7
Annie thinks that she has accurately measured
and labelled the diameter of the circle.
a) Is Annie correct?
Explain your answer.
b) Is the diameter greater or less than 3 cm?
Explain how you know.
3 cm

when running a line, in a right-triangle, from the 90° angle perpendicular to its opposite side, we will end up with three similar triangles, one Small, one Medium and a containing Large one.  Check the picture below.

[tex]\cfrac{y}{5}=\cfrac{11}{y}\implies y^2=55\implies y=\sqrt{55}\implies y\approx 7.4 \\\\[-0.35em] ~\dotfill\\\\ \cfrac{x}{5}=\cfrac{16}{x}\implies x^2=80\implies x=\sqrt{80}\implies x\approx 8.9[/tex]

If 1.5 inches on the map represent a distance of 8.1 miles in reality, we can use the proportion:

1.5 inches / 1 inch = 8.1 miles / x miles

where x is the actual distance represented by 1 inch on the map, and is the scale of the map we are looking for.

We can solve for x by cross-multiplying:

1.5x = 8.1

x = 8.1 / 1.5

x = 5.4

Therefore, the scale of the map is 1 inch represents 5.4 miles.

Greg receives $1100, Nigel receives $825, and Mike receives $1375 from the sale of the boat.

Greg lost $550 and Mike lost $825.

Nigel made the smallest loss, which was $275.

a) Let's first find the total amount paid by the three of them:

$1400 + $1050 + $1750 = $4200

Since they share the money from the sale of the boat in the same ratio as they paid for it, we need to find the total of the ratio:

$1400 + $1050 + $1750 = $4200

$3300/$4200 = $11/14

Now we can find how much each of them receives from the sale of the boat by multiplying the total amount by their ratio:

Greg: ($11/14) x $1400 = $1100

Nigel: ($11/14) x $1050 = $825

Mike: ($11/14) x $1750 = $1375

Therefore, Greg receives $1100, Nigel receives $825, and Mike receives $1375 from the sale of the boat.

b) To find out how much more money Mike lost from the sale of the boat than Greg, we need to calculate their individual losses.

We know that the boat was sold for $3300, and that the total cost was $4200. Therefore, their total loss is:

$4200 - $3300 = $900

To calculate their individual losses, we need to use the same ratio as before:

Greg: ($11/14) x $4200 - $1400 = $550

Mike: ($11/14) x $4200 - $1750 = $825

Therefore, Greg lost $550 and Mike lost $825.

To find out how much more money Mike lost than Greg, we subtract Greg's loss from Mike's loss:

$825 - $550 = $275

Therefore, Mike lost $275 more than Greg from the sale of the boat.

c) To find out who made the smallest loss from the sale of the boat, we need to calculate their individual losses again:

Greg: ($11/14) x $4200 - $1400 = $550

Nigel: ($11/14) x $4200 - $1050 = $275

Mike: ($11/14) x $4200 - $1750 = $825

Therefore, Nigel made the smallest loss, which was $275.

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The chi-square distributions are skewed in general, and they are skewed to the right.

The maximum number of logically independent values, or values with the ability to fluctuate, in the data sample is referred to as degrees of freedom.

Degrees of freedom are frequently used in statistics to describe many types of hypothesis testing, such as the chi-square test.

Degrees of freedom are important for finding critical cutoff values for inferential statistical tests. Depending on the type of the analysis you run, degrees of freedom typically relate the size of the sample.

Thus, chi-square distributions is right-skewed.

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The unit normal vector to the surface at the point (1, 2, 1) is u = (1/√24)i + (2/√24)j + (1/√24)k.

To find a unit normal vector to the surface at the given point x^2 + y^2 + z^2 = 6 at (1, 2, 1), we need to normalize the gradient vector ∇f(x, y, z).

The gradient vector ∇f(x, y, z) of the given surface is:

∇f(x, y, z) = 2xi + 2yj + 2zk

Substituting the given point (1, 2, 1) into ∇f(x, y, z), we get:

∇f(1, 2, 1) = 2i + 4j + 2k

To normalize this vector, we first find its magnitude:

|∇f(1, 2, 1)| = √(2^2 + 4^2 + 2^2) = √24

Then, we divide each component of the vector by its magnitude to obtain the unit vector:

u = (1/√24)i + (2/√24)j + (1/√24)k

Therefore, the unit normal vector to the surface at the point (1, 2, 1) is u = (1/√24)i + (2/√24)j + (1/√24)k.

