solve 3-x/2≥15
A.x ≤ -24
B.x ≤6
C.x ≥ -24
D.x ≥ -6

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.

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~~~

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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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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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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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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The amount of money in Marianna's annuity after 6 years will be approximately $300,516.

To calculate the amount of money in Marianna's annuity after 6 years, we can use the formula for compound interest on an annuity:

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

Where:

A = the final amount in the annuity

P = the regular contribution (each quarter) = $10,000

r = annual interest rate = 8% = 0.08

n = number of compounding periods per year = 4 (since it's compounded quarterly)

t = number of years = 6

Plugging in the values:

A = 10000 * ((1 + 0.08/4)^(4*6) - 1) / (0.08/4)

Calculating this expression:

A ≈ 10000 * ((1.02)^24 - 1) / 0.02

A ≈ 10000 * (1.601032449136241 - 1) / 0.02

A ≈ 10000 * 0.601032449136241 / 0.02

A ≈ 10000 * 30.05162245681205

A ≈ 300,516.22

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Answer:

304200

Step-by-step explanation:

To find the value of P6, use the savings annuity formula

PN=d((1+r/k)N k−1)r/k.

From the question, we know that r=0.08, d=$10,000, k=4 compounding periods per year, and N=6 years. Substitute these values into the formula gives

P6=$10,000 ((1+0.08/4)6⋅4−1)/(0.08/4).

Simplifying further gives P6=$10,000 ((1.02)24−1)/(0.02) and thus P6=$304,218.62.

Rounding as requested, our answer is 304200.

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.

a) To express FB in terms of b, we need to consider the relationship between FB and EF. Since EF is equal to 2a, we can substitute this value into the expression for FB:

FB = EF - FB

= (2a) - (2a)

= 0

Therefore, FB is equal to 0 in terms of b.

b) To express FD in terms of a and b, we can use the given relationship between ED and FD. ED is equal to 3b, so we can substitute this value into the expression for FD:

FD = ED - FB

= (3b) - (0)

= 3b

Therefore, FD is equal to 3b in terms of a and b.

c) To express CB in terms of a and b, we need to consider the relationship between CB and EF. Since EF is equal to 2a, we can substitute this value into the expression for CB:

CB = EF - EB

= (2a) - (FB + FD)

= (2a) - (0 + 3b)

= 2a - 3b

Therefore, CB is equal to 2a - 3b in terms of a and b.

Note that the roots of the equation Unequal and rational (Option D)

The roots of the equation   (x - 3)² = 4 can be found by taking the square root of both sidesof the equation.

x - 3 = ±√4

⇒ x - 3 = ±2

Solve for x

For the positive square root.

x - 3 = 2

x = 2 + 3

x = 5

For the negative square root.

x - 3 = -2

x = -2 + 3

x = 1

Since the equation has two roots, x = 5 and x = 1. These roots are unequal and rational. (Option D)

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Full Question:

Although part of your question is missing, you might be referring to this full question:

The roots of the equation (x - 3)² = 4 are

A.Unequal and irrational.

B.Equal and rational.

C.  Equal and irrational.

D.  Unequal and rational.

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]

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:

the train will travel 245 feet in 3 and 1/2 seconds

Step-by-step explanation:

To determine the distance the train will travel in 3 and 1/2 seconds, we can use a proportion based on the given information.

Let's set up the proportion:

70 feet / (1/10 second) = x feet / (3 1/2 seconds)

To solve this proportion, we can first convert the mixed number 3 1/2 to an improper fraction.

3 1/2 = 7/2

Now we can rewrite the proportion:

70 / (1/10) = x / (7/2)

To simplify the proportion, we can multiply the numerator and denominator of the right side by 10/1:

70 / (1/10) = (x * 10) / (7/2)

Simplifying further, we get:

70 * (10/1) = x * (10/7/2)

700 = x * (20/7)

To find x, we can divide both sides of the equation by (20/7):

x = 700 / (20/7)

x = 700 * (7/20)

x = 245 feet

Therefore, at the same speed, the train will travel 245 feet in 3 and 1/2 seconds.

