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

Answer 1

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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Related Questions

Which sequence of transformations produces R'ST' from RST?

-5

-4

-3

-2

-

1

2

3

4

5

X

a 90° clockwise rotation about the origin and then a translation 2 units left

a 90º counterclockwise rotation about the origin and then a translation 2 units right

a translation 2 units left and then a reflection over the y-axis

a translation 2 units right and then a reflection over the x-axis

Answers

Answer:

-3

Step-by-step explanation:

got it right on edg

The sequence of transformations that produces R'ST' from RST is reflection over the x axis and translation 2 units right.

What is transformation?

Transformation is the movement of a point from its initial location to a new location. Types of transformation are rotation, reflection, translation and dilation. Rotation, translation and reflection produce congruent images.

Triangle RST has vertices at R(0, 0), S(-2, 3) and T(-3, 1).

Triangle RST is reflected over the x axis to give R*(0, 0), S*(-2, -3) and T*(-3, -1). It is then translated 2 units right to get R'(2, 0), S'(0, -3) and T*(-1, -1).

The sequence of transformations that produces R'ST' from RST is reflection over the x axis and translation 2 units right.

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use the law of cosines to find the missing angle. find m∠a to the nearest tenth of a degree

Answers

The law of cosines can be used to find the missing angle when the lengths of three sides of a triangle are known. In this case, we want to find the measure of angle a, so we'll use the formula: cos(a) = (b² + c² - a²) / (2bc)where a is the side opposite angle a, b is the side opposite angle b, and c is the side opposite angle c.

To find angle a, we need to rearrange the formula: cos(a) = (b² + c² - a²) / (2bc)cos(a) = (7² + 13² - 10²) / (2 * 7 * 13)cos(a) = 0.81923077a = cos⁻¹(0.81923077)a ≈ 34.2°Therefore, m∠a is approximately 34.2 degrees.

When either the lengths of the two sides and the measure of the included angle (SAS) or the lengths of the three sides (SSS) are known, the Law of Cosines is used to determine the remaining parts of an oblique (non-right) triangle.

The square of a side of an oblique triangle is equal to the sum of the squares of the other two sides minus twice the product of the two sides if A, B, and C are the measures of the angles of the triangle and a, b, and c are the lengths of the sides opposite the corresponding angles.

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For a new study conducted by a fitness magazine, 260 females were randomly selected. For each, the mean daily calorie consumption was calculated for a September-February period. A second sample of 300 females was chosen independently of the first. For each of them, the mean daily calorie consumption was calculated for a March-August period. During the September-February period, participants consumed a mean of 2387.1 calories daily with a standard deviation of 210. During the March-August period, participants consumed a mean of 2412.9 calories daily with a standard deviation of 267.5. The population standard deviations of daily calories consumed for females in the two periods can be estimated using the sample standard deviations, as the samples that were used to compute them were quite large. Construct a 90% confidence interval for μ1−μ2, the difference between the mean daily calorie consumption of μ1 females in September-February and the mean daily calorie consumption of μ2 females in March-August.
Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. (If necessary, consult a list of formulas.)
What is the lower limit of the 90% confidence interval?
What is the upper limit of the 90% confidence interval?

Answers

the lower limit of the 90% confidence interval is approximately -65.25, and the upper limit is approximately 13.65.

To construct a 90% confidence interval for the difference between the mean daily calorie consumption of females in September-February (μ₁) and the mean daily calorie consumption of females in March-August (μ₂), we can use the formula:

CI = ([tex]\bar{X_1}[/tex] - [tex]\bar{X_2}[/tex]) ± Z * √((s₁² / n₁) + (s₂² / n₂))

Where:

[tex]\bar{X_1}[/tex] and [tex]\bar{X_2}[/tex] are the sample means of calorie consumption for the two periods.

s₁ and s₂ are the sample standard deviations of calorie consumption for the two periods.

n₁ and n₂ are the sample sizes for the two periods.

Z is the z-score corresponding to the desired confidence level.

Given data:

[tex]\bar{X_1}[/tex] = 2387.1 (mean daily calorie consumption for September-February)

[tex]\bar{X_2}[/tex] = 2412.9 (mean daily calorie consumption for March-August)

s₁ = 210 (standard deviation for September-February)

s₂ = 267.5 (standard deviation for March-August)

n₁ = 260 (sample size for September-February)

n₂ = 300 (sample size for March-August)

Confidence level = 90%

First, we need to find the z-score corresponding to a 90% confidence level. The z-score can be found using a z-table or a calculator. For a 90% confidence level, the z-score is approximately 1.645.

