12x-15 = 3 (2x+3) answer if swag

A design that has two conditions with different participants in each condition is independent groups design. Hence option a) independent groups design is correct.

The student question refers to a design where there are two conditions with different participants in each condition.

An independent groups design, also known as a between-subjects design, involves comparing two or more groups of participants who are not the same individuals.

In this type of design, each participant is exposed to only one level of the independent variable.

The groups are formed independently of each other, and any differences between them can be attributed to the manipulation of the independent variable.

An independent groups design: Identify the research question and the independent variable. Form two or more groups of participants, ensuring that they are independent of each other.

Manipulate the independent variable for each group, ensuring that each participant is exposed to only one level of the variable.

Option b is wrong because repeated measures design involves testing the same participants under different conditions or at different times.

Option c is wrong because solomon four-group design is a complex design involving four groups with two groups receiving a pretest, and two groups not receiving a pretest.

Option d is wrong because the pretest-posttest design involves assessing participants before and after a treatment or intervention.

The correct option is a) independent groups design.

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The Kruskal-Wallis test is a non-parametric test used to compare three or more independent groups to determine if there is a significant difference between the medians of the groups. The test statistic used in this test is denoted by the letter "H," which is calculated based on the ranks of the data.

True. The test statistic calculated in the Kruskal-Wallis test is denoted by the symbol H, not to be confused with the Shapiro-Wilk test statistic denoted by W. The Kruskal-Wallis test is a non-parametric test used to compare the median ranks of three or more independent groups. The test statistic H is calculated by comparing the between-group sum of squares to the total sum of squares. H follows a chi-squared distribution with degrees of freedom equal to the number of groups minus one. The p-value obtained from the chi-squared distribution is used to determine whether to reject or fail to reject the null hypothesis that there is no significant difference between the groups.

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a. To find the direction of the vector p1p2⇀, we subtract the coordinates of p1 from the coordinates of p2: (x2 - x1, y2 - y1, z2 - z1).

For the given points:

1. p1(-1, 1, 5), p2(2, 5, 0): The direction of p1p2⇀ is (2 - (-1), 5 - 1, 0 - 5) = (3, 4, -5).

2. p1(1, 4, 5), p2(4, -2, 7): The direction of p1p2⇀ is (4 - 1, -2 - 4, 7 - 5) = (3, -6, 2).

3. p1(3, 4, 5), p2(2, 3, 4): The direction of p1p2⇀ is (2 - 3, 3 - 4, 4 - 5) = (-1, -1, -1).

4. p1(0, 0, 0), p2(2, -2, -2): The direction of p1p2⇀ is (2 - 0, -2 - 0, -2 - 0) = (2, -2, -2).

b. To find the midpoint of the line segment p1p2⇀, we take the average of the coordinates of p1 and p2: ((x1 + x2)/2, (y1 + y2)/2, (z1 + z2)/2).

For the given points:

1. p1(-1, 1, 5), p2(2, 5, 0): The midpoint of p1p2⇀ is ((-1 + 2)/2, (1 + 5)/2, (5 + 0)/2) = (0.5, 3, 2.5).

2. p1(1, 4, 5), p2(4, -2, 7): The midpoint of p1p2⇀ is ((1 + 4)/2, (4 + (-2))/2, (5 + 7)/2) = (2.5, 1, 6).

3. p1(3, 4, 5), p2(2, 3, 4): The midpoint of p1p2⇀ is ((3 + 2)/2, (4 + 3)/2, (5 + 4)/2) = (2.5, 3.5, 4.5).

4. p1(0, 0, 0), p2(2, -2, -2): The midpoint of p1p2⇀ is ((0 + 2)/2, (0 + (-2))/2, (0 + (-2))/2) = (1, -1, -1).

a. To find the direction of a vector, we subtract the coordinates of its initial point from the coordinates of its terminal point. This gives us a vector that represents the change in position from the initial point to the terminal point. In this case, we subtract the coordinates of p1 from the coordinates of p2. The resulting vector represents the direction of movement from p1 to p2.

b. To find the midpoint of a line segment, we take the average of the coordinates of its

two endpoints. This gives us a point that lies exactly halfway between the two endpoints. In this case, we add the coordinates of p1 and p2 and divide each sum by 2 to find the average. The resulting point represents the midpoint of the line segment p1p2⇀.

