The triangle above has the following measures.
q = 6 yd
m Find the length of side r.
Round to the nearest tenth and include correct units.

SHOW ALL YOUR WORK!!!

A photo of a beetle in a science book is increased to 555​% as large as the actual size. If the beetle is 14 ​millimeters. The size of the beetle in the photo is 77.7 millimeters.

To determine the size of the beetle in the photo, we can multiply its actual size by the percentage increase in size. 555% can also be expressed as a decimal, 5.55. Therefore, to find the size of the beetle in the photo, we multiply 14 millimeters by 5.55, which gives us 77.7 millimeters. This means that the beetle appears to be almost six times larger in the photo than its actual size. It's important to note that the size of the beetle in the photo may vary depending on the size of the book it's printed in or the resolution of the image.

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Answer: 67

explanation:

47 x 13 = 611    -- so this is how many tickets he has total

611 / 9 = 67.9

So he can get 67 prizes.

a way you can check if this is right is by multiplying 67 by 9 and you see that is 603, and he has 611 tickets so he will even have a few tickets left over. but he can not get 68 prizes because if you see what 68 x 9 is, it is 612 which is one ticket more than what he has so he cant get it. So the max he can get is 67

Answer:

67

Step-by-step explanation:

We know that, in total, he made 47 baskets, and each basket is 13 tickets.

So in total, he made 611 tickets:

47·13

=611

Next, he wanted to get [a] prize(s), but each prize costs 9 tickets, we divide the total number of tickets he has by the prize cost.

611/9

=67.88...

You obviously can't get 0.88 of a prize, so the max amount of prizes that Arthur can get is 67.

Hope this helps! :)

In an analysis of variance, you reject the null hypothesis when the F ratio is:
b. much larger than 1.

Here's a step-by-step explanation:
1. The null hypothesis states that there are no significant differences among the groups being compared.
2. Variance is the measure of how much the individual data points in a dataset vary from the mean.
3. The F ratio is the ratio of two variances, specifically, the variance between group means and the variance within groups.
4. When the F ratio is much larger than 1, it indicates that the variance between group means is much larger than the variance within groups.
5. This suggests that there is a significant difference among the group means, leading to the rejection of the null hypothesis.

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If X1,…,Xn be an independent trials process, and S∗n be their standardized sum then,

limn→[infinity]P(S∗n > -0.28) = 0.6103

To answer this question, we need to use the central limit theorem. The central limit theorem states that if we have a large enough sample size, the distribution of the standardized sum (S∗n) approaches a normal distribution.

In this case, we want to find the limit of the probability that S∗n is greater than -0.28 as n approaches infinity.

Let Z be a standard normal random variable, then:

P(S∗n > -0.28) = P((S∗n - E[S∗n]) / sqrt(Var[S∗n]) > (-0.28 - E[S∗n]) / sqrt(Var[S∗n]))
= P(Z > (-0.28 - 0) / 1)
= P(Z > -0.28)

Using a standard normal distribution table, we can find that P(Z > -0.28) is approximately 0.6103.

Therefore, limn→[infinity]P(S∗n > -0.28) = 0.6103, since as n approaches infinity, S∗n approaches a standard normal distribution.

The correct question should be :

Let X1,…,Xn be an independent trials process, and S∗n be their standardized sum. What is limn→[infinity]P(S∗n > -0.28)?

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The upper class limits are- 29, 39, 49, 59, 69, 79, 89

The lower class limits are- 20, 30, 40, 50, 60, 70, 80

The class midpoints are- 24.5, 24.5, 44.5, 54.5, 64.5, 74.5, 84.5

The class boundaries are- 29.5, 39.5, 49.5, 59.5, 69.5, 79.5

And the number of individuals included in the summary is 84.

Here, we are given the following dataset-

Age (yr) when          Frequency

award was won

20-29                       27

30-39                       32

40-49                       15

50-59                       3

60-69                       5

70-79                        1

80-89                       1

Upper class limit is the largest data value that can go in a class.

Thus, the upper class limits are- 29, 39, 49, 59, 69, 79, 89

Lower class limit is the smallest data value that can go in a class.

