What is 1/6 as a percent.

Answer: D

Step-by-step explanation:

Answer:

 (b)  f(x) = -1/x⁹ and g(x) = -8x +4

  (d)  f(x) = x⁹ and g(x) = -1/(-8x +4) . . . . . alternate solution

Step-by-step explanation:

You want to decompose h(x) = -1/(-8x +4)⁹ into f(x) and g(x) such that h(x) = f(g(x)).

The composition h(x) = f(g(x)) means that the function g(x) will replace x in the definition of f(x).

It is often convenient to look at the order of operations when asked to decompose a function like this. Here, the parenthetical expression (-8x+4) is raised to the 9th power and its opposite reciprocal is found. This suggests that f(x) can be a function that finds the opposite reciprocal of a 9th power,  matching choice B. Thus, a reasonable choice is ...

  (b)  f(x) = -1/x⁹ and g(x) = -8x +4

We note that the reciprocal of a 9th power is also the 9th power of a reciprocal. A negative sign is preserved by the odd power. This means that another reasonable choice for the decomposition is ...

  (d)  f(x) = x⁹ and g(x) = -1/(-8x +4)

__

Additional comment

We list choice B first because that one is probably the one you're supposed to claim as the answer. However, this question has two correct decompositions among those listed. You may want to discuss this with your teacher.

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The length of side X in simple radical form with a rational denominator is 10/√3.

In Mathematics and Geometry, a 30-60-90 triangle is also referred to as a special right-angled triangle and it can be defined as a type of right-angled triangle whose angles are in the ratio 1:2:3 and the side lengths are in the ratio 1:√3:2.

This ultimately implies that, the length of the hypotenuse of a 30-60-90 triangle is double (twice) the length of the shorter leg (adjacent side), and the length of the longer leg (opposite side) of a 30-60-90 triangle is √3 times the length of the shorter leg (adjacent side):

Adjacent side = 5/√3

Hypotenuse, x = 2 × 5/√3

Hypotenuse, x = 10/√3.

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Missing information:

The question is incomplete and the complete question is shown in the attached picture.

Answer:

Please attach a photo to this so I can better answer your question

Answer:

To find the mean, median, mode, and range of the resulting data set after dividing each value by 5, we need to perform the calculations. Here are the steps:

Original data set: 9214709

Step 1: Divide each value by 5:

Resulting data set: 1842941.8

Step 2: Calculate the mean:

To find the mean, we sum up all the values in the resulting data set and divide by the total number of values:

Mean = (1842941.8) / 1 = 1842941.8

Therefore, the mean of the resulting data set is 1842941.8.

Please note that for the median, mode, and range calculations, we need more than one value in the data set. As the original data set only contains one value, we cannot proceed with those calculations.

Step-by-step explanation:

Answer:

The mean, also known as the average, is a measure of central tendency that is calculated by adding up all the values in a data set and then dividing by the number of values in the set. For example, if you have a data set with the values 1, 2, and 3, the mean would be calculated as (1 + 2 + 3) / 3 = 2.

The median is another measure of central tendency that represents the middle value in a data set when the values are arranged in ascending order. If the data set has an odd number of values, the median is the middle value. If the data set has an even number of values, the median is calculated as the average of the two middle values.

For example, if you have a data set with the values 1, 2, and 3, the median would be 2 because it is the middle value when the values are arranged in ascending order. If you have a data set with the values 1, 2, 3, and 4, the median would be calculated as (2 + 3) / 2 = 2.5 because there are an even number of values and the two middle values are 2 and 3.