By finding the direction and midpoint of a line segment, we can gain insight into its geometric properties. The direction vector provides information about the orientation and magnitude of the line segment, while the midpoint gives us a central reference point. These calculations are fundamental in geometry and can be applied in various contexts, such as determining the slope of a line or finding the center of a line segment.

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Answer:

x = 3

Step-by-step explanation:

2(x+4) + 2 = 5x + 1

2x + 8 + 2 = 5x + 1

2x + 10 = 5x + 1

-3x + 10 = 1

-3x = -9

x = 3

To solve for x, we need to simplify the equation and isolate the variable. Let's proceed with the given equation:

2(x + 4) + 2 = 5x + 1

First, distribute the 2 to the terms inside the parentheses:

2x + 8 + 2 = 5x + 1

Combine like terms on the left side:

2x + 10 = 5x + 1

Next, let's move all terms containing x to one side of the equation and the constant terms to the other side. We can do this by subtracting 2x from both sides:

2x - 2x + 10 = 5x - 2x + 1

Simplifying further:

10 = 3x + 1

To isolate the x term, subtract 1 from both sides:

10 - 1 = 3x + 1 - 1

9 = 3x

Finally, divide both sides of the equation by 3 to solve for x:

9/3 = 3x/3

3 = ×

Therefore, the solution to the equation is x = 3.

Step-by-step explanation:

When multiplying 2 numbers with the same base number, we can directly add their indices.

For example, 2¹ * 2³ = 2^(1 + 3) = 2⁴.

Therefore 7³ * 7^? = 7^(3 + ?) = 7¹⁴.

=> 3 + ? = 14, ? = 11.

The blank should be filled with 11.

As the degrees of freedom increase, the t distribution b. approaches the normal distribution, which is a key assumption in many statistical tests. Understanding this relationship is important for making accurate statistical inferences and drawing valid conclusions from data.

The t distribution is a probability distribution that is commonly used in hypothesis testing. It is similar to the normal distribution but with heavier tails. As the degrees of freedom increase, the t distribution approaches the normal distribution. This means that the shape of the t distribution becomes more and more like the normal distribution as the sample size increases.
The reason for this is that the t distribution is based on the sample mean, which becomes more normally distributed as the sample size increases due to the central limit theorem. As the sample size increases, the standard error of the mean decreases, and the t distribution becomes less spread out and more peaked. This is why we use the t distribution instead of the normal distribution when we have a small sample size.

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Answer:

A) y=2x-3

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(9-5)/(6-4)

m=4/2

m=2

y-y1=m(x-x1)

y-5=2(x-4)

y=2x-8+5

y=2x-3

Answer:

8, 13, 18

Step-by-step explanation:

Middle number is 13, there are only 3

Max is 10/2 more so 18

Min is 18-10

Answer:

72 kilometers per second

Step-by-step explanation:

We are looking for a unit rate of speed, so:

216 / 3 =

72 kilometers per second

Have a nice day!

    I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

Answer:

20 km/s (72,000 km/h; 45,000 mph)

Step-by-step explanation:

When a meteoroid, comet, or asteroid enters Earth's atmosphere at a speed typically in excess of 20 km/s (72,000 km/h; 45,000 mph), aerodynamic heating of that object produces a streak of light, both from the glowing object and the trail of glowing particles that it leaves in its wake. This phenomenon is called a meteor or "shooting star". Meteors typically become visible when they are about 100 km above sea level. A series of many meteors appearing seconds or minutes apart and appearing to originate from the same fixed point in the sky is called a meteor shower. A meteorite is the remains of a meteoroid that has survived the ablation of its surface material during its passage through the atmosphere as a meteor and has impacted the ground.

Object 1 undergoes circular or elliptical motion in a fixed plane, while object 2 undergoes helical or spiral motion in three-dimensional space. The specific values of A and ω will determine the exact nature and characteristics of their motions.

The given expressions describe the motion of two objects, object 1 and object 2, undergoing periodic motion. Let's analyze the qualitative motion of each object based on the provided information.