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To find the associated z-score for each standard normal area, you can use the Excel function =NORM.S.INV().

a. For the highest 10 percent, use the formula =NORM.S.INV(1-0.10), which gives a z-score of 1.2816.
b. For the lowest 50 percent, use the formula =NORM.S.INV(0.50), which gives a z-score of 0.0000.
c. For the highest 7 percent, use the formula =NORM.S.INV(1-0.07), which gives a z-score of 1.4758.


The Excel function =NORM.S.INV() calculates the inverse of the standard normal cumulative distribution function for a given probability. The function takes one argument, which is the probability of the area to the left of the desired z-score.

To find the z-score for the highest X percent, use the formula =NORM.S.INV(1-X). For the lowest X percent, simply input the probability X as the argument in the function. Finally, round the resulting z-scores to four decimal places.

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a. The expected number of major defects in a randomly selected gas stove of this type is 2.

b. The probability that there are at least 2 defects in a randomly selected stove is 0.75.

c. Var(X + 1) is 0.8.

a. The expected value of X is given by the formula:

E(X) = ∑ x*p(x) where the summation is taken over all possible values of X.

Using the given pmf, we have:

E(X) = 0(.10) + 1(.15) + 2(.45) + 3(.25) + 4(.05)

= 0 + 0.15 + 0.90 + 0.75 + 0.20

= 2

Therefore, the expected number of major defects in a randomly selected gas stove of this type is 2.

b. The probability that there are at least 2 defects in a randomly selected stove is given by the sum of the probabilities for X = 2, 3, and 4:

P(X ≥ 2) = P(X = 2) + P(X = 3) + P(X = 4)

= 0.45 + 0.25 + 0.05

= 0.75

Therefore, the probability that there are at least 2 defects in a randomly selected stove is 0.75.

c. To find Var(X + 1), we first need to find E(X + 1):

E(X + 1) = E(X) + E(1) = E(X) + 1 = 2 + 1 = 3

Now, using the formula for variance, [tex]Var(X + 1) = E[(X + 1 - E(X + 1))^2][/tex]:

[tex]Var(X + 1) = E[(X + 1 - 3)^2][/tex]

[tex]= E[(X - 2)^2][/tex]

= ∑[tex](x - 2)^2 * p(x)[/tex] where the summation is taken over all possible values of X.

Using the given pmf, we have:

Var(X + 1) = [tex](0 - 2)^2(.10) + (1 - 2)^2(.15) + (2 - 2)^2(.45) + (3 - 2)^2(.25) + (4 - 2)^2(.05)[/tex]

= 4(.10) + 1(.15) + 0(.45) + 1(.25) + 4(.05)

= 0.4 + 0.15 + 0.25

= 0.8

Therefore, Var(X + 1) is 0.8.

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The matrix A of the linear transformation is:

A = | -1/2 √3/2|

| -√3/2 -1/2 |

To find the matrix A of the linear transformation T that rotates any vector through an angle of 120 degrees in the clockwise direction, we can use the standard matrix formula for rotating a vector in two dimensions:

|cos θ - sin θ|

|sin θ cos θ|

where θ is the angle of rotation in radians. Since we want to rotate through an angle of 120 degrees clockwise, which is equivalent to -2π/3 radians, we can plug in θ = -2π/3 to get:

|cos(-2π/3) - sin(-2π/3)|

|sin(-2π/3) cos(-2π/3)|

We can simplify this using the trigonometric identities cos(-θ) = cos(θ) and sin(-θ) = -sin(θ), to get:

| -1/2 √3/2|

| -√3/2 -1/2 |

Therefore, the matrix A of the linear transformation T that rotates any vector through an angle of 120 degrees in the clockwise direction is:

A = | -1/2 √3/2|

| -√3/2 -1/2 |

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Step-by-step explanation:

2/3 = 66.666666666666.......

1/6= 16.66666666667

The value is less than 1, the excess pressure is decreased by a factor of approximately 0.526 when the radius is 1.9 times its original value.

