a rectangle with a length of 20 meters and a width of 11 meters is being dilated by a scale factor of 5. What is the length of the rectangle after the dilation?

Answer:

3/6

Step-by-step explanation:

1 2 and 3 are less than 4 so only 3 numbers

The first, second, and third quartiles from the set of data (3,4,5,6,7,37,100) include the following:

Q₁ = 4.

Q₂ = 6.

Q₃ = 37

Based on the data set, the first quartile (Q₁) can be calculated as follows;

Q₁ = [(n + 1)/4]th term

Q₁ = (7 + 1)/4

Q₁ = 2nd term

Q₁ = 4.

For the second quartile (Q₂), median, or 50th percentile, we have the following:

second quartile (Q₂) = 4th term

second quartile (Q₂) = 6.

For the third quartile (Q₃), we have:

Q₃ = [3(n + 1)/4]th term

Q₃ = 3 × 2nd term

Q₃ = 6th term

Q₃ = 37

In conclusion, a box plot for the given data set is shown in the image attached below.

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Answer:

At 2 pm, the temperature was -9 degrees. If the temperature drops 8 degrees, what is the new temperature?

Step-by-step explanation:

sorry I can't delete it

Answer:

-15

Step-by-step explanation:

-7 Dropping 8 would be -15 because 8+7 is 15 but its negative so its -15 (im terrible at explaining)

Answer:

22= 6(1x+5x)

Step-by-step explanation:

He could have 4 $5 bills and 2 $1 bills.

Answer:

he has 4 $5 bills

Step-by-step explanation:

22 - 2 = 20

20 divided by 4 = 5

so 4 $5 bills and 2 $1 bills

Euler's formula, v − e + f = 2, relates the number of vertices (v), the number of edges (e), and the number of faces (f) of a polyhedron.

Therefore, the correct option is (d) v = e - f + 2.

To solve Euler's formula for v, we'll have to isolate v on one side of the equation. The first step is to add e to both sides of the equation:

v − e + f + e = 2 + e

v + f = e + 2

Now subtract f from both sides of the equation:

v + f - f = e + 2 - f

v = e + 2 - f

Hence, the solution for Euler's formula for v is:

v = e + 2 - f

Therefore, the correct option is (d) v = e - f + 2.

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Answer:

Length = 3.84 feets (nearest hundredth)

Step-by-step explanation:

The length of section of wire can be obtyaiejd using the length of of a arc formular :

Length of arc = θ/360° * 2πr

Radius, r = 2.5 feets

Length of arc = (88/360) *2πr

Length of arc = (88/360) * 2π*2.5

Length of arc = 0.244444 * 15.707963

Length of arc = 3.8397

Length = 3.84 feets (nearest hundredth)

Answer:

Step-by-step explanation:

3.84

The derivative of f(x) = -x² + 4x - 9 is f'(x) = -2x + 4.  Evaluating f'(x) at x = 1, 2, and 3 gives f'(1) = 2, f'(2) = 0, and f'(3) = -2. To find the derivative of the function f(x) = -x² + 4x - 9, we will use the four-step process.

After applying the process, we obtain the derivative f'(x) = -2x + 4. Evaluating this derivative at x = 1, x = 2, and x = 3 gives us f'(1) = 2, f'(2) = 0, and f'(3) = -2.

The four-step process involves the following steps:

1. Begin by applying the power rule, which states that the derivative of [tex]x^n[/tex] is [tex]nx^{(n-1)[/tex], where n is a constant. In this case, we have -x², so the derivative becomes -2x.

2. Apply the power rule to the next term, which is 4x. The derivative of 4x is 4.

3. Since -9 is a constant term, its derivative is zero.

4. Combine the derivatives obtained in steps 1, 2, and 3 to find the overall derivative of the function f(x). In this case, f'(x) = -2x + 4.

To find the values of f'(1), f'(2), and f'(3), we substitute the corresponding values of x into the derivative function.

When x = 1, f'(1) = -2(1) + 4 = 2.

When x = 2, f'(2) = -2(2) + 4 = 0.

When x = 3, f'(3) = -2(3) + 4 = -2.

Therefore, the derivative of f(x) is f'(x) = -2x + 4, and the values of f'(1), f'(2), and f'(3) are 2, 0, and -2, respectively.

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Answer:

9

Step-by-step explanation:

Answer:

To find the volume of a rectangular prism, multiply its 3 dimensions: length x width x height. The volume is expressed in cubic units.

Answer:

x = 3

Step-by-step explanation:

Given that,

The mode of the data 2,3,3,5,7,7,6,5, x and 8 is 3.

We need to find the value of x.

We know that, Mode is the number in a data with Max frequency. x can be 3 or 7. If x = 7, mode becomes 7 and if x = 3, mode equals 3.