Now, we can substitute the values into the formula to calculate the confidence interval:

CI = (2387.1 - 2412.9) ± 1.645 * √((210² / 260) + (267.5² / 300))

Calculating the values inside the square root:

√((210² / 260) + (267.5² / 300)) ≈ √(342.461538 + 238.083333) ≈ √(580.544872) ≈ 24.107

Substituting the values into the formula:

CI = -25.8 ± 1.645 * 24.107

Calculating the limits of the confidence interval:

Lower limit = -25.8 - 1.645 * 24.107 ≈ -65.246

Upper limit = -25.8 + 1.645 * 24.107 ≈ 13.646

Rounding the values to two decimal places:

Lower limit ≈ -65.25

Upper limit ≈ 13.65

Therefore, the lower limit of the 90% confidence interval is approximately -65.25, and the upper limit is approximately 13.65.

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A silo is a composite of a cylindrical tower with a cone for a roof. What is the volume of the silo if the radius of the base is 40 feet, the height of the roof is 10 feet, and the height of the entire silo is 75 feet?

a. 393,746.3 ft
b. 343,480.8 ft
c. 326,725.6 ft
d. 376,991.1 ft

Answers

Answer:

The answer is b. 343,480.8ft

Step-by-step explanation:

I thought it was c at first because I forgot to add the volume of the cone.

The equation for the volume of a cylinder is V=π×r²×h

The solution would look like V=π40²×65=326725.8

The equation for the volume of a cone is V=1/3πr²×h

The solution would look like V=1/3π×40²×10=16755

Adding the two volumes would equal 326725.8±16755= 343480.8

a passive, first-order, high pass filter has the following transfer function: please answer questions 17 to 20 based on the above transfer function.

Answers

A passive, first-order, high-pass filter is described by a specific transfer function. Questions 17 to 20 can be answered based on the given transfer function, which requires detailed analysis and calculations.

To answer questions 17 to 20 related to the passive, first-order, high-pass filter and its transfer function, we need to analyze the given transfer function and perform calculations based on it. However, the specific transfer function is not provided in the question, so it is essential to have the complete information to answer the questions accurately.

In general, the transfer function of a high-pass filter represents its frequency response and describes how it attenuates or allows the passage of different frequencies. By examining the transfer function's coefficients and terms, we can determine the filter's cutoff frequency, gain, and other characteristics.

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look at screenshot for question

Answers

Answer:

I think it is the first one as you add the 4 like from the other

Choose the phrase that correctly completes the statement. If r(x) is a rational function in simplest form where the degree of the numerator is 1 and the degree of the denominator is 3, then Or(x) has a horizontal asymptote at y=0 Or(x) has no horizontal asymptote or(x) has a nonzero horizontal asymptote

Answers

To predict a linear regression score, you first need to train a linear regression model using a set of training data.

Once the model is trained, you can use it to make predictions on new data points. The predicted score will be based on the linear relationship between the input variables and the target variable,

A higher regression score indicates a better fit, while a lower score indicates a poorer fit.

To predict a linear regression score, follow these steps:

1. Gather your data: Collect the data p

points (x, y) for the variable you want to predict (y) based on the input variable (x).

2. Calculate the means: Find the mean of the x values (x) and the mean of the y values (y).

3. Calculate the slope (b1): Use the formula b1 = Σ[(xi - x)(yi - y)]  Σ(xi - x)^2, where xi and yi are the individual data points, and x and y are the means of x and y, respectively.

4. Calculate the intercept (b0): Use the formula b0 = y - b1 * x, where y is the mean of the y values and x is the mean of the x values.

5. Form the linear equation: The linear equation will be in the form y = b0 + b1 * x, where y is the predicted value, x is the input variable, and b0 and b1 are the intercept and slope, respectively.

6. Predict the linear regression score: Use the linear equation to predict the value of y for any given value of x by plugging in the x value into the equation. The resulting y value is your predicted linear regression score.

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6. How do you find the slope of a line?
A Divide the difference of the y values by the differences of the x values.
B Divide the difference of the x values by the differences of the y values.
C Divide the rise of the line by the run of the line.
D Both A and C.