The limit of the given function as x approaches 1 is 0.

To find the limit of the given function as x approaches 1, we need to evaluate the expression by substituting x = 1. Let's break it down step by step:

1. Begin by substituting x = 1 into the numerator:

[tex]\[12\sin\left(\frac{\pi}{2}\cdot 1\right)\ln(1) = 12\sin\left(\frac{\pi}{2}\right)\ln(1) = 12(1)\cdot 0 = 0\][/tex]

2. Now, substitute x = 1 into the denominator:

(1³ + 5)(1 - 1) = 6(0) = 0

3. Finally, divide the numerator by the denominator:

0/0

The result is an indeterminate form of 0/0, which means further analysis is required to determine the limit. To evaluate this limit, we can apply L'Hôpital's rule, which states that if we have an indeterminate form 0/0, we can take the derivative of the numerator and denominator and then evaluate the limit again. Applying L'Hôpital's rule:

4. Take the derivative of the numerator:

[tex]\[\frac{d}{dx}\left(12\sin\left(\frac{\pi}{2}x\right)\ln(x)\right) = 12\left(\cos\left(\frac{\pi}{2}x\right) \cdot \left(\frac{\pi}{2}\right) \cdot \frac{-1}{x} + \frac{\sin\left(\frac{\pi}{2}x\right)\ln(x)}{x}\right)\][/tex]

5. Take the derivative of the denominator:

[tex]\[\frac{d}{dx}\left((x^3 + 5)(x - 1)\right) = \frac{d}{dx}\left(x^4 - x^3 + 5x - 5\right) = 4x^3 - 3x^2 + 5\][/tex]

6. Substitute x = 1 into the derivatives:

Numerator: [tex]\[12\left(\cos\left(\frac{\pi}{2}\right) \cdot \left(\frac{\pi}{2}\right) \cdot \frac{-1}{1} + \sin\left(\frac{\pi}{2}\right) \cdot \frac{\ln(1)}{1}\right) = 0\][/tex]

Denominator: 4(1)³ - 3(1)² + 5 = 4 - 3 + 5 = 6

7. Now, reevaluate the limit using the derivatives:

lim as x approaches 1 of [tex]\[\frac{{12\left(\cos\left(\frac{\pi}{2}x\right) \cdot \left(\frac{\pi}{2}\right) \cdot \frac{{-1}}{{x}} + \sin\left(\frac{\pi}{2}x\right) \cdot \frac{{\ln(x)}}{{x}}\right)}}{{4x^3 - 3x^2 + 5}}\][/tex]

= 0 / 6

= 0

Therefore, the limit of the given function as x approaches 1 is 0.

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Car A is 2 times Faster than Car B during the first hour.

The graph of Car A is a straight line, indicating that it is traveling at a constant speed.

The graph shows that Car A is traveling 100 miles in 2 hour .The table of Car B shows that it travels 50 miles in 1 hour, 100 miles in 2 hours, and 150 miles in 3 hours. Thus, the rate of Car B is increasing, as it travels at a faster speed during each hour compared to the previous hour.To find the rate of each car, we need to divide the distance by the time. For Car A, rate = distance ÷ time = 100 miles ÷ 2 hours = 50 miles per hour.

For Car B, we can find the average rate for each hour by dividing the distance traveled during that hour by the time. Thus, the rates are: First hour: 50 miles per hour Second hour: 50 miles ÷ 1 hour = 50 miles per hour Third hour: 50 miles ÷ 1 hour = 50 miles per hour By comparing the rates, we see that both cars are traveling at the same speed during the second and third hours. However, during the first hour, Car A is traveling faster than Car B.

Thus, Car A has the greatest speed.To determine how many times faster Car A is compared to Car B during the first hour, we can divide their rates. The rate of Car A is 50 miles per hour, while the rate of Car B is 50 miles per hour. Therefore, Car A is traveling at the same speed as Car B during the second and third hours. During the first hour, Car A is traveling twice as fast as Car B. Thus, Car A is 2 times faster than Car B during the first hour.