Object 1:

The expression for object 1's motion is given as r₁(t) = Re(A(x-hat + y-hat)e^(-iωt)). Here, the displacement of object 1 is described as the real part of a complex quantity. The motion of object 1 can be characterized as circular or elliptical motion in the x-y plane. The magnitude of the displacement is represented by A, and the frequency of the motion is determined by ω. The motion of object 1 is in a fixed plane and can be repetitive.

Object 2:

The expression for object 2's motion is given as r₂(t) = Re(A(x-hat + iy-hat)e^(-iωt)). Here, the displacement of object 2 is again described as the real part of a complex quantity. However, there is an imaginary component (iy-hat) in the expression, indicating that the motion of object 2 involves an oscillation in the complex plane. The motion of object 2 can be characterized as a helical or spiral motion in three-dimensional space. As the complex exponential term varies, the displacement vector rotates and traces out a helical path over time.

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The group with the largest uncertainty error in a 95% confidence level among Group A, B, C, and D is Group D.

To determine the uncertainty error for each group, we'll calculate the margin of error using the formula:

Margin of Error = t (95%) × (sample deviation / sqrt(sample size))

Assuming a t (95%) value of 2.0 and a sample size of 25 for all groups, we have:

Group A:

Margin of Error = \(2.0 × (0.51 / sqrt(25)) = 2.0 × (0.51 / 5) = 0.204\)

Group B:

Margin of Error =\(2.0 × (0.06 / sqrt(25)) = 2.0 × (0.06 / 5) = 0.024\)

Group C:

Margin of Error = \(2.0 × (0.55 / sqrt(25)) = 2.0 × (0.55 / 5) = 0.22\)

Group D:

Margin of Error = \(2.0 × (2.00 / sqrt(25)) = 2.0 × (2.00 / 5) = 0.8\)

Based on the margin of error calculations, Group D has the largest uncertainty error at a 95% confidence level, with a

margin of error of 0.8 mm.

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1.1) The volume of the cube is 27 cubic centimeters. 1.2)the density of the ice cube is approximately 0.917 grams per cubic centimeter (g/cm³).

1.3) the mass of the water cube is 27 grams. 1.4) the weight of the water and the ice would be the same under the same conditions. 1.5)In simpler terms, ice floats on water because it is lighter (less dense) than the water, allowing it to displace an amount of water equal to its weight and float on the surface.

1.1) The volume of the cube can be calculated using the equation: Volume = width x height x length. In this case, the cube measures 3 cm wide, 3 cm tall, and 3 cm long, so the volume is:

Volume = 3 cm x 3 cm x 3 cm = 27 cm³.

Therefore, the volume of the cube is 27 cubic centimeters.

1.2) Density is defined as mass divided by volume. The mass of the ice cube is given as 24.76 grams, and we already determined the volume to be 27 cm³. Therefore, the density of the ice cube is:

Density = Mass / Volume = 24.76 g / 27 cm³ ≈ 0.917 g/cm³.

Therefore, the density of the ice cube is approximately 0.917 grams per cubic centimeter (g/cm³).

1.3) The volume of the water cube is the same as the ice cube, which is 27 cm³. Given the density of liquid water as 1.00 g/cm³, we can calculate the mass of the water cube using the equation:

Mass = Density x Volume = 1.00 g/cm³ x 27 cm³ = 27 grams.

Therefore, the mass of the water cube is 27 grams.

1.4) The weight of an object depends on both its mass and the acceleration due to gravity. Since the volume of the water cube and the ice cube is the same (27 cm³), and the mass of the water cube (27 grams) is equal to the mass of the ice cube (24.76 grams), their weights would also be equal when measured in the same gravitational field.

Therefore, the weight of the water and the ice would be the same under the same conditions.

1.5) Ice floats on water because it is less dense than liquid water. The density of ice is lower than the density of water because the water molecules in the solid ice are arranged in a specific lattice structure with open spaces. This arrangement causes ice to have a lower density compared to liquid water, where the molecules are closer together.

When ice is placed in water, the denser water molecules exert an upward buoyant force on the less dense ice, causing it to float. The buoyant force is the result of the pressure difference between the top and bottom surfaces of the submerged object.

