a triangle with an area of 23 cm² is dilated by a factor of 6. what is the area of the dilated triangle?

The total number of flops required to solve the system is approximately 2ns.

Let R be an n x n upper-triangular matrix with semiband width s. Show that the system Rx = y can be solved by back substitution in about 2ns flops. An analogous result holds for lower-triangular systems.Back substitution is an efficient technique for solving systems of linear equations in matrix form.

This is because back substitution only works on upper- or lower-triangular matrices, which have certain features that make solving systems of equations easier.The back substitution algorithm starts by solving the first equation of the system and obtaining a solution for the first variable. It then uses this value to solve the second equation and obtain a solution for the second variable.

This process is continued until all the variables are solved for.Let R be an n x n upper-triangular matrix with semiband width s. The semiband width of a matrix is the maximum number of nonzero entries in any row or column of the matrix. This means that all entries below the diagonal of R are zero. Let y be a vector of length n.

We want to solve the system Rx = y using back substitution.Since R is upper-triangular, we can solve for the last variable x_n first. This only requires one multiplication and one subtraction. We can then use the value of x_n to solve for the second-to-last variable x_{n-1}, which requires two multiplications and two subtractions.

Continuing in this way, we can solve for all the variables x_1, x_2, ..., x_n, each time requiring one more multiplication and subtraction than the previous step.In total, the number of flops required to solve the system Rx = y using back substitution is approximately 1 + 2 + 3 + ... + n, which is equal to n(n+1)/2.

Since R has semiband width s, this means that each row of R has at most s nonzero entries, so each variable requires at most s multiplications and s-1 subtractions.

Therefore, the total number of flops required to solve the system is approximately 2ns.

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Answer:

$2000

Step-by-step explanation:

Simple Interest = $70

Rate = 3.5%

Time = 1

Principal = ?

Simple Interest = (Principal × Rate × Time)/100

Principal = (Simple Interest × 100)/(Rate × Time)

Principal = (70 × 100)/(3.5 × 1)

Principal = 7000/3.5

Principal = 14000/7

Principal = 2000

Answer:

y=45u

Step-by-step explanation:

4 with a 5 try adding / bc its supposed to be like a check sin

The area of the banner is 562.5 square units.

To calculate the area of the banner, we can divide it into two triangles and then find the sum of their areas.

First, let's calculate the base and height of each triangle:

Triangle 1: Vertices (-15,10), (0,-5), and (30,-5)

The base of Triangle 1 is the distance between (-15,10) and (30,-5), which is 30 - (-15) = 45 units.

The height of Triangle 1 is the distance between (-15,10) and (0,-5), which is 10 - (-5) = 15 units.

Triangle 2: Vertices (0,-5), (30,-5), and (15,10)

The base of Triangle 2 is the distance between (0,-5) and (15,10), which is 15 units.

The height of Triangle 2 is the distance between (0,-5) and (30,-5), which is 30 - 0 = 30 units.

Now, let's calculate the area of each triangle using the formula for the area of a triangle: Area = (base * height) / 2.

Area of Triangle 1 = (45 units * 15 units) / 2 = 337.5 square units

Area of Triangle 2 = (15 units * 30 units) / 2 = 225 square units

Finally, to find the total area of the banner, we sum the areas of the two triangles:

Total Area = Area of Triangle 1 + Area of Triangle 2

Total Area = 337.5 square units + 225 square units

Total Area = 562.5 square units

Therefore, the area of the banner is 562.5 square units.

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Answer:(1,0)

Step-by-step explanation:

Given:

The quadratic equation is:

[tex]x^2+13x+k=0[/tex]

[tex]x_1=-9[/tex]

To find:

The value of k and [tex]x_1[/tex].

Solution:

We have,

[tex]x^2+13x+k=0[/tex]                ...(i)

Putting [tex]x=-9[/tex], we get

[tex](-9)^2+13(-9)+k=0[/tex]

[tex]81-117+k=0[/tex]

[tex]-36+k=0[/tex]

[tex]k=36[/tex]

Putting [tex]k=36[/tex] in (i), we get

[tex]x^2+13x+36=0[/tex]

Splitting the middle term, we get

[tex]x^2+9x+4x+36=0[/tex]

[tex]x(x+9)+4(x+9)=0[/tex]

[tex](x+9)(x+4)=0[/tex]

[tex]x=-9,-4[/tex]

Here, [tex]x_1=-9[/tex] and [tex]x_2=-4[/tex].