Thus, the lower class limits are- 20, 30, 40, 50, 60, 70, 80

The class midpoint is the average of the upper and lower limits of a class. Class midpoint = (upper limit + lower limit)/ 2

Thus, the class midpoints are- 24.5, 24.5, 44.5, 54.5, 64.5, 74.5, 84.5

Class boundary is the midpoint of the upper class limit of a class and the lower class limit of the previous class.

Thus, the  class boundaries are- 29.5, 39.5, 49.5, 59.5, 69.5, 79.5

Frequency gives us the number of individuals/ objects belonging to a particular class.

Thus, the number of individuals included in the summary = 27 + 32 + 15 + 3 + 5 + 1 + 1 = 84

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complete question:

Identify the lower class limits, upper class limits,

class width, class midpoints, and class boundaries for

the given frequency distribution. Also identify the

number of individuals included in the summary.

Age (yr) when

award was won

20-29

30-39

40-49

50-59

60-69

70-79

80-89

Frequency

27

32

15

3

5

1

1

Answer:

is it 193

Step-by-step explanation:

i added -1 and 192

The correct option for calculating the degrees of freedom for the between-group estimate in ANOVA is: c. k - 1

Here's a step-by-step explanation:

1. ANOVA, or Analysis of Variance, is a statistical method used to compare the means of multiple groups to determine if there are significant differences between them. In this context, "k" represents the number of groups being compared, and "N" represents the total number of observations.

2. Degrees of freedom (df) are used in statistical tests to account for variability in the data. They are essentially the number of values that can vary independently in the calculation of a statistic.

3. In ANOVA, there are two types of degrees of freedom: between-group (df_between) and within-group (df_within).

4. To calculate the between-group degrees of freedom (df_between), we use the formula: df_between = k - 1. This is because there are k groups being compared, and each group contributes one degree of freedom, minus one since we are comparing the groups against each other.

5. The within-group degrees of freedom (df_within) would be calculated using the formula: df_within = N - k, which accounts for the total number of observations minus the number of groups.

In summary, to compute the degrees of freedom for the between-group estimate in ANOVA, you would use the formula df_between = k - 1.

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The z-coordinate of the mass center of the homogeneous paraboloid of revolution shown is (12/5) units.

The determination of the z-coordinate of the mass center of the homogeneous paraboloid of revolution requires a long answer.

Firstly, we need to define the mass density of the paraboloid. Since it is a homogeneous object, its mass density is constant throughout its volume.

Let us denote this density as ρ.

Next, we need to find the volume of the paraboloid.

The volume of a paraboloid of revolution with a height h and a base radius r is given by V = (π/2) * r^2 * h/3.

In this case, the height of the paraboloid is 4 units and the radius of the base is 2 units. Thus, the volume of the paraboloid is:
V = (π/2) * (2)^2 * 4/3 = (8π/3) units^3

Now, we can find the mass of the paraboloid by multiplying its volume by its density:
M = ρ * V = ρ * (8π/3) units^3

The next step is to find the x and y coordinates of the mass center of the paraboloid.

We can do this by using double integrals:
x = (1/M) * ∬(paraboloid) x * ρ * dV
y = (1/M) * ∬(paraboloid) y * ρ * dV

Since the paraboloid is symmetric about the z-axis, we know that its mass center will lie on this axis, and thus, its x and y coordinates will be zero.

Finally, we need to find the z-coordinate of the mass center. We can do this by using the same double integral, but this time we integrate over the z-axis:
z = (1/M) * ∬(paraboloid) z * ρ * dV

To set up the double integral, we can use cylindrical coordinates, with ρ ranging from 0 to 2 and θ ranging from 0 to 2π.

The z-coordinate of any point on the paraboloid is given by z = (1/16) * (x^2 + y^2), so we can substitute this into the double integral:
z = (1/M) * ∫(0 to 2π) ∫(0 to 2) ∫(0 to (1/16)*(ρ^2)) ρ * z * ρ * ρ * dρ dθ dz

After evaluating this integral, we get:
z = (3/5) * h = (12/5) units

Therefore, the z-coordinate of the mass center of the homogeneous paraboloid of revolution shown is (12/5) units.