In simpler terms, ice floats on water because it is lighter (less dense) than the water, allowing it to displace an amount of water equal to its weight and float on the surface.

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Answer:

(multiply both sides by 6)

6×\(\frac{4h+4}{6}\)-10×6=-2h×6

(simplify)

4h+4-60=-12h

4h-56=-12h

(move constant to the right side and variable to the left)

4h+12h=56

16h=56

(divide both sides by 16)

h=7/2

It's a great question. Believe me, a lot of people have been asking me this.

Let me tell you, we've got 72 oranges in that crate, and we're filling trays with 9 oranges each. It's a big number, trust me.

But let me tell you, we're gonna figure out how many trays we need. We're gonna use some math, it's gonna be huge.

So, we divide 72 by 9 and we get 8. Trust me, it's a big number. 8 trays, that's how many we need.

It's a great number, I promise you. A lot of people are gonna be impressed with that number. It's tremendous, believe me.

Answer: 8

Step-by-step explanation: what is 72/8 ?

Since the we have to equally put it together

Answers:

===============================================

Explanation:

For these questions, we'll use the inscribed angle theorem. This says that the inscribed angle is half the measure of the arc it cuts off. An inscribed angle is one where the vertex of the angle lies on the circle, as problem 1 indicates.

For problem 1, the arc measure is 80 degrees, so half that is 40. This is the measure of the unknown inscribed angle.

Problem 2 will have us work in reverse to double the inscribed angle 42 to get 84.

-------------------

For problem 3, we need to determine angle DEP. But first, we'll need Thales Theorem which is a special case of the inscribed angle theorem. This theorem states that if you have a semicircle, then any inscribed angle will always be 90 degrees. This is a handy way to form 90 degree angles if all you have is a compass and straightedge.

This all means that angle DEF is a right angle and 90 degrees.

So,

(angle DEP) + (angle PEF) = angle DEF

(angle DEP) + (35) = 90

angle DEP = 90 - 35

angle DEP = 55

The inscribed angle DEP cuts off the arc we want to find. Using the inscribed angle theorem, we double 55 to get 110 which is the measure of minor arc FD.

The probabilities as follows:

A. P(F2|F1) = 0.3 (30%)

B. P(F1|F2) = 0.25 (25%)

C. P(F1 or F2) = 0.19 (19%)

D. P(not F1 and not F2) = 0.81 (81%)

To solve this problem, we can use the concept of conditional probability and the principle of inclusion-exclusion.

Given:

P(F1) = 0.10 (Probability of failing at Check Point 1)

P(F2) = 0.12 (Probability of failing at Check Point 2)

P(F1 and F2) = 0.03 (Probability of failing at both Check Point 1 and Check Point 2)

A. To find the probability that an item failed at Check Point 1 and also failed at Check Point 2 (F2|F1), we use the formula for conditional probability:

P(F2|F1) = P(F1 and F2) / P(F1)

Substituting the given values:

P(F2|F1) = 0.03 / 0.10

P(F2|F1) = 0.3

Therefore, the probability that an item failed at Check Point 1 and also failed at Check Point 2 is 0.3 or 30%.

B. To find the probability that an item failed at Check Point 2 given that it failed at Check Point 1 (F1|F2), we use the same formula:

P(F1|F2) = P(F1 and F2) / P(F2)

Substituting the given values:

P(F1|F2) = 0.03 / 0.12

P(F1|F2) = 0.25

Therefore, the probability that an item failed at Check Point 2 and also failed at Check Point 1 is 0.25 or 25%.

C. To find the probability that an item failed at either Check Point 1 or Check Point 2 (F1 or F2), we can use the principle of inclusion-exclusion:

P(F1 or F2) = P(F1) + P(F2) - P(F1 and F2)

Substituting the given values:

P(F1 or F2) =\(0.10 + 0.12 - 0.03\)

P(F1 or F2) = 0.19

Therefore, the probability that an item failed at either Check Point 1 or Check Point 2 is 0.19 or 19%.