The excess pressure inside a bubble is given by the equation ΔP = (2γ)/r, where ΔP is the excess pressure, γ is the surface tension, and r is the radius of the bubble. After finding the pi terms, we can rewrite this equation as ΔP = f(γ,r/R), where R is a reference length scale.
Using the Buckingham Pi Theorem, we can determine that the two pi terms for this problem are π1 = γ^(1/2) and π2 = r/R. Therefore, the excess pressure can be expressed as a function of these pi terms: ΔP = f(π1, π2).
To determine the variation in excess pressure if the radius r = 1.9R, we need to first calculate the value of the pi terms for this condition. Since γ and R are constants, we can see that π1 is also a constant. For π2, we have:
π2 = r/R = (1.9R)/R = 1.9
Therefore, the excess pressure for this condition can be calculated as:
ΔP = f(π1, π2) = f(γ^(1/2), 1.9)
Since we don't know the specific function relating excess pressure to the pi terms, we can't determine the exact value of ΔP. However, we can determine the factor by which it changes relative to the case where r = R:
ΔP(1.9R)/ΔP(R) = f(γ^(1/2), 1.9)/f(γ^(1/2), 1)
Without knowing the specific function, we can't calculate this ratio, but we can make some general observations. Since π1 is constant, we know that it won't affect the ratio. The only factor that changes is π2, which increases by a factor of 1.9. Since the excess pressure is inversely proportional to r, we would expect it to decrease as r increases. Therefore, we might expect the ratio to be less than 1 (i.e. the excess pressure decreases), but we can't say for certain without knowing the specific function.
Excess Pressure (P) = (4 * Surface Tension (S)) / Radius (r)
Now, let's compare the excess pressure in the given scenario (r = 1.9 r):
P1 = (4 * S) / r
P2 = (4 * S) / (1.9 * r)
Now, let's find the ratio P2/P1 to determine the factor by which the excess pressure is increased or decreased:
P2/P1 = ((4 * S) / (1.9 * r)) / ((4 * S) / r)
By simplifying, we get:
P2/P1 = r / (1.9 * r) = 1 / 1.9 ≈ 0.526
Since the value is less than 1, the excess pressure is decreased by a factor of approximately 0.526 when the radius is 1.9 times its original value.

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For 1+6x - 2x^3, the linear differential operator that annihilates the given function is  D^3 - 6D - 2. For e^-x + 2xe^x - x^2e^x, the linear differential operator that annihilates the given function is  D^2 - 3D + 2.

To find a linear differential operator that annihilates a given function, we need to apply the differential operator D to the function and obtain a linear combination of the function and its derivatives that is equal to zero. In other words, we need to find a differential operator of the form a_nD^n + a_(n-1)D^(n-1) + ... + a_1D + a_0, where a_n, a_(n-1), ..., a_0 are constants, that makes the given function vanish when applied to it.

In the first example, we apply D^3 - 6D - 2 to the function 1 + 6x - 2x^3, and obtain the expression 0, which means that this linear differential operator annihilates the given function.

In the second example, we apply D^2 - 3D + 2 to the function e^-x + 2xe^x - x^2e^x, and obtain the expression 0, which means that this linear differential operator annihilates the given function.

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The domain of f(x,y) = √x + √y is the set of all points (x,y) in the first quadrant of the xy-plane, where both x and y are greater than or equal to zero. So, correct option is E.

The domain of a function is the set of all possible values of the input variables for which the function is defined. In the case of the function f(x,y) = √x + √y, the values of x and y must be such that the expression under the square root is non-negative. This means that x and y must be greater than or equal to zero.

Therefore, the domain of f(x,y) is the set of all points (x,y) in the first quadrant, where both x and y are greater than or equal to zero. The first quadrant is the region of the xy-plane where x and y are both positive.

Therefore, the correct answer is (e) the first quadrant.

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Complete question is:

The domain of the function f(x,y) = √x + √y  is

a. all of the xy-plane

b. the area inside a parabola

c. the union of two intervals

d. the first and third quadrants

e. the first quadrant

The region in the plane for points with polar coordinates 2 < r < 3 and 5π/3 ≤ θ ≤ 7π/3.


Identify the range of r and θ: In this case, 2 < r < 3 and 5π/3 ≤ θ ≤ 7π/3. Draw the polar coordinate plane (with an origin, labeled "O") and radial lines representing the angles θ.Mark the angle 5π/3 on the plane, which is located in the fourth quadrant (5π/3 = 300°). Draw a radial line from the origin to represent this angle.

Mark the angle 7π/3 on the plane, which is also located in the fourth quadrant (7π/3 = 420°, but since 360° brings us back to the origin, it is equivalent to 60°). Draw a radial line from the origin to represent this angle.Draw two concentric circles around the origin with radii 2 and 3. These represent the range of r values.

Shade the region in the plane that is bounded by the radial lines, as well as the circles with radii 2 and 3. This is the region consisting of points whose polar coordinates satisfy the given conditions.

Your sketch should now show the region in the plane for points with polar coordinates 2 < r < 3 and 5π/3 ≤ θ ≤ 7π/3.

