-6) = -880, f(5) = 222

Therefore, the absolute minimum value of the function f(x) is -880.

To find the absolute maximum value of the function f(x), we follow the same process. The critical points are still x=-6, x=-4, and x=4, but now we need to evaluate f(x) at each of these points and at the endpoints of the interval to determine which corresponds to a maximum value:

f(-6) = -880, f(-4) = -184, f(4) = -136, f(-6) = -880, f(5) = 222

Therefore, the absolute maximum value of the function f(x) is 222.

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The possible values of x that satisfy the equation |x+5| = c are x = c - 5 and x = -c - 5.

what is algebra?

Algebra is a branch of mathematics that deals with mathematical operations and symbols used to represent numbers and quantities in equations and formulas.

Assuming you meant to write |x+5|= some value, I can give you a general method to solve equations involving absolute values.

If |a| = b, then either a = b or a = -b. Thus, to solve the equation |x+5| = c, where c is some given value, we can split it into two cases:

Case 1: x+5 = c

Solving for x, we get x = c - 5.

Case 2: -(x+5) = c

Solving for x, we get x = -c - 5.

So, the possible values of x that satisfy the equation |x+5| = c are x = c - 5 and x = -c - 5.

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Andrew and his mother will have wrapped the same number of candy pieces in 21.6 minutes.

We need to find out how many minutes (s) Andrew and his mother wrapped the same number of candy pieces.

Given data:

The number of pieces that Andrew’s mother can wrap is y = 73.

Andrew wraps at a rate of y = 3x + 8.

To find the number of minutes (s) at which Andrew and his mother have wrapped the same number of candy pieces, we need to equate both equations and then find the value of x the equation is given as,

73 = 3x + 8

65 = 3x

x = 21.6

Therefore, Andrew and his mother will have wrapped the same number of candy pieces after 21.6 minutes.

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The total number of gifts is given as follows:

439 gifts.

The Fundamental Counting Theorem states that if there are m ways to do one thing and n ways to do another, then there are m x n ways to do both.

This can be extended to more than two events, where the number of ways to do all the events is the product of the number of ways to do each individual event, according to the equation presented as follows:

[tex]N = n_1 \times n_2 \times \cdots \times n_n[/tex]

For a single gift, the number of options is given as follows:

10 + 4 + 7 = 21 gifts.

For two gifts, the number of options is given as follows:

10 x 4 + 10 x 7 + 7 x 4 = 138 gifts.

For three gifts, the number of options is given as follows:

10 x 4 x 7 = 280 gifts.

Hence the total number of gifts is obtained as follows:

280 + 138 + 21 = 439 gifts.

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The total area of the wrapping paper is 72.5 in².

In the given figure (attached below), we have two shapes one is a triangle and the other one is a rectangle. To find the total area of the wrapping paper we have to add the area of the rectangle part and the area of the trianglular part.

Total area = Area of the rectangular part + area of the triangular part.

Area of the rectangular part = length x breadth

from the below figure, length = 15 in

                                     breadth = 4 in

So, area of the rectangular part = 15 in x 4 in = 60 in²

Similarly, area of the triangular part = 1/2 x base x height

from the below figure, base of the triangle = 15 in -10 in = 5 in

                                      height of the triangle = 9 in - 4 in = 5 in

So, area of the triangular part = 1/2 x 5 in x 5in =  12.5 in²

Now, the total area of the wrapping paper = area of the rectangular part + area of the triangular part = 60 in² + 12.5 in² = 72.5 in².

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The area of the sector, to the nearest hundredth, is 45.87 cm^2.

The formula for the length of an arc of a circle is L = rθ, where L is the arc length, r is the radius, and θ is the angle in radians subtended by the arc.

We  solve for θ by dividing both sides by r: θ = L/r.

In this case, r = 5 cm and L = 18.5 cm, so θ = 18.5/5 = 3.7 radians.

The formula for the area of a sector of a circle is A = (1/2)r^2θ.