Answers

Answer:

D

Step-by-step explanation:

A group of 12 students participated in a dance competition. Their scores are below:


Score (points)

1


2


3


4


5


Number of Students

1


2


4


3


2


Would a dot plot or a histogram best represent the data presented here? Why? (3 points)


Group of answer choices


Histogram, because a large number of scores are reported as ranges


Histogram, because a small number of scores are reported individually


Dot plot, because a large number of scores are reported as ranges

Answers

Answer:

put a picture please so i can understand better

Answer:

The answer is A

Step-by-step explanation:

i have done this and got 100 :0


Prove that in n , a single point {} is a closed
sets
please write down your answer in detail.

Answers

To prove that a single point {a} is a closed set in Rⁿ, we need to show that its complement, denoted as Rⁿ \ {a}, is open.

Let's consider an arbitrary point x in the complement Rⁿ \ {a}. Since x is not equal to a, there exists a positive radius r such that the open ball B(x, r) centered at x with radius r does not contain a.

Now, let's show that B(x, r) is entirely contained within Rⁿ\ {a}. We need to demonstrate that for any point y in B(x, r), y is also in Rⁿ \ {a}.

If y is equal to x, then y is not equal to a since a is excluded from Rⁿ \ {a}. Therefore, y is in Rⁿ\ {a}.

If y is not equal to x, then we can consider the distance between y and a. Since y is in B(x, r), we have:

d(y, x) < r

However, since a is not in B(x, r), we have:

d(a, x) ≥ r

Now, let's consider the distance between y and a:

d(y, a) ≤ d(y, x) + d(x, a) < r + (d(a, x) - r) = d(a, x)

Since d(y, a) is strictly less than d(a, x), it follows that y is not equal to a. Therefore, y is in Rⁿ \ {a}.

This shows that for every point x in Rⁿ \ {a}, there exists an open ball B(x, r) that is entirely contained within Rⁿ \ {a}. Hence, Rⁿ\ {a} is open.

By the definition of a closed set, if the complement of a set is open, then the set itself is closed. Therefore, a single point {a} is a closed set in Rⁿ.

The question should be:

Prove that in Rⁿ , a single point {a} is a closed sets.please write your answer in detail

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QUESTION 4
Factorise fully:
4.1.1
12x² – x - 6​

Answers

Answer:

y = (4x - 3)(3x + 2)

Step-by-step explanation:

Factorization is the process of taking like terms out of number such that one can rewrite an expression by its factors, or as the products of serval smaller expressions.

The first step to factoring an equation is to rewrite the linear term as the sum or difference of two other linear terms. One must do this in such as way that it shares a common factor with one of the other terms in the expression.

[tex]12x^2 -x - 6[/tex]

[tex]12x^2 -9x + 8x - 6[/tex]

Take out the common factor,

[tex]3x(4x-3)+2(4x-3)[/tex]

Factor,

[tex](3x+2)(4x-3)[/tex]

SQUARE ROOT OF 21316 WITH STEP BY STEP SOLUTION THANKS

Answers

Answer:

146

Step-by-step explanation:

[tex] = \sqrt{21316 } [/tex]

[tex] = \sqrt{4 \times 5329} [/tex]

[tex] = 2 \sqrt{5329} [/tex]

[tex] = 2 \sqrt{73 \times 73} [/tex]

[tex] = 2 \times 73[/tex]

[tex] = 146[/tex]

Evaluate the following integrals. a. if R is the rectangle R = [0,3] x [0,1/2). = [ xsinº y dA R b. if D is the region bounded by the y-axis, yox, and yu4 If y²er dA.