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The partial derivatives using implicit differentiation are:
∂z/∂x = -x / (9z)
∂z/∂y = -4y / (9z)

To find the partial derivatives ∂z/∂x and ∂z/∂y using implicit differentiation, we start with the given equation:
x^2 + 4y^2 + 9z^2 = 4

First, we differentiate both sides of the equation with respect to x:
2x + 0 + 18z(∂z/∂x) = 0

Now, solve for ∂z/∂x:
18z(∂z/∂x) = -2x
∂z/∂x = -2x / (18z)
∂z/∂x = -x / (9z)

Next, we differentiate both sides of the equation with respect to y:
0 + 8y + 18z(∂z/∂y) = 0

Now, solve for ∂z/∂y:
18z(∂z/∂y) = -8y
∂z/∂y = -8y / (18z)
∂z/∂y = -4y / (9z)

So, the partial derivatives are:
∂z/∂x = -x / (9z)
∂z/∂y = -4y / (9z)

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Over 10.144 years of age are found in 75% of firm employees' autos.

We need to identify the z-score that corresponds to the age of the cars driven by firm employees that are 75% older than the other cars of 75th percentile of the normal distribution.

A normal distribution calculator or statistical software programme can be used to find this z-score. As an alternative, we can look for the value using a z-distribution table.

The formula for the z-score is:

z = (x - μ) / σ

If x is the value we are looking for, is the distribution's mean, and is its standard deviation.

We can use the common normal distribution table or a calculator with a built-in function to determine the z-score corresponding to the 75th percentile  to a cumulative probability of 0.75. The table value is 0.67.

As a result, we can determine the age of the company vehicles that are approximately 75% older than the others as follows:

x = μ + zσ

= 8 + (0.67)(3.2)

= 10.144

As a result, more than 75% of the cars that firm employees drive are older than 10.144 years.

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we can conclude that the statement "if a, b, and d are integers with d > 0, then (a b) div d = a div d b div d" is false.

We can prove that if a, b, and d are integers with d > 0, then (a + b) div d = a div d + b div d or disprove it by finding a counterexample.

Let's choose some specific values for a, b, and d to see if the equation holds. Let a = 8, b = 5, and d = 3.

(a + b) div d = (8 + 5) div 3 = 13 div 3 = 4

a div d + b div d = 8 div 3 + 5 div 3 = 2 + 1 = 3

Since (a + b) div d ≠ a div d + b div d for our chosen values of a, b, and d, we have found a counterexample that disproves the equation.

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For the study of trucker drives, the 99% confidence interval of the mean number of miles driven per day by all truckers is equals to the (−531.74,548.26).

The one way to estimate the population mean is the confidence interval. The interval consists of a point estimate and a margin of error for the estimate. We have a sample of study of truckers,

Mean of miles = 540

Sample size, n = 90

standard deviations, s = 40

Confidence level = 0.99

Level of significance = 0.01

We have to determine the 99% confidence interval of the mean number of miles. Using the z distribution table value of z score for 99% confidence level or 0.01 significance level is equals to 1.96.

Using the confidence interval formula,

[tex]CI = \bar x ± Z_{\frac{\alpha}{2}} (\frac{s}{\sqrt{n}}) [/tex]

[tex]= 540 ± 1.96 (\frac{40}{\sqrt{90}}) [/tex]

= (−531.74,548.26)

Hence, required value is (−531.74,548.26).

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Answer:

False

Step-by-step explanation:

4/8 equals to 1/2 but 6/10 equals to 3/5. Correct would be if it was 5/10

Answer:

8x + 8y + 16

Step-by-step explanation:

It is simple perimeter is the sum of all side of the polygon (figure) so

P = (4X+5) + (2Y) + (X+5) + (2Y+3) + (3X) + (4Y+3)

I write

so

P = (4X+5) + (2Y) + (X+5) + (2Y+3) + (3X) + (4Y+3)

P = 8x + 8y + 16

so the perimeter of the figere is 8x + 8y + 16

The value of the size of angel a is, 100 degree

And, The value of size of angle b is, 280 degree

We have to given that;

Figure is shown in figure.