Plugging in the values, we get A = (1/2)(5^2)(3.7) ≈ 45.87 cm^2.

Therefore, the area of the sector, to the nearest hundredth, is 45.87 cm^2.

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Answer:

60°

Step-by-step explanation:

A hexagon has 6 angles

The sum of the measures of the exterior angles of a hexagon is equal to 360°

So, measure of each exterior angle = 360∘ / 6 = 60∘

The probability that a student chosen randomly from the class passed the test or completed the homework is 20/27.

The probability that a student chosen randomly from the class passed the test or completed the homework is calculated as follows:

Let the probability that a student completed the homework be P(B).

Also, let the probability that a student passed the test be P(A)

P(A or B) = P(A) + P(B) - P(A * B)

From the data table:

The number of students who passed the test = 18

The number of students who completed the homework = 17

The number of students who both passed the test and completed the homework = 15.

Total number of students = 27

P(A) = 18/27

P(B) = 17/27

P(A*B) = 15/27

Therefore,

P(A or B) = 18/27 + 17/27 - 15/27

P(A or B) = 20/27

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let's recall that two tangent lines to the same circle meeting outside it, will have the same length, so all those pair of tangent lines are equal in length.

Check the picture below.

The function 1/[tex]r^2[/tex] can be expressed in terms of the spherical Bessel functions of the first kind, which are a family of solutions to the spherical Bessel differential equation.

The expansion involves a combination of the delta function and the first two spherical Bessel functions, j_0(r) and j_1(r). Specifically, the expansion can be written as (1/2)*[pi * delta(r) + (1/r)*d/d(r)(r * j_0(r)) + (1/[tex]r^2[/tex])*d/d(r)[[tex]r^2[/tex] * j_1(r)]]. This expansion is valid for all values of r except for r=0, where the first term dominates. The spherical Bessel functions are commonly used in physics, particularly in the context of scattering problems and wave propagation.

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The speed of the train is 60 miles per hour.

We calculate in two steps:

To calculate the speed of the train, we need to use the formula:

Speed = Distance / Time

Here, the distance travelled by the train is 480 miles, and the time taken is 8 hours (from 10:30 A.M. to 6:30 P.M.). So, we can calculate the speed of the train as:

Speed = 480 miles / 8 hours

Speed = 60 miles per hour

Therefore, the speed of the train is 60 miles per hour.

Sabine rode on a passenger train for 480 miles between 10:30 A.M. and 6:30 P.M.

To calculate the speed of the train, we used the formula Speed = Distance / Time, where Distance is 480 miles and Time is 8 hours (since the journey was between 10:30 A.M. and 6:30 P.M.).

Substituting the values, we get the speed of the train as 60 miles per hour.

This means that the train travelled at a speed of 60 miles per hour throughout the journey, covering a distance of 480 miles in 8 hours.

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Answer:

Angle C also measures 64°.

The value of k = 2

Substituting into the integral:

k ∫₀^(π/2k) sec²(u) du

= k [tan(u)]₀^(π/2k)

= k [tan(π/2k) - tan(0)]

= k [∞ - 0]

= ∞

This means that the integral diverges unless k = 0.

∫₀^(π/2k) sec²(x/k) dx

= ∫₀^(π/2k) (1 + tan²(x/k)) dx

= [x + k tan(x/k)]₀^(π/2k)

= π/2

So we have:

π/2 = [π/2k + k tan(π/2k)] - [0 + k tan(0)]

= π/2k + k tan(π/2k)

Multiplying through by k:

π/2 = π/2 + k² tan(π/2k)

Subtracting π/2 from both sides:

0 = k² tan(π/2k)

The only way for this equation to hold for k > 0 is if tan(π/2k) = 0. This occurs when π/2k is an integer multiple of π/2, i.e., when k is an even integer.

Therefore, the value of k that satisfies the original integral is k = 2.