Answers

a. The value of the integral ∬R xsin(θ) dA over the rectangle R = [0,3] x [0,1/2) is (9/4) sin(θ).

b. The value of the integral ∬D y^2 dA over the region bounded by the y-axis, y = 0, and y = 4 is 64.

a. To evaluate the integral ∬R xsin(θ) dA over the rectangle R = [0,3] x [0,1/2), we can use Cartesian coordinates. The integral becomes:

∬R xsin(θ) dA = ∫[0,3] ∫[0,1/2] x sin(θ) dy dx

Integrating with respect to y first:

∫[0,3] ∫[0,1/2] x sin(θ) dy dx = ∫[0,3] [x sin(θ) y] [0,1/2] dx

                                = ∫[0,3] (x sin(θ) (1/2 - 0)) dx

                                = ∫[0,3] (x sin(θ)/2) dx

                                = (sin(θ)/2) ∫[0,3] x dx

                                = (sin(θ)/2) [x^2/2] [0,3]

                                = (sin(θ)/2) (9/2)

                                = (9/4) sin(θ)

b. To evaluate the integral ∬D y^2 dA over the region bounded by the y-axis, y = 0, and y = 4, we can use Cartesian coordinates. The integral becomes:

∬D y^2 dA = ∫[0,4] ∫[0,y] y^2 dx dy

Integrating with respect to x first:

∫[0,4] ∫[0,y] y^2 dx dy = ∫[0,4] [y^2 x] [0,y] dy

                       = ∫[0,4] (y^3 - 0) dy

                       = ∫[0,4] y^3 dy

                       = [y^4/4] [0,4]

                       = 4^4/4 - 0

                       = 64

Therefore, the value of the integral ∬D y^2 dA is 64.

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match the answer to the equation

Answers

1: no solution

2: infinitely many solutions

3: one solution

Norma builds a 1/8 scale Model of her own house.Her living room measures 12 feet by 28 feet. What are the dimensions of the models living room?

Answers

Answer:

1.5 feets by 3.5 feets

Step-by-step explanation:

Scale model = 1/8

Dimension of living room = 12 feets by 28 feets

The dimension of the model using the scale model given :

Scale model * dimension

1/8 * (12 feets by 28feets)

= 1.5 feets by 3.5 feets

A principal was buying T-shirts for his school's chess club and found that the total cost
in dollars could be found by the function f(x) = 9x + 19, where x is the number of
members in the club. If there are at least 8 members on the team but not more than
11, then which of the following statements describes the function?

A. The value of x must be a whole number between 8 and 11 and the value of f(x) must
be a whole number between 72 and 99.

B. The value of x must be a whole number between 8 and 11 and the value of f(x) must
be a whole number between 91 and 118.

C. The values of x and f(x) must both be whole numbers between 91 and 118.

D. The values of x and f(x) must both be whole numbers between 9 and 19.

Answers

I think the answer is C

(10-4)²(12+10/2)=_________

Answers

Answer:

should be 612

Step-by-step explanation:

Answer:

(10-4)²(12+10/2)=

(6)²(12+5)=

(36) (17) = 612

Evaluate (-2xy^3)^3 when x = 4 and y = -1

Answers

Step-by-step explanation:

(-2xy^3)^3

(-2(4)(-1)^3)^3

(-2(4)(-1))^3

(8)^3

512

​​​​If the unit rate is constant, what are the total sales for 12 pounds of asparagus?

Answers

How many r sold per day

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The dimensions of two pyramids formed of sand are shown. How much more sand is in the pyramid with the greater volume?

There are __ more cubic inches of sand in the pyramid with the greater volume.​

Answers

Answer:

V=75 in 3 9 in. 17 in. There are 1 more cubic inches of sand in the pyramid with the greater volume.

Step-by-step explanation:

As discussed in class (and in Hartmann Chapter 4) the exchange of heat and momentum between the atmosphere and the earth surface can be computed using the aerodynamic formula: T = = PcU? CPCU.(T. -T.) (5) SH = LE = LPCU.(9-9-) where UT, and q, are the wind speed, temperature and specific humidity respectively at a reference height (usually 10 m) above the earth surface. T. and q, are respectively the temperature and specific humidity at the surface, co is the aerodynamic exchange coefficient, and L is the latent heat of vaporization of water. (a) Compute the wind stress T, sensible heat flux (SH) and latent heat of evaporation (LE) from the land surface when T. = 30°C, T, = 28°C, 45 = 1.6 x 10-29, = 1.5 x 10-- and U. = 5 ms! Let's assume that the atmospheric boundary layer is unstable (as during the daytime when the land surface is warmer than the air above) so that co = 4 x 10-2. In each case state whether the direction of the flur is toward or away from the surface, and provide the appropriate units for t, SH and LE. (10 points) = = (6) Now repeat the calculations in (a) but over the ocean for T, = 28°C, T, = 30°C, 9. 1.6 x 10-29, = 1.5 x 10-2 and U, = 5 m st. In this case the atmospheric boundary layer is generally stable because the surface is cooler than the air above so co = 1 x 10- In each case state whether the direction of the flux is toward or away from the surface, and provide the appropriate units for T, SH and LE. (5 points) (C) State what kind of turbulence to you would expect to be dominant in the atmospheric boundary layer in each case. (4 points). If the value of a constant or parameter is not given, you will need to look it up in the textbook or online.