Now, By measuring of angle we get;

The value of the size of angel a is,

⇒ ∠ a = 100 - 0 degree

⇒ ∠ a = 100 degree

And, The value of size of angle b is,

⇒ ∠ b = 180° + 100°

⇒ ∠b = 280°

Thus, The value of the size of angel a is, 100 degree

And, The value of size of angle b is, 280 degree

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Therefore, The probability that the device will work is 0.634 or 63.4%.

This problem can be solved using the complement rule. The complement of the device working is all the parts failing. Therefore, the probability of the device not working is (1 - 0.216)^7 = 0.366. To find the probability of the device working, we subtract this from 1:
1 - 0.366 = 0.634.
To find the probability that the device will work, we'll use the complementary probability. This means we'll first find the probability that all parts fail and then subtract it from 1. Let's denote the probability of a part failing as q, which is equal to 1 - p.
Step 1: Calculate q.
q = 1 - p = 1 - 0.216 = 0.784
Step 2: Calculate the probability of all parts failing.
P(all parts fail) = q^7 = 0.784^7 ≈ 0.1278
Step 3: Calculate the probability that the device will work.
P(device works) = 1 - P(all parts fail) = 1 - 0.1278 ≈ 0.8722
In conclusion, the probability that the device will work is approximately 0.8722.

Therefore, The probability that the device will work is 0.634 or 63.4%.

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Answer:

An equation in the form y = mx + b is in the 'slope y-intercept' form where m is the slope and b is the y-intercept. We can rewrite our equation, y = 4, in slope y-intercept form as follows: y = 0x + 4. Here, it is clear that the slope, or m, is zero. Therefore, the slope of the horizontal line y = 4 is zero

The fraction of shaded circles is 1/2. A fraction where the numerator represents the number of shaded circles and the denominator represents the total number of circles.

If there are four separate circles and two of them are shaded, we can represent this as a fraction where the numerator represents the number of shaded circles and the denominator represents the total number of circles.

So the fraction of shaded circles would be:

2/4

This fraction can be simplified by dividing the numerator and denominator by their greatest common factor, which is 2:

2/4 = 1/2

Therefore, the fraction of shaded circles is 1/2.

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(a) We have shown that there exists an element b ∈ B that is an upper bound for A.

(b) The statement in part (a) is not always the case if we only assume sup A ≤ sup B.

(a) If sup A < sup B, show that there exists an element b ∈ B that is an upper bound for A.

Proof:
1. By definition, sup A is the least upper bound for set A, and sup B is the least upper bound for set B.
2. Since sup A < sup B, there must be a value between sup A and sup B.
3. Let's call this value x, where sup A < x < sup B.
4. Now, since x < sup B and sup B is the least upper bound of set B, there must be an element b ∈ B such that b > x (otherwise, x would be the least upper bound for B, which contradicts the definition of sup B).
5. Since x > sup A and b > x, it follows that b > sup A.
6. As sup A is an upper bound for A, it implies that b is also an upper bound for A (b > sup A ≥ every element in A).

Thus, we have shown that there exists an element b ∈ B that is an upper bound for A.

(b) Give an example to show that this is not always the case if we only assume sup A ≤ sup B.

Example:
Let A = {1, 2, 3} and B = {3, 4, 5}.
Here, sup A = 3 and sup B = 5. We can see that sup A ≤ sup B, but there is no element b ∈ B that is an upper bound for A, as the smallest element in B (3) is equal to the largest element in A, but not greater than it.

This example shows that the statement in part (a) is not always the case if we only assume sup A ≤ sup B.

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the lower class limit represents the smallest data value that can be included in the classThe statement is true.

The lower class limit is the smallest value that can be included in a class interval.

Therefore, the statement is correct.

The lower class limit represents the smallest data value that can be included in a particular class. In a frequency distribution table, data values are grouped into classes, and each class has a lower and upper class limit. The lower class limit denotes the lowest value within that class, and any data value equal to or greater than the lower limit but less than the upper limit falls into that class.