Therefore, the required values are [tex]k=36[/tex] and [tex]x_2=-4[/tex].

Answer:

297 sq mm

Step-by-step explanation:

Area of Rectangle: 22 x 18 = 396

Area of Triangle:  (1/2)(22)(9) = 99

Area of Rectangle - Area of Triangle = Area of Shaded area

396 - 99 = 297

Answer:8.75

Step-by-step explanation:

Answer: it’s B

Step-by-step explanation:

Answer:

Rotation about 180*

Translation about 7 units to the right and 8 down

Step-by-step explanation:

Answer:

66ft

Step-by-step explanation:

I took the quiz

Answer:

the last one

Step-by-step explanation:

A box-and-whisker plot needs to be created. To have a mean distance of 6 miles on the 10th day, Chris needs to ride a specific distance. If Chris rides 20 miles on the 10th day, it will change the mean distance.

Part A: To create a box-and-whisker plot, we need to arrange the given data in ascending order: 1, 2, 3, 4, 5, 6, 8, 8, 9. The plot will consist of a box representing the interquartile range (from the first quartile to the third quartile), a line within the box representing the median, and whiskers extending to the minimum and maximum values (excluding outliers if any).

Part B: To determine how far Chris needs to ride on the 10th day to have a mean distance of 6 miles, we need to consider the current total sum of distances and the total number of days. By calculating the difference between the desired mean and the current mean, we can determine the additional distance Chris needs to ride on the 10th day.

Part C: If Chris rides 20 miles on the 10th day, it will change the mean distance. The extent of the change in the mean depends on the initial data. To calculate the new mean, we need to include the additional distance (20 miles) and recalculate the mean using the updated total sum of distances and the total number of days.

Note: Without knowing the total number of days and the current sum of distances, precise calculations for Parts B and C cannot be provided.

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Answer:

Step-by-step explanation:

Answer:

12 in^2

Step-by-step explanation:

Answer: 12in^2

Step-by-step explanation:

The Residual for an observed value of (2, 402) is -4.

The regression analysis and the least squares regression line was found to be ŷ = 400 3x.

The observed value is (2, 402).To find the residual for an observed value of (2, 402),

we need to use the formula for residual

Residual = Observed value - Predicted value

where Observed value = (2, 402) , Predicted value = ŷ = 400 + 3x , Putting x = 2 in the above equation

we get,

ŷ = 400 + 3(2) = 406

Now, Residual = Observed value - Predicted value= 402 - 406= -4

Therefore, the residual for an observed value of (2, 402) is -4.

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Answer: 11 times

Step-by-step explanation:

Have a wonderful day!

Answer:

A. [tex] log_{8}(64) = x[/tex]

Step-by-step explanation:

[tex] {8}^{x} = 64 \\ \\ \implies log_{8}(64) = x[/tex]

The sewing kits bought by Ms, Clark is 17 in number.

The equation model is defined as the model of the given situation in the form of an equation using variables and constants.

Here,

As given in the question, Ms. Clark spent $89.85 on sewing kits that cost $5 each plus $4.85 tax on the total bill.

Let the number of sewing kits be x,
According to the question,
5x + 4.85 = 89.85
5x = 89.85 - 4.85
5x = 85
x = 17

Thus, the sewing kits bought by Ms, Clark is 17 in number.

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Answer:

please what is the exact question

Answer:

161.56 ft^2

Step-by-step explanation:

base area = (leg 1 x leg 2)/2 = (5 x 5)/2 = 25/2 = 12.5 ft^2

base perimeter = 5 + 5 + 7.07 = 17.07 ft

lateral surface = (perimeter x height) = 17.07 x 8 = 136.56 ft^2

surface area = base area x 2 + lateral surface = (12.5 x 2) + 136.56 = 161.56 ft^2

Answer:

3

Step-by-step explanation:

3x2 +x −14 = 0 12 −4(3)(−14) = 1+168 =169 = 132

The roots of 3x² + x = 14 are real, irrational and unequal

A quadratic equation is a second-order polynomial equation in a single variable x, ax² + bx +c=0 with a ≠ 0 .

Given equation is :

3x² + x = 14

3x² + x - 14=0

we have, a=3, b=1 c=-14

D= b²-4ac

  = 1²-4*3*(-14)

  = 1+168

  = 169

As, D>0

Hence, the roots are real.