Answers

(a) T = 24 N/m² (toward surface), SH = 120 W/m² (away from surface), LE = 800 W/m² (away from surface).

(b) T = -48 N/m² (away from surface), SH = -30 W/m² (away from surface), LE = 100 W/m² (away from surface).

(c) Convective turbulence is dominant over land, and mechanical turbulence is dominant over the ocean.

To compute the wind stress (T), sensible heat flux (SH), and latent heat of evaporation (LE) from the land surface, we'll use the given equations and provided values:

(a) Land Surface Calculations:

T. = 30°C, T = 28°C, 45 = [tex]1.6 * 10^{-2}[/tex], Ω = 1.5 x [tex]10^{-2}[/tex], U = 5 m/s, and co = 4 x [tex]10^{-2}[/tex].

Using Equation (5) for T:

T = ρcU²(T. - T) = (1.2 kg/m³)(4 x 10^-2)(5 m/s)²(30°C - 28°C) = 24 N/m² (toward surface)

Using Equation (5) for SH:

SH = LPCU(T. - T) = ([tex]1.5 x 10^3[/tex]J/kg)([tex]4 * 10^{-2}[/tex])(30°C - 28°C) = 120 W/m² (away from surface)

Using Equation (5) for LE:

LE = LPCU(q*. - q) = (2.5 x [tex]10^6[/tex] J/kg)([tex]4 x 10^{-2}[/tex])(1.6 x [tex]10^{-2}[/tex] - 1.5 x [tex]10^{-2}[/tex]) = 800 W/m² (away from surface)

(b) Ocean Surface Calculations:

T. = 28°C, T = 30°C, 45 = 1.6 x [tex]10^{-2}[/tex], Ω = 1.5 x [tex]10^{-2}[/tex], U = 5 m/s, and co = 1 x [tex]10^{-2}[/tex].

Using Equation (5) for T:

T = ρcU²(T. - T) = (1.2 kg/m³)(1 x [tex]10^{-2}[/tex])(5 m/s)²(28°C - 30°C) = -48 N/m² (away from surface)

Using Equation (5) for SH:

SH = LPCU(T. - T) = (1.5 x [tex]10^3[/tex] J/kg)(1 x [tex]10^{-2}[/tex])(28°C - 30°C) = -30 W/m² (away from surface)

Using Equation (5) for LE:

LE = LPCU(q*. - q) = (2.5 x [tex]10^6[/tex] J/kg)(1 x [tex]10^{-2}[/tex])(1.6 x [tex]10^{-2}[/tex] - 1.5 x [tex]10^{-2}[/tex]) = 100 W/m² (away from surface)

(c) Dominant Turbulence:

In the unstable atmospheric boundary layer over land, convective turbulence is expected to be dominant due to the warmer surface heating the air above.

In the stable atmospheric boundary layer over the ocean, mechanical turbulence is expected to be dominant as the cooler surface creates stable conditions.

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A. 3
B. 10
C. 20
D. 30​

Answers

Answer:

D. 30

Explanation:

292/100×9.9 = 2.92×9.9 = 28.908

30 is closest to 28.908

Triangle ABC below has a right angle C and a height CD. (A height in a triangle connects its vertex and the opposite side, and is perpendicular to that side.)



If AC is 3 units long and BC is 4 units long, what is the length of AD?

Answers

can you show the picture please?

Find the surface area of the triangular prism. The base of the prism is an isosceles triangle. The answer is _ cm^2

Thanks in advance!

Answers

Answer:

3408cm²

Step-by-step explanation:

If the figure shown on the grid below is dilated by a scale factor of 2/3 with the center of dilation at (-4,4), what is the coordinate of point M after the dilation?

Answers

Step-by-step explanation:

To determine the coordinate of point M after the dilation, we need to apply the scale factor and center of dilation to the original coordinates.