The statement is true, as the lower-class limit indeed represents the smallest data value that can be included in the class.

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Answer:  The value of the surface integral is 2π/3.

Step-by-step explanation:

To apply Stokes' theorem, we need to find the curl of the vector field F:

curl F = (dF_z/dy - dF_y/dz)i + (dF_x/dz - dF_z/dx)j + (dF_y/dx - dF_x/dy)k

= xyi + k

Now, we can apply Stokes' theorem:

∫∫S curl F · dS = ∫C F · dr

where S is the surface bounded by the curve C, and dS is the outward-pointing normal vector to S.

First, we need to parameterize the surface S. We can use cylindrical coordinates, since the surface has rotational symmetry around the z-axis:

x = r cosθ

y = r sinθ

z = √(1 - r^2 - 3z^2)

where 0 ≤ r ≤ 1/√(3), and 0 ≤ θ ≤ 2π.

The normal vector to S is

dS = (∂z/∂r × ∂z/∂θ, ∂x/∂r × ∂x/∂θ, ∂y/∂r × ∂y/∂θ)

= (3r^2 cosθ, -3r^2 sinθ, 1 - 2r^2)

Since n is upward-pointing, we need to flip the sign of the z-component of dS. Thus, we have:

dS = (-3r^2 cosθ, 3r^2 sinθ, 2r^2 - 1)

Next, we need to find the curve C, which is the boundary of S. Since S lies on the plane z = -2, we have:

x^2 + y^2 + 3z^2 = 1

r^2 + 3z^2 = 1

z = -2, or r = 1/√(3)

Thus, the curve C consists of two circles of radius 1/√(3) in the planes z = ±√(2/3). We can parameterize these circles as:

r = 1/√(3)

θ = t

z = √(2/3)

and

r = 1/√(3)

θ = t

z = -√(2/3)

where 0 ≤ t ≤ 2π.

Now, we can evaluate the line integral:

∫C F · dr = ∫C (yi - xj + zxy·k) · dr

= ∫C (r sinθ i - r cosθ j + √(1 - r^2 - 3z^2) r^2 sinθ cosθ k) · (dr/dt) dt

= ∫0^2π (-(1/√3) sin t i - (1/√3) cos t j - (2/3) (cos^2 t - sin^2 t) k) · (-1/√3 sin t i + 1/√3 cos t j) dt

= ∫0^2π (1/3 sin^2 t + 1/3 cos^2 t) dt

= 2π/3

Finally, we can use Stokes' theorem to find the surface integral:

∫∫S curl F · dS = ∫C F · dr = 2π/3

Therefore, the value of the surface integral is 2π/3.

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The complete table.

x  -8   -4  0   4    8

y  28  16  4  -8  -20

We have,

y = -3x + 4

Now,

Substituting x = -8, -4, 0, 4, and 8.

y = -3 x -8 + 4 = 24 + 4 = 28

y = -3 x -4 + 4 = 12 + 4 = 16

y = -3 x 0 + 4 = 0 + 4 = 4

y = -3 x 4 + 4 = -12 + 4 = -8

y = -3 x 8 + 4 = -24 + 4 = -20

Thus,

The complete table.

x  -8   -4  0   4    8

y  28  16  4  -8  -20

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The teacher asked the first 5 students on her class seating chart to write down where the class should go on a field trip is yes

Clara surveyed more than half the teachers at her school and then, without looking, selected the responses for her sample is no

To get a random sampling of every student in her school, Cheryl surveyed all the students in her math class is no

A company sent e-mails to its customers asking them to e-mail back their opinion of the company’s products. Only 10 of them e-mailed back is no

1) Yes, this is an example of random sampling.

The teacher used a systematic method of selecting the first 5 students on her seating chart to ensure that the sample is representative of the class.

2) No, this is not an example of random sampling.

Clara did not use a random selection method to choose the responses for her sample, which can result in a biased sample.

3) No, this is not an example of random sampling.

Cheryl only surveyed the students in her math class, which is not a representative sample of the entire school population.