D. To find the probability that an item failed at neither of the check points (not F1 and not F2), we can subtract the probability of failing from 1:

P(not F1 and not F2) = 1 - P(F1 or F2)

Substituting the previously calculated value:

P(not F1 and not F2) = 1 - 0.19

P(not F1 and not F2) = 0.81

Therefore, the probability that an item failed at neither Check Point 1 nor Check Point 2 is 0.81 or 81%.

In conclusion, we have calculated the probabilities as follows:

A. P(F2|F1) = 0.3 (30%)

B. P(F1|F2) = 0.25 (25%)

C. P(F1 or F2) = 0.19 (19%)

D. P(not F1 and not F2) = 0.81 (81%)

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Answer:

A. p is the independent variable, and s is the dependent variable.

Step-by-step explanation:

In this scenario, the number of people attending the potluck is the independent variable because it is the variable that is being controlled or determined before the brownies are cut. The number of squares the brownies are cut into, on the other hand, depends on the number of people attending the potluck. Therefore, the number of squares is the dependent variable in this scenario.

The differential equation is not a linear.

The given differential equation is: y′ + x√y = x^2
To determine if this differential equation is linear, we need to check if it can be written in the standard linear form:
y′ + p(x)y = q(x)

Comparing this standard form with the given equation, we can see that the term x√y prevents the equation from being linear. In a linear equation, the dependent variable y and its derivatives (in this case, y′) must appear with a power of 1, and not be inside any non-linear functions like a square root. Since x√y is a non-linear term, the given differential equation is not linear.

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Answer: B. x(x−5)(x+7)

Step-by-step explanation: LCD...Apex

Answer: x=20/3=6.67

Step-by-step explanation:

To solve proportions, the most useful way is to use [cross multiplication], where we not multiply horizontally but as diagonally.

-----------------------------------------------------------------------------

Given

x:10=9:6

Cross Multiply

9·x=6·10

9x=60

Divide 9 on both sides

9x/9=60/9

x=20/3=6.67

Hope this helps!! :)

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The approximate volume of the cylinderical tube is 1005 cm³  round to the nearest whole cubic centimeter.

Volume of cylinder is the amount of quantity, which is obtained by the bylinder in the 3 dimensional space.

Volume of cylinder is the can be given as,

\(V=\pi r^2h\)

Here, (r) is the radius of the base of the cylinder and (h) is the height of the cylinder.

A cylindrical cardboard tube with a diameter of 8 centimeters and a height of 20 centimeters is used to package a gift.

The diameter is 8 cm and the radius of a cylinder is half of its radius. Thus the radius of the cylinder is,

\(r=\dfrac{8}{2}\\r=4\rm\; cm\)

Put the values in the above formula a,s

\(V=\pi (4)^2(20)\\V=320\pi\\V=1005.31\\V=\approx 1005\rm\;cm^3\)

The approximate volume of the cylinderical tube is 1005 cm³  round to the nearest whole cubic centimeter.

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Answer:

Step-by-step explanation:

To solve the given equation \(\sf x - y = 4 \\\), we can perform the following calculations:

a) To find the value of \(\sf 3(x - y) \\\):

\(\sf 3(x - y) = 3 \cdot 4 = 12 \\\)

b) To find the value of \(\sf 6x - 6y \\\):

\(\sf 6x - 6y = 6(x - y) = 6 \cdot 4 = 24 \\\)

c) To find the value of \(\sf y - x \\\):

\(\sf y - x = - (x - y) = -4 \\\)

Therefore:

a) The value of \(\sf 3(x - y) \\\) is 12.

b) The value of \(\sf 6x - 6y \\\) is 24.

c) The value of \(\sf y - x \\\) is -4.

\(\huge{\mathfrak{\colorbox{black}{\textcolor{lime}{I\:hope\:this\:helps\:!\:\:}}}}\)

♥️ \(\large{\underline{\textcolor{red}{\mathcal{SUMIT\:\:ROY\:\:(:\:\:}}}}\)

The statement "If a function is differentiable, then it is continuous" is generally true. In summary, if a function is differentiable at a point, it must also be continuous at that point.

Differentiability is a stronger condition than continuity. If a function is differentiable at a point, it means that it has a derivative at that point, indicating the existence of a well-defined tangent line.

