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

Answers

Answer 1

When a triangle is dilated by a scale factor, the area of the dilated triangle is equal to the scale factor squared times the area of the original triangle. The area of the dilated triangle is 828 cm².

In this case, the original triangle has an area of 23 cm². The triangle is dilated by a factor of 6, so the scale factor is 6.

To find the area of the dilated triangle, we use the formula:

Area of Dilated Triangle = (Scale Factor)^2 * Area of Original Triangle

Plugging in the values:

Area of Dilated Triangle = 6^2 * 23 cm²

                       = 36 * 23 cm²

                       = 828 cm²

Therefore, the area of the dilated triangle is 828 cm².

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Related Questions

Find the perimeter of 13.2 yd, 6.2 yd, 11yd

Answers

Answer: 900.24

Step-by-step explanation:

Perimeter=L•W•H

plssss help

answer it plssssssssss​

Answers

k= 28/5
yeah yeah yeah

Surface areas 98 Find the shaded area. Round to the nearest tenth if necessary. 22 mm 18 mm 9 mm​

Answers

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

How many different ways can you have 55¢ in change using only quarters, dimes and nickels?

A. 1
B. 5
C. 11
D. 15

Answers

Answer: 11 times

Step-by-step explanation:

Have a wonderful day!

C 11 times. Hope this helps

How do I write [tex]\sqrt[4]{5}[/tex] using rational exponets?

Answers

This the answer 5^1/4

Hope this help

Answer:

y=45u

Step-by-step explanation:

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

Help!! please. will mark brainstest​

Answers

Answer:(1,0)

Step-by-step explanation:

y=mx+b
y=4/2x+1
Or simplify
y=2x+1

Just help me please?!?!?!

Answers

Answer:

the last one

Step-by-step explanation:

someone please help!

Answers

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

How would I do this??

Answers

There on opposite sides of the transversal and there inside the two lines crossed by the transversal.

What is the area of this tile

Answers

Answer:

12 in^2

Step-by-step explanation:

Answer: 12in^2

Step-by-step explanation:

Find the value of k and x2
x^2+ 13x + k = 0, x1=-9

Answers

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].

Ms. Clark spent $89.85 on sewing kits that cost $5 each plus $4.85 tax on the total bill. How many kits did she buy?

Answers

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

What are equation models?

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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Exercise 1.5.9 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.

Answers

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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The roots of 3x2 + x = 14 are
1. imaginary
2. real,rational,equal
3.real,rational,unequal
4.real,irrational,unequal

Answers

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

What is Quadratic equation?

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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Which logarithmic equation correctly rewrites this exponential equation?

Answers

Answer:

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

Step-by-step explanation:

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

Identify each scatterplot below with an appropriate value of r.

Answers

Answer:

A would be the answer

Step-by-step and

The augmented matrix of a system of linear equations AX = B was reduced to upper-triangular form so that 2 1 0 1 2 [AB] 0 -1 31 0 0 mln where m and n are real numbers. State all values of m and/or n such that the following statements are true. (a) Matrix A is invertible. (b) The system AX = B has no solutions. (c) The system AX = B has an infinite number of solutions. (d) Columns of the augmented matrix (AB) are linearly independent. (e) The system AX = 0 has a unique solution. (f) At least one eigenvalue of the matrix A is zero. (g) Columns of the matrix A form a basis in R3.

Answers

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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How much money would produce $70 as simple interest at 3.5% per
annum?

Answers

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

Solve: 4x^2 = 32 thanks

Answers

Answer:

D.

Step-by-step explanation:

It is difficult to describe, but you just need to follow through the steps acorddingly.

A square ceiling has a diagonal of 23 ft. Shelton wants to put
molding around the perimeter of the ceiling. The molding is sold
by the foot
What is the minimum amount of molding he needs?

66 ft
l65 ft
17 ft
16 ft

Answers

66ft is the answer :))))

Answer:

66ft

Step-by-step explanation:

I took the quiz

Find the GCF of the monomials: 18x² and 21x²y
A)3x
B)3x²
C)3xy
D)3x²y
PLEASE HELP MEEE

Answers

It’s b because they both have x squared and are both divisible by 3

The list below shows the number of miles Chris rode his bike on each of nine consecutive days. 9, 3, 1, 4, 8, 2, 6, 8, 5
Read the questions and write your answer for each part. Make sure to label each part: Part A, Part B and Part C. Write your answers in complete sentences.
Part A: Create a box-and-whisker plot with the above data. You may use the snipping tool to use the number line shown, or you may use a different number line. Upload your box-and-whisker plot using the "insert" or "+" option.
Part B: How far does Chris need to ride on the 10th day to have a mean distance of 6 miles? Show or explain your work.
Part C: On the 10th day, if Chris rides 20 miles, how will this change the mean?

Answers

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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Let : R² R2 given by (r,0) = (r cos(0), r sin(0)), 0≤ r ≤ R, 0≤0 ≤ 2m (this is a disk of radius R centered at (0,0)). Compute ∫ fdx .

Answers

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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help please! i dont understand how i'm supposed to fill the table if i dont have all the information

Answers

Use the ticket price in the first column in the formula 400-7p to get the answer for the second column (number of tickets sold)

Then multiple the number to f tickets in the second column by the price in the first to determine the revenue.

Determine a series of transformations that would map Figure C onto Figure D plz help asap

Answers

Answer:

Rotation about 180*

Translation about 7 units to the right and 8 down

Step-by-step explanation:

I need help please

What is the area?

____ Square millimeters

Answers

the area is 210 square millimeters

On a coordinate grid, a scale drawing of a banner is shaped like a parallelogram with verticals at (-15,10), (0,-5), (30,-5), and (15,10. Each square on the grid represents 1 square inch. What is the area of the banner?

Answers

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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regression analysis was applied and the least squares regression line was found to be ŷ = 400 3x. what would the residual be for an observed value of (2, 402)?

Answers

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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Can someone please help!!!
Ill give brainliest!!

Answers

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

Jacob distributed a survey to his fellow students asking them how many hours they'd spent playing sports in the past day. He also asked them to rate their mood on a scale from 0 to 10, with 10 being the happiest. A line was fit to the data to model the relationship.

Which of these linear equations best describes the given model?

Choose 1 answer:

5x+1.5

1.5x+5

−1.5x+5


Based on this equation, estimate the mood rating for a student that spent 2.52, point, 5 hours playing sports.

Round your answer to the nearest hundredth.

Answers

Answer:8.75

Step-by-step explanation:

Answer: it’s B

Step-by-step explanation:

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