Hence, the value of x is equal to 3.

Answer:

[tex]48cm^{2}[/tex]

Step-by-step explanation:

Split the shape up into two rectangles

(1) rectangle would be 2 x 10 = 20

(2) rectangle would be 4 x 7 = 28

Add them up = [tex]48cm^{2}[/tex]

Answer:

4

Step-by-step explanation:

12(1/3)

so it is just 12/3 which is 4

Hope that helps :)

Answer:

52

Step-by-step explanation:

Its simple, its just division because she is asking how many for EACH(key word) so 1,248/24 equals 52 giving your answer.

   mark brainliesttt??

Answer:

Given sequence:

Common difference is:

Find the first term:

Recursive formula:

Correct choice is A

Answer:

25

Step-by-step explanation:

(y+2)^2=x

(-7+2)^2=x

(-5)^2=x

(-5)(-5)=x

25=x

Hope that helps :)

Answer:

x=25

Step-by-step explanation:

Plug it innnn plug it innnn

a) Bisection Method MATLAB code for equation [tex]x^3 + 4x^2 - 10 = 0[/tex] in the interval [0,5]:

function root = bisection_method()

   f = [tex]x^3 + 4*x^2 - 10[/tex];

   a = 0;

   b = 5;

   tol = 1e - 6;

   while (b - a) > tol

       c = (a + b) / 2;

       if f(c) == 0

           break;

       elseif f(a) * f(c) < 0

           b = c;

       else

           a = c;

       end

   end

   root = (a + b) / 2;

end

b) Bisection Method MATLAB code for equation [tex]x^3 - 6x^2 + 10x - 4 = 0[/tex] in the interval [0,4]:

function root = bisection_method()

   f = [tex]x^3 - 6*x^2 + 10*x - 4[/tex];

   a = 0;

   b = 4;

   tol = 1e-6;

   while (b - a) > tol

       c = (a + b) / 2;

       if f(c) == 0

           break;

       elseif f(a) * f(c) < 0

           b = c;

       else

           a = c;

       end

   end

   root = (a + b) / 2;

end

c) Newton's Method MATLAB code for equation [tex]x^3 + 3x + 1 = 0[/tex] in the interval (-2,0]:

function root = newton_method()

   f = [tex]x^3 + 3*x + 1[/tex];

   df =  [tex]3*x^2 + 3[/tex];

   [tex]x_0[/tex] = -1;

   tol = 1e-6;

   while abs(f([tex]x_0[/tex])) > tol

       [tex]x_0 = x_0 - f(x_0) / df(x_0)[/tex];

   end

   root = [tex]x_0[/tex];

end

d) Fixed-Point Method MATLAB code for equation [tex]x^3 - 2x - 1 = 0[/tex] in the interval (1.5,2]:

function root = fixed_point_method()

   g = [tex](x^3 - 1) / 2[/tex];

   [tex]x_0 = 2[/tex];

   tol = 1e-6;

   while abs([tex]g(x_0) - x_0[/tex]) > tol

       [tex]x_0 = g(x_0)[/tex];

   end

   root = [tex]x_0[/tex];

end

e) Secant Method MATLAB code for equation 1 - 2*exp(-x) - sin(x) = 0 in the interval (0,4]:

function root = secant_method()

   f = 1 - 2*exp(-x) - sin(x);

   [tex]x_0[/tex] = 0;

   [tex]x_1[/tex] = 1;

   tol = 1e-6;

   while abs(f([tex]x_1[/tex])) > tol

       [tex]x_2 = x_1 - f(x_1) * (x_1 - x_0) / (f(x_1) - f(x_0))[/tex];

       [tex]x_0 = x_1[/tex];

       [tex]x_1 = x_2[/tex];

   end

   root = [tex]x_1[/tex];

end

f) Secant Method MATLAB code for equation [tex]2 - x^3 + 4*x^2 - 10 = 0[/tex] in the interval [0,4]:

function root = secant_method()

   f = [tex]2 - x^3 + 4*x^2 - 10[/tex];

   [tex]x_0 = 0[/tex];

   [tex]x_1 = 1[/tex];

   tol = 1e-6;

   while abs(f([tex]x_1[/tex])) > tol

       [tex]x_2 = x_1 - f(x_1) * (x_1 - x_0) / (f(x_1) - f(x_0))[/tex];

       [tex]x_0 = x_1[/tex];

       [tex]x_1 = x_2[/tex];

   end

   root = [tex]x_1[/tex];

end

MATLAB provides several numerical methods for solving equations. In this case, we have used the Bisection Method, Newton's Method, Fixed-Point Method, and Secant Method to solve different equations.

The Bisection Method starts with an interval and iteratively narrows it down until the root is found within a specified tolerance. It relies on the intermediate value theorem.