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The solutions of the equation are y = (-3/5)x - 59/5 and x = (-5/3)y - 59/3

To solve for one variable in terms of the other, we can rearrange the equation to isolate one of the variables. For example, solving for y in terms of x:

3x + 5y = -59

5y = -3x - 59

y = (-3/5)x - 59/5

So the solution of the equation is:

y = (-3/5)x - 59/5

Alternatively, we could solve for x in terms of y:

3x + 5y = -59

3x = -5y - 59

x = (-5/3)y - 59/3

So another possible solution of the equation is:

x = (-5/3)y - 59/3

Note that both solutions represent the same line in the xy-plane, since they are equivalent equations.

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The simplified rational expression for this problem is given as follows:

[tex]\frac{3\sqrt{21} + 13}{12}[/tex]

The rational expression in the context of this problem is defined as follows:

[tex]\frac{\sqrt{7} + \sqrt{3}}{2\sqrt{3} - \sqrt{7}}[/tex]

The first step in simplifying the expression is removing the root from the denominator, multiplying numerator and denominator by the conjugate, as follows:

[tex]\frac{\sqrt{7} + \sqrt{3}}{2\sqrt{3} - \sqrt{7}} \times \frac{2\sqrt{3} + \sqrt{7}}{2\sqrt{3} + \sqrt{7}}[/tex]

Applying the subtraction of perfect squares, the denominator is given as follows:

2² x 3 - 7 = 12.

The numerator is:

[tex](\sqrt{7} + \sqrt{3})(2\sqrt{3} + \sqrt{7}) = 2\sqrt{21} + 7 + 6 + \sqrt{21} = 3\sqrt{21} + 13[/tex]

Thus the simplified expression is:

[tex]\frac{3\sqrt{21} + 13}{12}[/tex]

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Answer:

In triangle ABC, m∠BAC = 50°. If m∠ACB = 30°, then the triangle is triangle. If m∠ABC = 40°, then the triangle is triangle. If triangle ABC is isosceles, and AB = 6 and BC = 4, then AC =

Answer:

52 degrees

Step-by-step explanation: because i looked and they looked the same so i put 52 and it was right

To find the derivative of y = cot(sen x/X + 14), we need to use the chain rule and the derivative of cot(x) which is -csc^2(x).

First, we let u = sen x/X + 14.
Then, we can rewrite y as y = cot(u).
Using the chain rule, the derivative of y with respect to x is:

dy/dx = dy/du * du/dx

To find dy/du, we need to use the derivative of cot(u) which is -csc^2(u).
So,

dy/du = -csc^2(u)

To find du/dx, we need to use the quotient rule.
Let v = X, so u = sen x/v + 14.
Then,

du/dx = (v*cos x - sen x * 0)/(v^2)
du/dx = cos x/v

Now we can substitute the values of dy/du and du/dx:

dy/dx = dy/du * du/dx
dy/dx = (-csc^2(u)) * (cos x/v)

But u = sen x/X + 14, so we substitute this in:

dy/dx = (-csc^2(sen x/X + 14)) * (cos x/X)

Therefore, the derivative of y = cot(sen x/X + 14) is

dy/dx = (-csc^2(sen x/X + 14)) * (cos x/X).
To find the derivative of y = cot(sen(x)/(x + 14)), we will use the quotient rule and the chain rule.

Let u = sen(x) and v = x + 14, then y = cot(u/v).

First, find the derivatives of u and v:
du/dx = cos(x) (since the derivative of sen(x) is cos(x))
dv/dx = 1 (since the derivative of x is 1, and the derivative of a constant is 0)

Now, apply the quotient rule for cotangent:
d(cot(u/v))/dx = -1/(sin^2(u/v)) * (du/dv - u*dv/dx) / (v^2)

Substitute the expressions for u, v, du/dx, and dv/dx:
dy/dx = -1/(sin^2(sen(x)/(x + 14))) * ((cos(x)*(x + 14) - sen(x)*1) / (x + 14)^2)

This is the derivative of y with respect to x.

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