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Answer:The inequality represented by the equation y = -11/7x - 4 can be written as:

y ≤ -11/7x - 4

This represents a less than or equal to inequality, indicating that the values of y are less than or equal to the expression -11/7x - 4.

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 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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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:

(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]

Answer:

Step-by-step explanation: 5√2

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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We need to find the z-score for Hank using the above formula.z = (x - μ) / σ= (23 - 20) / 5= 0.6So, the z-score for Hank is 0.6.

The z-score measures the number of standard deviations a particular value is away from the mean. A positive z-score indicates that Hank's score is above the mean, while a negative z-score would indicate a score below the mean. In this case, a z-score of 0.6 suggests that Hank's score is 0.6 standard deviations above the average.

The z-score is a measure of the number of standard deviations that a value is above or below the mean of a distribution. It is calculated using the formula z = (x - μ) / σ, where x is the value being evaluated, μ is the mean of the distribution, and σ is the standard deviation.

The given problem states that the test scores of a local DMV had an average of 20 and a standard deviation of 5.

So, the mean μ = 20 and the standard deviation σ = 5.Hank scored a 23. This means that his score is 0.6 standard deviations above the mean of the distribution.

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Denis's and Dasha's methods use different operations to simplify the expressions in different orders.

Denis's method uses addition, subtraction, and multiplication operations, but Dasha's method does not.

Multiplication operations are described as using mathematical operation that indicates how many times a number is added to itself.

Denis's Assignment

2* 6+2w-54

12+2w = 54

12-12+2w= 54-12

2w=42

w= 21  

Dasha's Assignment

54-2* 6

54-12= 42

The two widths add to 42 cm

42/ 2-21

We can see that Dasha's method also uses addition and subtraction operations, but she simplifies the expression by performing the operations in a different order.

She subtracts the product of 2 and 6 from 54.

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A digit to make the number divisible by 3 is by adding the digit 2 to the number 1234.

To make the number 1234 divisible by 3, we can find the sum of its digits and determine if it is divisible by 3. If the sum of the digits is divisible by 3, then the original number is also divisible by 3.

Let's calculate the sum of the digits in 1234:

1 + 2 + 3 + 4 = 10

The sum of the digits is 10. Since 10 is not divisible by 3, we need to add or subtract a digit to make the sum divisible by 3.

To find the digit we need to add or subtract, we can use the fact that the difference between the original sum and the next multiple of 3 is the required digit.

The next multiple of 3 greater than 10 is 12 (12 - 10 = 2). Therefore, we need to add 2 to the number 1234 to make it divisible by 3.

1234 + 2 = 1236

we obtain the number 1236, which is divisible by 3.

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The  two pairs of congruent angles are determined as angle XWY and angle YZX.

Congruent angles are the angles that have equal measure. So all the angles that have equal measure will be called congruent angles.

So congruent angles are two or more angles that are identical or equal to each other.

From the given diagram , the pair of angles are congruent to each other.

Angle XWY is congruent to angle YZX, this is because vertical opposite angles in a cyclic quadrilateral are equal in measure.

Thus, the  two pairs of congruent angles are determined as angle XWY and angle YZX.

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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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We have a total of seven 4-tuples that satisfy the given relation.The given relation is {(a,b,c,d) | a,b,c,d∈!+ , a+b+c+d = 6}. It can be understood as the set of 4-tuples (a, b, c, d) such that a, b, c, and d are positive integers and their sum is equal to 6.

Let's now list all the possible 4-tuples that satisfy the given relation. The possible combinations are as follows: (1, 1, 1, 3), (1, 1, 2, 2), (1, 2, 1, 2), (2, 1, 1, 2), (1, 2, 2, 1), (2, 1, 2, 1), and (2, 2, 1, 1).

Here's a brief explanation on how these 4-tuples were obtained. Let a, b, c, and d be positive integers such that a+b+c+d = 6. The least possible value that each variable can take is 1.

So, we start with a=1 and find all possible values of (b, c, d) that satisfy the given equation. Then, we move to a=2 and repeat the process. Finally, we list all the possible 4-tuples that we obtained.

Thus, we have a total of seven 4-tuples that satisfy the given relation.

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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.

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]