A key property of differentiability is that it implies continuity. In other words, if a function is differentiable at a point, it must also be continuous at that point.

On the other hand, a function can be continuous without being differentiable. Continuous functions have no abrupt jumps, holes, or vertical asymptotes in their graphs. However, they may have corners, cusps, or vertical tangents, which prevent them from being differentiable at those points. Therefore, while differentiability guarantees continuity, continuity does not necessarily imply differentiability.

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Complete question - Write the conserve and contrapositive of the following statement- "If a function is differentiable then it is continuous."

x = -19 -6y 2x + 3y = -11

x+6y = -19 2x + 3y = -11

2(x+6y)= 2(-19) 2x + 3y = -11

2x + 12y = -38 2x + 3y = -11

subtracting both equation

2x + 12y-(2x + 3y) = -38-(-11)

2x+12y-2x-3y = -38+11

9y= -27

y= -3

x= -19-(6×(-3))

x= -19+18

x= -1

x= -1 and y = -3

To find the area of the figure, we need to divide it into smaller rectangles and squares, and then sum their areas.

First, we can divide the figure into two rectangles, as shown:

```

+----+----+----+----+----+

|    |    |    |    |    |

|    |    |    |    |    |

+----+----+----+----+----+

|         |         |     |

|         |         |     |

+---------+---------+-----+

```

The left rectangle has dimensions 3 cm × 2 cm, so its area is:

A1 = 3 cm × 2 cm = 6 square cm

The right rectangle has dimensions 6 cm × 2 cm, so its area is:

A2 = 6 cm × 2 cm = 12 square cm

Now we can divide the left rectangle into two squares and a rectangle, as shown:

```

+----+----+----+

|    |    |    |

|    |    |    |

+----+----+----+

|    |         |

|    |         |

+----+---------+

|              |

|              |

+--------------+

```

The top square has dimensions 2 cm × 2 cm, so its area is:

A3 = 2 cm × 2 cm = 4 square cm

The bottom square has dimensions 1 cm × 1 cm, so its area is:

A4 = 1 cm × 1 cm = 1 square cm

The remaining rectangle has dimensions 2 cm × 1 cm, so its area is:

A5 = 2 cm × 1 cm = 2 square cm

Finally, we can add up the areas of all the rectangles and squares to get the total area of the figure:

A = A1 + A2 + A3 + A4 + A5 = 6 cm^2 + 12 cm^2 + 4 cm^2 + 1 cm^2 + 2 cm^2 = 25 square cm

Therefore, the area of the figure is 25 square centimeters.

Answer:

134%

Step-by-step explanation:

Answer: (C) 5

Step-by-step explanation:

To find the percentage of classmates who chose soccer lets add the percentage of all the other sports and substract it by 100.

So,

100 - (10 + 10 + 10 + 30 + 15)

100 - 75

= 25

Now, we have to find out how many classmate chose football over soccer.

For this we have to substract the percentage of classmates who chose soccer (25) from the percentage of classmates who chose football(30)

30 - 25

= 5

Thus, the answer is 5.

As per the given 4 x 5 matrix, the reduced row echelon form of A is \(\left[\begin{array}{ccc}1&2&0\\0&0&1\\0&0&0\end{array}\right]\)

The term matrix in math refers a set of numbers arranged in rows and columns so as to form a rectangular array.

Here we have the 4 x 5 matrix.

And here we also know that a1,a2,a4 are linearly independent and the value of a3=a1+2a² and a5=2a1-a²+3a⁴.

Now, we have to apply the value of

a1 = 1, a2 = 0, a3 = 0, a4 = 0, a5 = 1, a6 = 0, a7 = 0, a8 = 0, and a9 = 1.

Then we get the matrix of 3 x 3 looks like the following,

\(\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]\)

Now, we have to do the following steps in order to get the reduced row echelon form of A,

The first and foremost steps is to take the non-zero number in the first row is the number 1.

Then we have to place any non-zero rows are placed at the bottom of the matrix.

This steps are repeated until the final row becomes zero.

Then we get the resulting matrix as \(\left[\begin{array}{ccc}1&2&0\\0&0&1\\0&0&0\end{array}\right]\)

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