Now,

x= -b±√b²-4ac/2a

 = -1±√169/2*3

 =-1±13/6

x= -1-13/6  and x= -1+13/6

x= -7/3  and x= 12/6=2

Hence, the roots are real, irrational and unequal.

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To compute the integral ∫ fdx over the disk D of radius R centered at (0,0), we need to express the function f in terms of the given coordinate transformation.

In polar coordinates, a point (r, θ) in the disk D can be represented as (r cos(θ), r sin(θ)).

Now, let's substitute these polar coordinates into the integral. The differential element dx becomes r cos(θ)dr, and the integral becomes:

∫ fdx = ∫ f(r cos(θ), r sin(θ)) r cos(θ)dr dθ

We can now evaluate this integral by integrating over the range of r and θ. The range for r is from 0 to R, and the range for θ is from 0 to 2π (since we are integrating over the entire disk).

Thus, the integral becomes:

∫ fdx = ∫[0 to R] ∫[0 to 2π] f(r cos(θ), r sin(θ)) r cos(θ)dr dθ

By evaluating this double integral, we can find the value of ∫ fdx over the given disk D.

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Answer: 900.24

Step-by-step explanation:

Perimeter=L•W•H

Answer:

12 units

Step-by-step explanation:

1st, I found the distance for the two parallel sides on the hexagon. I counted the lines to be 2 units each, which makes 4 units. And since all sides of a hexagon are equal. All the sides make 12 units, or centimeters. Therefore, the perimeter is 12 units

Answer:

A would be the answer

Step-by-step and

a. Matrix A is invertible when |A| = -m ≠ 0 then statement true.

b. The system AX = B has no solution when m = 0 and n ≠ 0 has a real number then statement true.

c. The system AX = B has an infinite number of solutions when m = n = 0 then statement true.

d. Columns of the augmented matrix (AB) are linearly independent when m ≠ 0 and n= 0 then statement true.

e. The system AX = 0 has a unique solution when m ≠ 0 then statement true.

f. At least one eigenvalue of the matrix A is zero when m = 0 then statement true.

g. Columns of the matrix A form a basis in R³ when m ≠ 0 then statement true.

Given that,

The augmented matrix of a system of linear equations AX = B was reduced to upper-triangular form so that

[A|B] = [tex]\left[\begin{array}{ccc}2&1&0 \ | \ 2\\0&-1&3 \ | \ 1 \\0&0&m \ | \ n\end{array}\right][/tex]

Where m and n are real numbers.

We know that,

a. We have to prove matrix A is invertible.

For A to be invertible.

|A| ≠ 0

|A| is the determinant of the matrix A.

|A| = 2(-m) -1(0) + 0(0) = -m

Here, m is the real number.

So, |A| = -m ≠ 0

Therefore, Matrix A is invertible when |A| = -m ≠ 0 then statement true.

b. We have to prove the system AX = B has no solution.

When Rank[A|B] > Rank[A]

m = 0 and n ≠ 0 has a real number

Therefore, The system AX = B has no solution when m = 0 and n ≠ 0 has a real number then statement true.

c. We have to prove the system AX = B has an infinite number of solutions.

When m = n = 0, and Rank[A] < 3

Therefore, The system AX = B has an infinite number of solutions when m = n = 0 then statement true.

d. We have to prove columns of the augmented matrix (AB) are linearly independent.

When m ≠ 0 and m∈R and n= 0

Therefore, Columns of the augmented matrix (AB) are linearly independent when m ≠ 0 and n= 0 then statement true.

e. We have to prove the system AX = 0 has a unique solution.

When [tex]\left[\begin{array}{ccc}2&1&0 \\0&-1&3 \\0&0&m \end{array}\right]\left[\begin{array}{ccc}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}0\\0\\0\end{array}\right][/tex]

The equation are 2x + y = 0, -y + 3z = 0 and mz = 0

m ≠ 0 should be any real number except zero.

Therefore, The system AX = 0 has a unique solution when m ≠ 0 then statement true.

f. We have to prove at least one eigenvalue of the matrix A is zero.

When λ = 2, 1, m

m = 0 then eigen value is zero

Therefore, At least one eigenvalue of the matrix A is zero when m = 0 then statement true.

g. We have to prove columns of the matrix A form a basis in R³.

When m ≠ 0

Therefore, Columns of the matrix A form a basis in R³ when m ≠ 0 then statement true.

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