Answer:

(x - 8)² + (y - 5)² = 400

Step-by-step explanation:

the equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k ) are the coordinates of the centre and r the radius

the radius is the distance from the centre to a point on the circle

use the distance formula to calculate r

r = [tex]\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }[/tex]

with (x₁, y₁ ) = (8, 5 ) and (x₂, y₂ ) = (- 4, 21 )

r = [tex]\sqrt{(-4-8)^2+(21-5)^2}[/tex]

 = [tex]\sqrt{(-12)^2+16^2}[/tex]

 = [tex]\sqrt{144+256}[/tex]

 = [tex]\sqrt{400}[/tex]

 = 20

then with (h, k ) = (8, 5 ) and r = 20, the equation of the circle is

(x - 8)² + (y - 5)² = 20² , that is

(x - 8)² + (y - 5)² = 400

Answer:

[tex](x-8)^2+(y-5)^2=400[/tex]

Step-by-step explanation:

The standard equation of a circle is:

[tex]\boxed{(x-h)^2+(y-k)^2=r^2}[/tex]

where:

The given center of the circle is (8, 5).

To find the value of r², substitute the circle and the given point (-4, 21) into the equation and solve for r².

[tex]\begin{aligned}(-4-8)^2+(21-5)^2&=r^2\\(-12)^2+(16)^2&=r^2\\144+256&=r^2\\400&=r^2\end{aligned}[/tex]

Finally, substitute the center and r² into the formula to create an equation of the circle with the given parameters:

[tex]\boxed{(x-8)^2+(y-5)^2=400}[/tex]

The equation of line A is y = 4.

The equation of line B is y = 0.

The equation of lines A and B is calculated by applying the general equation of line as follows;

Mathematically, the formula for the general equation of lines is given as;

y = mx + b

where;

For line A, the equation is determined as;

y = 0x +  4

the slope of the this line is zero

y = 4

For line B, the equation is determined as

y = 0x + 0

the slope and y intercept of the this line is zero

y = 0

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r = r2

​1 − r=(−2 i^ −2 j^ +0 k^ )−(4 i^ −4 j^ +0 k^ )

⇒ r =−6 i^ +2 j^ +0 k^

∴∣ r ∣= (−6)2 +(2) 2 +0 2

= 36+4

​ = 40

​ =210

​

A quantity or phenomenon with independent qualities for both magnitude and direction is called a vector. The term can also refer to a quantity's mathematical or geometrical representation. Velocity, momentum, force, electromagnetic fields, and weight are a few examples of vectors in nature. The above option D is appropriate.

Computer graphics known as vector graphics allow for the direct creation of visual pictures using geometric structures such as points, lines, curves, and polygons that are defined on a Cartesian plane.

Any pathogen that conveys and spreads an infectious agent into other living things is referred to be a vector. These vectors could be bacteria or parasites.

The above option D is correct.

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Answer:0.00007593302

Step-by-step explanation:

The total amount of interest Ron paid over the course of the loan is approximately $12,220.63.

To calculate the total amount of interest paid on Ron's student loan, we need to use the formula for compound interest. The formula for compound interest is:

A = P(1 + r/n)^(nt) - P

Where:

A is the total amount including principal and interest.

P is the principal amount (loan amount).

r is the annual interest rate (as a decimal).

n is the number of times the interest is compounded per year.

t is the number of years.

In this case, Ron's loan amount is $35,000, the annual interest rate is 3.8% (or 0.038 as a decimal), the loan term is 8 years, and the interest is compounded monthly (n = 12).

Plugging in these values into the formula:

A = $35,000(1 + 0.038/12)^(12*8) - $35,000

Calculating the exponent:

A = $35,000(1 + 0.00316667)^(96) - $35,000

Using a calculator to evaluate the expression within parentheses:

A = $35,000(1.00316667)^(96) - $35,000

Calculating the exponent:

A = $35,000(1.34898777) - $35,000

Subtracting the principal amount:

A = $47,220.63 - $35,000

A = $12,220.63

Therefore, the total amount of interest Ron paid over the course of the loan is approximately $12,220.63.

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The volume of the cone is 13 cm³. option D

To determine the volume of the cone, we have that;

The formula for calculating the volume of a cone is expressed as;

Volume = (1/3)πr ²√(L ² - r ²).