Newton's Method, also known as Newton-Raphson Method, approximates the root using the tangent line at an initial guess. It iteratively refines the guess until the desired accuracy is achieved.

The Fixed-Point Method transforms the equation into an equivalent fixed-point iteration form. It repeatedly applies a function to an initial guess until convergence.

The Secant Method is a modification of the Newton's Method that uses a numerical approximation of the derivative. It does not require the derivative function explicitly.

By implementing these methods in MATLAB, we can numerically solve various equations and find their roots within specified intervals.

These numerical methods are powerful tools for solving equations when analytical solutions are not feasible or not known.

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Solve the inequality 8 + 4.5sin(πt) ≥ 5 to find the times when it is safe for Kairvi to sail into Matheshan's Cove.

To determine the times when it will be safe for Kairvi to sail into Matheshan's Cove, we need to solve the inequality 8 + 4.5sin(πt) ≥ 5, where t represents time in hours.

Here are the steps to solve the inequality:

Subtract 8 from both sides of the inequality:

  4.5sin(πt) ≥ 5 - 8

  4.5sin(πt) ≥ -3

Divide both sides of the inequality by 4.5 to isolate the sine function:

  sin(πt) ≥ -3/4.5

  sin(πt) ≥ -2/3

To find the values of t that satisfy this inequality, we need to consider the inverse sine (arcsine) function. Taking the inverse sine of both sides, we get:

  πt ≥ arcsin(-2/3)

Solve for t by dividing both sides by π:

  t ≥ (1/π)arcsin(-2/3)

The resulting expression (1/π)arcsin(-2/3) represents the minimum time at which the water depth will be at least 5 meters. To determine the specific times, you can use a calculator or reference table to evaluate the arcsin function and calculate the corresponding time values.

Note: Keep in mind that this solution assumes a 24-hour time format and that the given water depth equation and conditions are accurate.

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108 inches

hope this helped <3

Answer:

A. Survey

Step-by-step explanation:

Answer:

Option A. Survey

Step-by-step explanation:

What is survey?

Survey is defined as the act of examining a process or questioning a selected sample of individuals to obtain data about a service, product, or process.

Nick collected samples and then concluded his answer which is a survey he conducted.

Correct answer is Option A.

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Real-World Application: Image Compression: Image compression is a fundamental concept in the field of computer graphics and image processing.

Linear algebra plays a crucial role in various image compression techniques. Let's consider a complete solution for image compression using concepts of linear algebra.

Linear algebra concepts, such as matrix operations and transformations, are fundamental to every step of the image compression process outlined above. By applying these techniques, we can achieve efficient storage and transmission of images while balancing the trade-off between image quality and compression ratio.

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The expected number of red marbles is 197/240.

Given that a box contains 3 red and 7 green marbles. If 9 marbles are drawn without replacement, we need to determine the expected number of red marbles.

The total number of marbles in the box = 3 + 7 = 10The probability of selecting a red marble at the first draw = 3/10

The probability of selecting a red marble at the second draw = 2/9

The probability of selecting a red marble at the third draw = 1/8

∴ The expected number of red marbles = Probability of selecting a red marble at the first draw + Probability of selecting a red marble at the second draw + Probability of selecting a red marble at the third draw= 3/10 + 2/9 + 1/8= (216 + 240 + 135) / (10 * 9 * 8)= 591/720= 197/240

Hence, the expected number of red marbles is 197/240.

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The given information is: A box contains 3 red and 7 green marbles.  If 9 marbles are drawn without replacement, what is the expected number of red marbles?The expected number of red marbles can be found as follows:

Let us assume that X is a random variable, that represents the number of red marbles drawn from the box.

If we want to calculate the expected value of X, then we can use the formula:E(X) = Σ(xP(x))Where Σ(xP(x)) is the sum of all possible outcomes, multiplied by their respective probabilities.

The probability of drawing a red marble on the first draw is 3/10.

The probability of drawing a red marble on the second draw is 2/9, since there are only 2 red marbles left out of 9 marbles total.

The probability of drawing a red marble on the third draw is 1/8, since there are only 1 red marble left out of 8 marbles total.

Since we are drawing 9 marbles without replacement, we need to multiply all these probabilities together to get the probability of drawing a certain sequence of red and green marbles.

P(X = k) represents the probability of drawing k red marbles out of 9 marbles.Using the formula: E(X) = Σ(xP(x))E(X) = Σ(kP(X = k))k ranges from 0 to 3.

So, the expected value of red marbles is:E(X) = 0P(X = 0) + 1P(X = 1) + 2P(X = 2) + 3P(X = 3)E(X) = 0 + 1(3/10)(6/9)(5/8) + 2(3/10)(2/9)(5/8) + 3(3/10)(2/9)(1/8)E(X) = 0 + 0.27917 + 0.0625 + 0.01406E(X) = 0.35573

Therefore, the expected number of red marbles is approximately 0.356.