Given:

Scale factor = 2/3

Center of dilation = (-4, 4)

Let's assume the coordinates of point M in the original figure are (x, y). To find the new coordinates after dilation, we can use the following formula:

New x-coordinate = Center of dilation x-coordinate + (Original x-coordinate - Center of dilation x-coordinate) * Scale factor

New y-coordinate = Center of dilation y-coordinate + (Original y-coordinate - Center of dilation y-coordinate) * Scale factor

Substituting the given values, we have:

New x-coordinate = (-4) + (x - (-4)) * (2/3)

New y-coordinate = 4 + (y - 4) * (2/3)

Since we are specifically looking for the coordinate of point M after dilation, we can substitute M's original coordinates into the formulas. Let's assume the original coordinates of point M are (xM, yM):

New x-coordinate = (-4) + (xM - (-4)) * (2/3)

New y-coordinate = 4 + (yM - 4) * (2/3)

Now we have the coordinates of point M after the dilation.

Please provide the values of xM and yM to calculate the specific coordinate of point M after the dilation.

Answer:

What does it mean to dilate by a scale factor of 3?

The key thing is that the dilation value affects the distance between two points. As in the first example (dilation by a factor of 3), A is originally 1 unit down from P and 2 units to the left of P. 1*3 = 3, so A' (the dilated point) should be 3 unit

Step-by-step explanation:

help me pls & thank you:)

Answers

Answer:the answer is 53 degrees

160 overall degrees 3 numerals divide 160 by 3 and it's 53.33 but it's degrees so round to the nearest tenth so it stays as 53 degrees

Step-by-step explanation:

Give the similarity ratio for the following figures (small to large).

Answers

Answer: I think the answer would be 1/6

Step-by-step explanation:

If you notice, all of the numbers on the left model are 1/6 less than the one on the left. I don't know what type of answer they are looking for but the one on the left is 1.6 smaller than the one on the right:)

Please help me I will mark...BRAINLIEST
I need volume and surface area

Answers

Answer:

1. V=60m³ SA=94m²

2. V=32m³ SA=88m²

3. V=2000mm³ SA=1000mm²

4. SA=376.8cm² V=552.64cm³

5. SA=1130.4cm² V=1356.48mm³

Step-by-step explanation:

1. Volume=5*4*3=60m³

Surface Area=2*(3*4)+2*(5*3)+2*(4*5)=24+30+40=94m²

2. Volume=8*4*1=32cm³

Surface Area=2*(8*1)+2*(8*4)+2*(4*1)=16+64+8=88cm²

3. Volume=10*10*20=2000mm³

Surface Area=2*(10*10)+2*(20*10)+2*(20*10)=200+400+400=1000mm²

4. SA of the cylinder=(2πr*h)+(2*πr²)=(2*(3.14)*4*11)+(2*3.14*16)

=(276.32)+(100.48)=376.8cm²

Volume=hπr²=11*(3.14)*16=552.64cm³

5. SA of cylinder=(2πr*h)+(2*πr²)=(2*(3.14)*12*3)+(2*3.14*144)

=(904.32)+(226.08)=1130.4mm²

Volume=hπr²=3*(3.14)*144=1356.48mm³

mr. and mrs. simpson went to two movies. the first movie lasted 2 ⅓ hours and the second one lasted 1 ⅘ hours. how much longer was the first than the second movie?

Answers

The first movie was ⅔ hours longer than the second one.

The difference in length between the two movies, we need to subtract the length of the second movie from the first.

we need to convert the mixed numbers to improper fractions to make the subtraction easier.

2 ⅓ hours can be written as 7/3 hours.

1 ⅘ hours can be written as 8/5 hours.

Now, we can subtract them

7/3 - 8/5

we need a common denominator. The least common multiple of 3 and 5 is 15.

(7/3) * (5/5) - (8/5) * (3/3) = 35/15 - 24/15 = 11/15

Therefore, the first movie was 11/15 hours (or approximately 44 minutes) longer than the second one

To learn more about hours click brainly.com/question/26345517

#SPJ11

6 and 186 its divison i dont know division
[tex]\sqrt{x}[/tex]

Answers

Answer:

31

Step-by-step explanation:

u divide 186 by 6. u divide by pretty much seeing how many times 6 can go into 186.

Answer:

31

Step-by-step explanation:

assuming your asking "what is 186 divided by 6?" ...right?

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