Such that;

Substitute the values, we have;

Volume = 1/3 × 3.14 ² × √(25 - 9)

Find the squares, we get;

Volume, V = 1/3 × 9. 86 × √16

Find the square root

Volume, V = 1/3 × 9.86 × 4

Volume, V = 13 cm³

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Answer:

S = π(12)(20) + π(12^2) - (π(3)(5) + π(3^2))

= π(240 + 144 - (15 + 9))

= 360π = about 1,130.97 cm^2

To find the area of a parallelogram, you need to multiply the base by the height. In this case, the given dimensions are 60ft (base) and 67ft (height), and you need to subtract 52ft from the height.

New height = 67ft - 52ft = 15ft

Area of the parallelogram = Base * Height = 60ft * 15ft = 900 square feet.

Therefore, the area of the parallelogram is 900 square feet.

~~~Harsha~~~

Answer:

The answer is down below

Step-by-step explanation:

v-u=◇velocity

[tex] = 9 - 5[/tex]

◇velocity =4ms‐¹

acceleration =◇velocity/time

[tex]a = \frac{4}{5} [/tex]

[tex]a = 0.8m {s}^{ - 2} [/tex]

Change in velocity:

[tex] \sf \:final \: velocity - initial \: velocity = change \: in \: velocity[/tex]

[tex] \therefore \tt \delta \: v = 9 - 5 \\ \tt = 4m {s}^{ - 1} [/tex]

To find acceleration:

[tex] \rm \: a = \frac{ \triangle \: v}{\triangle \: t} [/tex]

[tex] \rm \: a = \frac{ 4}{5 - 0} = \frac{4}{5} = 0.8m {s}^{ - 1} [/tex]

Answer:

$100

Step-by-step explanation:

Since 10 vases cost $1000, you can find the cost vase by dividing $1000 by 10.

$1000 ÷ 10 = $100

Each vase costs $100.

Chandler ended up 8 feet below sea level. The negative sign indicates that his final position is below sea level. This means that he has descended further into the lake compared to the starting point on the cliff.

Chandler started on the cliff at 32 feet above sea level. When he jumped for 40 feet, we need to determine the final position in relation to sea level.

Since Chandler jumped down, the distance below sea level will be calculated as a negative value. To find how far below sea level Chandler ended up, we subtract the jump distance (40 feet) from the starting height (32 feet above sea level):

32 feet - 40 feet = -8 feet

It's important to note that negative values are used here to represent the direction and magnitude of Chandler's descent relative to sea level

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Answer:

-8 feet

Step-by-step explanation:

Using Pythagorean theorem, the length of the ladder is 10ft

In mathematical terms, if y and z are the lengths of the two shorter sides (also known as the legs) of a right triangle, and x is the length of the hypotenuse, the Pythagorean theorem can be expressed as:

x² = y² + z²

In the questions given, the only one we can use Pythagorean theorem to solve is the one with ladder since it's forms a right-angle triangle.

To calculate the length of the ladder, we can write the formula as;

x² = 8² + 6²

x² = 64 + 36

x² = 100

x = √100

x = 10

The length of the ladder is 10 feet

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The preparation of the branch's income statement for Santosh & Co. Chennai is as follows:

Branch of Santosh & Co. Chennai

For the year ended December 31, 2018

Sales revenue            $40,300

Cost of goods sold         4,200

Gross profit                 $36,100

Expenses:

Office expenses $36,000

Salaries                    3,600

Total expenses         $39,600

Loss                             $3,500

An income statement is a financial statement prepared at the end of an accounting period to determine the profit or loss generated by a business or branch.

The profit or loss is the difference between the total revenue and the total expenses for the accounting period.

Credit sales at branch 40,000

Office expenses by Head office 36,000

Cash remittance to branch for petty cash 9,000

Goods sent to Branch 7,200

Salaries paid by head office 3,600