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Answer:

dunno.

Step-by-step explanation:

duno

Answer:

SA = 48 cm²

Step-by-step explanation:

SA = (3x12) + (4x3) = 48 cm²

Answer:

[tex]\frac{x+20}{x+4}[/tex]

Step-by-step explanation:

Can't cancel out terms like that

need to factor out the top and bottom

[tex]\frac{x^{2} +16x-80}{x^{2} -16}[/tex] = [tex]\frac{(x+20)(x-4)}{(x-4)(x+4)}[/tex]   now cancel out the (x-4) from the top and bottom

= [tex]\frac{x+20}{x+4}[/tex]

Answer & Explanation:

Error: individual terms in an equation in a fraction cannot be directly canceled out.

Correction:

the easy way (solve the quadratic equation in the numerator and complete the square in the denominator):

(x^2 + 16x - 80) / (x^2 - 16)

(x+20)(x-4) / (x+4)(x-4)

x+20 / x+4

the complicated way (manipulate the exponents and common factors):

(x^2 + 16x - 80) / (x^2 - 16)

(x^2 - 4x + 20x - 80) / x^2 - 2^4

x(x^2-1 - 2^2)+4*5(x - 2^4-2) / (x - 2^2)(x + 2^2)

x(x-4)+20(x-4) / (x-4)(x+4)

x(x-4)+(2^2 (5))(x-4) / (x-4)(x+4)

(x-4)(x + 2^2 (5)) / (x-4)(x+4)

(x-4)(x+20) / (x-4)(x+4)

x+20 / x+4

Answer:

Ok first of all you start out splitting the shape into seperate rectangles, in this case 2. take those 2 rectangles and find the area of them. Say we start from the smaller rectangle. 2 x 4 = 8 so the area of the smaller rectangle is 8. Keep that in mind. Next we have to multiply 10 x 5 because that is the labeled number of meters and that equals 50. So add 50 + 8 = 58

Step-by-step explanation:

Your answer is 58 meters

Really hope i helped and have a wonderful day! :)

Answer:

58 m³

Step-by-step explanation:

To find the area of irregular shapes like this, it is easy to split the shape into 2 individual shapes, find the area of both, then add the areas to find the total.

Now, we'll split it into 2 rectangles with areas of:

2 x 4

10 x 5

To find area, you multiply the length of one side by the other. The first rectangle's area is (2 x 4) and the other rectangle's area is (10 x 5).

10 x 5 = 50

2 x 4 = 8

50 + 8 = 58

The area of the figure is 58 m³.

The frequency of oranges in the classroom can be calculated as follows:Frequency of oranges in the classroom = Total number of oranges / Total number of fruits= 16/42 = 0.3809 (rounded to 4 decimal places)Thus, the frequency of oranges in the classroom is approximately 0.3809.

To determine the frequency of oranges in the classroom, we need to calculate the proportion of students who brought oranges out of the total number of students.

First, let's calculate the total number of students:

Total students = Number of students with 2 apples + Number of students with 1 apple and 1 orange + Number of students with 2 oranges

Total students = 9 + 4 + 8 = 21

Next, let's calculate the number of students who brought oranges:

Number of students with oranges = Number of students with 1 apple and 1 orange + Number of students with 2 oranges

Number of students with oranges = 4 + 8 = 12

Finally, we can calculate the frequency of oranges:

Frequency of oranges = Number of students with oranges / Total students

Frequency of oranges = 12 / 21 ≈ 0.5714 (rounded to 4 decimal places)

Therefore, the frequency of oranges in the classroom is approximately 0.5714.

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To determine the frequency of oranges in the classroom, it is important to add up the total number of fruits brought to the classroom. The given information is as follows:

2 Apples: 9 students

1 Apple and 1 Orange: 4 students

2 Oranges: 8 students

Thus, the frequency of oranges in the classroom is approximately 0.2105.

Total number of fruits = 2(9) + 1(4) + 2(8)

= 18 + 4 + 16

= 38 fruit in total

So, total number fruits are 38.

Frequency of oranges = Number of oranges / Total number of fruits

There are 8 oranges brought to class, so the frequency of oranges is:

8 / 38 ≈ 0.2105 (rounded to 4 decimal places)

Hence, the frequency of oranges in the classroom is approximately 0.2105.

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Answer:

21 packs of gatorade would cost $36.88.

Step-by-step explanation:

Mathematically:

8.78 / 5 = 1.756

1.756 * 21 = 36.876 ~= $36.88

Answer:

16.76 inch³

Step-by-step explanation:

volume of cone is 1/3 πr²h

h=4

r=2

then 1/3* 22/7* 2*2 *4

=352/21

=16.76