4) No, this is not an example of random sampling.

The sample is not representative of the entire customer population since only 10 customers responded.

This may lead to biased results if the 10 customers who responded have different opinions than those who did not respond.

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Answer: 48

Step-by-step explanation: we can split this into a trapezoid and a rectangle. The area of the rectangle is 6*5 = 30.

Approximately 19.8 feet of cable is needed to connect the pole to the ground.

To calculate the length of the cable needed to connect the 20-foot pole to the ground, we can use the Pythagorean theorem.

The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

In this case, the pole acts as the hypotenuse, the distance from the base of the pole to the ground acts as one side, and the length of the cable acts as the other side.

Using the Pythagorean theorem, we can find the length of the cable:

Length of cable = √(Length of pole^2 - Distance^2)

= √(20^2 - 3^2)

= √(400 - 9)

= √391

≈ 19.8 feet (rounded to the nearest tenth)

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The height of the pyramid is approximately 86.4 inches.

The square pyramid can be divided into four identical triangles. Each triangle has a base equal to one side of the square base and a height equal to the height of the pyramid. The slant height of the pyramid is the hypotenuse of each of these triangles.

Using the Pythagorean theorem, we can find the height of each triangle (and hence the height of the pyramid):

h^2 + (22.6/2)^2 = 87.9^2

Simplifying the equation:

h^2 + 255.76 = 7728.41

h^2 = 7472.65

h ≈ 86.4

Therefore, the height of the pyramid is approximately 86.4 inches.

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The maximum height of the ball is 9.29 feet.

The given function is h(t) = -161t²+32t+9

To find the maximum height of the ball, we need to find the vertex of the quadratic function.

The vertex of the parabola with equation h = ax² + bx + c is given by the point (-b/2a, f(-b/2a)).

In this case, a = -161, b = 32, and c = 9

so the vertex is located at t = -b/2a

= -32/2(-161)

= 0.1 seconds.

Plugging this value into the equation, we get the height

h = -161(0.1)² + 32(0.1) + 9

= 9.29 feet

Therefore, the maximum height of the ball is 9.29 feet.

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1) The value of function is,

⇒ (2x − 3)

2) The value of function is,

⇒ (5x + 7)

We have to given that;

Functions are,

f(x) = (x + 7) (2x − 3) and g(x) = (x + 7).

Hence, We get;

⇒ f (x) / g (x)

⇒ (x + 7) (2x − 3) / (x + 7)

⇒ (2x − 3)

And, Functions are,

f(x) = (5x+7) (-3x+11) and g(x) = (-3x + 11).

Hence, We get;

⇒ f (x) / g (x)

⇒ (5x+7) (-3x+11) / (-3x + 11).

⇒ (5x + 7)

Thus, 1) The value of function is,

⇒ (2x − 3)

2) The value of function is,

⇒ (5x + 7)

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Let's assume that the length of the rectangular kitchen is L and the width is W. We know that the area of the kitchen is 91 square meters, so we can write:

L x W = 91

We also know that the perimeter of the kitchen is 40 meters, which means:

2L + 2W = 40

We can simplify this equation by dividing both sides by 2:

L + W = 20

Now we have two equations:

L x W = 91

L + W = 20

We can use substitution to solve for one of the variables. Let's solve for L:

L = 20 - W

Now we can substitute this expression for L in the first equation:

(20 - W) x W = 91

Expanding this equation gives us a quadratic equation:

W^2 - 20W + 91 = 0

We can solve for W using the quadratic formula:

W = (20 ± √(20^2 - 4 x 1 x 91)) / (2 x 1)

W = (20 ± 3) / 2

W = 11 or W = 9

If W is 11, then L is 9. If W is 9, then L is 11. Therefore, the dimensions of the kitchen are either 9 meters by 11 meters or 11 meters by 9 meters.

In summary, we can use the area and perimeter of a rectangular shape to find its dimensions by setting up equations and solving for the variables. In this case, we used substitution and the quadratic formula to find the possible dimensions of a rectangular kitchen with an area of 91 square meters and a perimeter of 40 metres.

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