Debtors 31.12.2018 30,000

Petty cash on 31.12.2018 16,200

Cash sales at branch 300

Stock of 31.12.2018 3,000

Sales revenue   $40,300 ($40,000 + $300)

Cost of goods sold $4,200 ($7,200 - $3,000)

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The 36th derivative of f(x) = cos(2x) is [tex]2^{17}[/tex]* cos(2x).

To find the 36th derivative of the function f(x) = cos(2x), we can apply the chain rule repeatedly. The chain rule states that if we have a composite function y = f(g(x)), then its derivative is given by dy/dx = f'(g(x)) * g'(x).

Let's start by finding the first few derivatives of f(x) = cos(2x):

f'(x) = -2sin(2x)

f''(x) = -4cos(2x)

f'''(x) = 8sin(2x)

f''''(x) = 16cos(2x)

We observe a pattern where the derivatives of cos(2x) alternate between sin(2x) and cos(2x), with the signs changing accordingly.

Based on this pattern, we can see that the 36th derivative will be:

f^(36)(x) =[tex](-1)^{17} * 2^{17} *[/tex] cos(2x)

Simplifying this expression, we have:

f^(36)(x) = [tex]2^{17} * cos(2x)[/tex]

Therefore, the 36th derivative of f(x) = cos(2x) is[tex]2^{17[/tex] * cos(2x).

It's important to note that in this case, the number 36 is even, and since the derivatives of cos(2x) follow a repeating pattern every 4 derivatives, the sign (-1) raised to the power of 17 accounts for the change in sign in the 36th derivative.

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Answer:
The factors correlate with the x-intercepts. For example, with the factor (x - 3), there is an x-intercept at (3, 0). Likewise, with the factor (x + 2), there is an x-intercept at (-2, 0). This suggests that the opposite sign of the factors are the x-intercepts in the graph.

Your overall grade, if the percentage scored in the categories are given, is 86. 35.

To calculate your overall grade, you'll need to multiply each individual grade by its respective weight and then add all these products together.

Tests:

= 85 % * 15%

= 12. 75

Quizzes:

= 90% * 10%

= 9. 0

Projects:

= 92% * 35%

= 32 .2

Participation:

= 80% * 35%

= 28. 0

Final Exam:

= 88% * 5%

= 4. 4

The overall grade is:

= 12.75 + 9.0 + 32.2 + 28.0 + 4.4

= 86. 35

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The scores are :
Tests: 85%

Quizzes: 90%

Projects: 92%

Participation: 80%

Final Exam: 88%

The correct answer is B. 2, as the scale factor used to create the dilated polygon is 2.

To determine the scale factor used to create the image, we can compare the corresponding side lengths of the original polygon ABCD and the dilated polygon A'B'C'D'.

Let's calculate the lengths of the corresponding sides:

Side AB: The length of side AB is 3 - 1 = 2 units in the original polygon. In the dilated polygon, the length of side A'B' is 6 - 2 = 4 units.

Side BC: The length of side BC is -2 - (-1) = -1 units in the original polygon. In the dilated polygon, the length of side B'C' is -4 - (-2) = -2 units.

Side CD: The length of side CD is 1 - 3 = -2 units in the original polygon. In the dilated polygon, the length of side C'D' is 2 - 6 = -4 units.

Side DA: The length of side DA is -1 - (-2) = 1 unit in the original polygon. In the dilated polygon, the length of side D'A' is -2 - (-4) = 2 units.

Now, let's compare the corresponding side lengths:

AB: A'B' = 4 units / 2 units = 2

BC: B'C' = -2 units / -1 units = 2

CD: C'D' = -4 units / -2 units = 2

DA: D'A' = 2 units / 1 unit = 2

The scale factor used to create the image is the ratio of the corresponding side lengths.

In this case, all the corresponding side lengths have a ratio of 2.

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so the idea being, we have a system of equations of two variables and 4 equations, each one rendering a line, for this case these aren't equations per se, they're INEquations, so pretty much the function will be the same for an equation but we'll use > or < instead of =, but fairly the function is basically the same, the behaviour differs a bit.

we have a line passing through (-6,0) and (0,8), side one

we have a line passing through the x-axis and -6, namely (-6,0) and the y-axis and -4, namely (0,-4), side two

we have a line passing through (0,-4) and (6,4), side three

now, side four is simply the line connecting one and three.

the intersection of all four lines looks like the one in the picture below, so what are those lines with their shading producing that quadrilateral?

well, we have two points for all four, and that's all we need to get the equation of a line, once we get the equation, with its shading like that in the picture, we'll make it an inequality.

[tex](\stackrel{x_1}{-6}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{8}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{8}-\stackrel{y1}{0}}}{\underset{\textit{\large run}} {\underset{x_2}{0}-\underset{x_1}{(-6)}}} \implies \cfrac{8 -0}{0 +6} \implies \cfrac{ 8 }{ 6 } \implies \cfrac{4}{3}[/tex]

[tex]\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{ \cfrac{4}{3}}(x-\stackrel{x_1}{(-6)}) \implies y -0 = \cfrac{4}{3} ( x +6) \\\\\\ y=\cfrac{4}{3}x+8\hspace{5em}\stackrel{\textit{side one} }{\boxed{y < \cfrac{4}{3}x+8}}[/tex]

[tex]\rule{34em}{0.25pt}\\\\ (\stackrel{x_1}{-6}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{-4}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{-4}-\stackrel{y1}{0}}}{\underset{\textit{\large run}} {\underset{x_2}{0}-\underset{x_1}{(-6)}}} \implies \cfrac{-4 -0}{0 +6} \implies \cfrac{ -4 }{ 6 } \implies - \cfrac{2}{3}[/tex]

[tex]\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{- \cfrac{2}{3}}(x-\stackrel{x_1}{(-6)}) \implies y -0 = - \cfrac{2}{3} ( x +6) \\\\\\ y=-\cfrac{2}{3}x-4\hspace{5em}\stackrel{\textit{side two} }{\boxed{y > -\cfrac{2}{3}x-4}} \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]

[tex]\stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{4}-\stackrel{y1}{(-4)}}}{\underset{\textit{\large run}} {\underset{x_2}{6}-\underset{x_1}{0}}} \implies \cfrac{4 +4}{6 -0} \implies \cfrac{ 8 }{ 6 } \implies \cfrac{4}{3}[/tex]

[tex]\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-4)}=\stackrel{m}{ \cfrac{4}{3}}(x-\stackrel{x_1}{0}) \implies y +4 = \cfrac{4}{3} ( x -0) \\\\\\ y=\cfrac{4}{3}x-4\hspace{5em}\stackrel{ \textit{side three} }{\boxed{y > \cfrac{4}{3}x-4}} \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]

[tex](\stackrel{x_1}{6}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{8}) ~\hfill~ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{8}-\stackrel{y1}{4}}}{\underset{\textit{\large run}} {\underset{x_2}{0}-\underset{x_1}{6}}} \implies \cfrac{ 4 }{ -6 } \implies - \cfrac{2}{3}[/tex]

[tex]\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{4}=\stackrel{m}{- \cfrac{2}{3}}(x-\stackrel{x_1}{6}) \\\\\\ y=-\cfrac{2}{3}x+8\hspace{5em}\stackrel{ \textit{side four} }{\boxed{y < -\cfrac{2}{3}x+8}}[/tex]

now, we can make that quadrilateral a trapezoid by simply moving one point for "side four", say we change the point (0 , 8) and in essence slide it down over the line to  (-3 , 4).  Notice, all we did was slide it down the line of side one, that means the equation for side one never changed and thus its inequality is the same function.

now, with the new points for side for of (-3,4) and (6,4), let's rewrite its inequality

[tex](\stackrel{x_1}{-3}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{4}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{4}-\stackrel{y1}{4}}}{\underset{\textit{\large run}} {\underset{x_2}{6}-\underset{x_1}{(-3)}}} \implies \cfrac{4 -4}{6 +3} \implies \cfrac{ 0 }{ 9 } \implies 0[/tex]

[tex]\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{4}=\stackrel{m}{ 0}(x-\stackrel{x_1}{(-3)}) \implies y -4 = 0 ( x +3) \\\\\\ y=4\hspace{5em}\stackrel{ \textit{side four changed} }{\boxed{y < 4}}[/tex]

The PQ is 2.265 cm and PX is 3.73 cm.An isosceles triangle of equal sides where PQ = PR and base angles of 65⁰. QX = XR= 2.64cm

Let's solve the problem step by step.

a) PQ: Since the triangle is isosceles and PQ = PR, we can conclude that angle PQR = angle PRQ. We also know that the sum of the angles in a triangle is 180 degrees.

Given that the base angles are 65 degrees each, we can calculate angle PQR as follows:

180 - 65 - 65 = 50 degrees

Now, let's consider triangle PQR. It is an isosceles triangle, with PQ = PR and angle PQR = angle PRQ = 50 degrees.

We are given that QX = XR = 2.64 cm. Using this information, we can apply the Law of Cosines to find PQ.

The Law of Cosines states:

c^2 = a^2 + b^2 - 2ab * cos(C)

In triangle PQR, a = PQ, b = PQ, and C = 50 degrees. Let's plug in the values:

(PQ)^2 = (2.64)^2 + (2.64)^2 - 2 * 2.64 * 2.64 * cos(50)

(PQ)^2 = 6.9696 + 6.9696 - 2 * 2.64 * 2.64 * 0.64278760968

(PQ)^2 = 6.9696 + 6.9696 - 8.81269008562

(PQ)^2 = 5.12650991438

Taking the square root of both sides, we get:

PQ = √5.12650991438

PQ ≈ 2.265 cm

b) PX: To find PX, we can use the Pythagorean theorem in triangle PXR.

In triangle PXR, we have the right angle at X. PX is the hypotenuse, and QX (or XR) is one of the legs.

Using the Pythagorean theorem, we have:

(PX)^2 = (QX)^2 + (XR)^2

(PX)^2 = (2.64)^2 + (2.64)^2

(PX)^2 = 6.9696 + 6.9696

(PX)^2 = 13.9392

Taking the square root of both sides, we get:

PX = √13.9392

PX ≈ 3.73 cm

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Markdowns are usually applied before any additional discounts or promotions. In this case, the markdown reduced the regular price by 33%, resulting in the final price of $26.80.

Jules Girard bought 1 gallon of paint at a 33% markdown from its regular price of $39.99. To calculate the final price after the markdown, we can apply the percentage decrease to the regular price.

The markdown of 33% means that the price is reduced by 33% of the regular price. To calculate the amount of the markdown, we can multiply the regular price by the percentage decrease:

Markdown amount = 33% of $39.99 = 0.33 * $39.99

Next, we can subtract the markdown amount from the regular price to find the final price:

Final price = Regular price - Markdown amount = $39.99 - (0.33 * $39.99)

Calculating the expression within parentheses:

Final price = $39.99 - $13.19

Final price = $26.80

Therefore, Jules Girard bought 1 gallon of paint at a 33% markdown, resulting in a final price of $26.80.

Markdowns are usually applied before any additional discounts or promotions. In this case, the markdown reduced the regular price by 33%, resulting in the final price of $26.80.

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Answer:

Step-by-step explanation: 5√2

X is an inductive set, then {X [tex]\in[/tex] x is transitive} is also an inductive set. Consequently, every n [tex]\in[/tex] N is transitive.

To prove the statement, we need to demonstrate that if X is an inductive set, then the set {[tex]X \in x[/tex]is transitive} is also an inductive set.

Let's break down the proof into two parts:

If X is an inductive set, then {[tex]X \in x[/tex] is transitive} is a subset of X:

To show that {[tex]X \in x[/tex]is transitive} is a subset of X, we need to prove that every element in {[tex]X \in x[/tex] is transitive} is also an element of X.

If X is an inductive set, it means that X contains the empty set (∅) and for every element x in X, the successor of x (denoted as S(x)) is also in X. Now, consider an arbitrary element y in {[tex]X \in x[/tex] is transitive}. By definition, y is a transitive set.

Since X is inductive, it contains the empty set and for every element in X, its successor is also in X. Thus, y must also be in X, and {[tex]X \in x[/tex] is transitive} is a subset of X.

{[tex]X \in x[/tex] is transitive} is an inductive set:

To show that {[tex]X \in x[/tex] is transitive} is an inductive set, we need to demonstrate that it satisfies the properties of an inductive set.

First, we prove that the empty set (∅) is an element of {EX: x is transitive}. Since the empty set is transitive (it vacuously satisfies the definition of transitivity), it belongs to {[tex]X \in x[/tex] is transitive}.

Second, we prove that for every element y in {[tex]X \in x[/tex] is transitive}, its successor S(y) is also in {[tex]X \in x[/tex]  is transitive}. Let y be an arbitrary element in {[tex]X \in x[/tex] is transitive}.

By definition, y is a transitive set. We need to show that S(y) is also a transitive set. Since X is inductive, it means that for every element x in X, its successor S(x) is also in X. Applying this property to y, we conclude that S(y) is in X. Since S(y) is in X, it is also in [tex]X \in x[/tex] is transitive}. Hence, {[tex]X \in x[/tex] is transitive} satisfies the property of an inductive set.

By proving both parts, we have shown that if X is an inductive set, then {[tex]X \in x[/tex] is transitive} is also an inductive set. Consequently, every [tex]n \in N[/tex] is transitive.

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These are a legitimate probability distribution:

To have a legitimate probability distribution, the probabilities for all possible values of X must satisfy two conditions:

Each probability value must be between 0 and 1, inclusive.

The sum of all probability values must equal 1.

Check if the given probabilities satisfy these conditions:

P(X=-3) = 0.14 (valid)

P(X=1) = 0.1 (valid)

P(X=3) = 0.135 (valid)

P(X=5) = 0.306 (valid)

Now, calculate the remaining probability to check if it satisfies the second condition:

P(X=6) = 1 - (P(X=-3) + P(X=1) + P(X=3) + P(X=5))

P(X=6) = 1 - (0.14 + 0.1 + 0.135 + 0.306)

P(X=6) = 1 - 0.681

P(X=6) = 0.319

Since the calculated probability is valid (between 0 and 1), There is a legitimate probability distribution for the discrete random variable X:

P(X=-3) = 0.14

P(X=1) = 0.1

P(X=3) = 0.135

P(X=5) = 0.306

P(X=6) = 0.319

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Answer:

To solve the equation 12^(x + 1) = 79 for x, we need to isolate the exponent.

Step 1: Subtract 1 from both sides of the equation:

12^(x + 1) - 1 = 79 - 1

12^(x + 1) - 1 = 78

Step 2: Rewrite 12^(x + 1) as (12^x)(12^1) using the exponent property:

(12^x)(12) - 1 = 78

Step 3: Simplify the left side of the equation:

12(12^x) - 1 = 78

Step 4: Add 1 to both sides of the equation:

12(12^x) = 78 + 1

12(12^x) = 79

Step 5: Divide both sides of the equation by 12:

(12(12^x))/12 = 79/12

12^x = 79/12

Step 6: Take the logarithm (base 12) of both sides of the equation:

log12(12^x) = log12(79/12)

x = log12(79/12)

Therefore, the solution for x is x = log12(79/12)

Hope this helps!

The Billy's net pay is K646

The given information are; Billy's gross pay is K850.00, he is taxed 12% on his taxable income, and contributes 7% to Nusfund. His loan deduction is 5%.

To find out his net pay, the following steps must be taken:Firstly, calculate Billy's tax amount, Nusfund contribution, and loan deduction by using their respective percentages and the gross pay.

 tax amount= 12/100 × 850= K102 Refund contribution= 7/100 × 850= K59.50Loan deduction= 5/100 × 850= K42.50 Secondly, the sum of all the deductions made from the gross pay is calculated to obtain the total deductions.

Total deductions= K102 + K59.50 + K42.50= K204The final step is to subtract the total deductions from the gross pay to find Billy's net pay. Net pay= K850 − K204= K646.

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