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Answer:

7. the solutions to the equation 2x^2 + x - 21 = 0 are x = -7/2 and x = 3

To factor the quadratic equation, we need to find two numbers that multiply to -42 and add up to 1. These numbers are 7 and -6. We can write:

2x^2 + x - 21 = 0

2x^2 + 7x - 6x - 21 = 0

(2x^2 + 7x) - (6x + 21) = 0

x(2x + 7) - 3(2x + 7) = 0

(2x + 7)(x - 3) = 0

Setting each factor equal to zero gives:

2x + 7 = 0 or x - 3 = 0

Solving for x, we get:

2x = -7,

x = -7/2 or x = 3

8. It took the receiver about 1.1 seconds to catch the ball.

Rearrange the equation to standard form, which is in the form of ax^2 + bx + c = 0:

-4.9t^2 + 7.5t - 0.3 = 0

Now we can use the quadratic formula to solve for t:

t = (-b ± sqrt(b^2 - 4ac)) / 2a

Substituting in the values, we get:

t = (-7.5 ± sqrt(7.5^2 - 4(-4.9)(-0.3))) / 2(-4.9)

t = (-7.5 ± sqrt(56.25 - 5.88)) / (-9.8)

t = (-7.5 ± sqrt(50.37)) / (-9.8)

t ≈ 1.1 seconds or t ≈ 0.06 seconds

Since the time can't be negative, we can discard the negative solution.

Answer: Number selected;  26

% score = 86.67%

Grade is a B

Step-by-step explanation:

Assuming a person selects a number from 21-29

Number seleted; 26

Say that person A gets 26/30 on his test

SN= selected number , 26

SN/Total score x 100%

26/30 x 100%

= 0.8667 x 100%

% score is 86.67%

The probability of the event occuring is 29/51

When we talk about probability, we are essentially trying to measure the likelihood of an event occurring. One way to do this is by using the ratio of the number of favorable outcomes to the total number of outcomes.

In this case, the odds for a certain event are given as 29 to 22. This means that for every 29 favorable outcomes, there are 22 unfavorable outcomes. So, the total number of outcomes can be calculated as the sum of the favorable and unfavorable outcomes, which is 29 + 22 = 51.

To calculate the probability of the event occurring, we need to divide the number of favorable outcomes by the total number of outcomes. In this case, the number of favorable outcomes is 29, so the probability can be calculated as:

Probability = 29 / 51

This can be simplified by dividing both the numerator and denominator by the greatest common factor, which is 1 in this case. So the final answer is:

Probability = 29/51

This means that if we were to repeat the event many times, we would expect the event to occur approximately 29/51 or 56.86% of the time.

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To create an array of 6 shape references named "shapes" and fill it with an assortment of shape, circle and square objects, 1) Define the base `Shape` class along with derived classes `Circle` and `Square`. 2) Create an array of `Shape` references named `shapes` with a size of 6. 3) Instantiate various `Circle` and `Square` objects and assign them to the elements in the `shapes` array.

To create an array of 6 shape references named "shapes" and fill it with an assortment of shape, circle and square objects, you can use the following code in a programming language like Java:
Shape[] shapes = new Shape[6]; // create an array of 6 shape references
shapes[0] = new Shape(); // add a generic shape object to the array
shapes[1] = new Circle(); // add a circle object to the array
shapes[2] = new Square(); // add a square object to the array
shapes[3] = new Circle(); // add another circle object to the array
shapes[4] = new Square(); // add another square object to the array
shapes[5] = new Circle(); // add a third circle object to the array
This code first declares an array of 6 shape references named "shapes" using the syntax `Shape[] shapes = new Shape[6];`. This creates an array that can hold 6 references to objects of type Shape.
Next, the code assigns different objects to each of the 6 array elements using the syntax `shapes[index] = new Object();`. In this case, we add a generic shape object to the first element, a circle object to the second, a square object to the third, another circle object to the fourth, another square object to the fifth, and a third circle object to the sixth.
By using this code, you have created an array of shape references and filled it with an assortment of circle and square objects.
To create an array of 6 shape references named `shapes` and fill it with an assortment of shape, circle, and square objects, you can follow these steps:
1. Define the base `Shape` class along with derived classes `Circle` and `Square`.
2. Create an array of `Shape` references named `shapes` with a size of 6.
3. Instantiate various `Circle` and `Square` objects and assign them to the elements in the `shapes` array.
Here's an example in Java:
```java
// Define the base Shape class
abstract class Shape {
   // Shape properties and methods
}
// Define the Circle class derived from Shape
class Circle extends Shape {
   // Circle properties and methods
}
// Define the Square class derived from Shape
class Square extends Shape {
   // Square properties and methods
}
public class Main {
   public static void main(String[] args) {
       // Create an array of 6 Shape references named shapes
       Shape[] shapes = new Shape[6];
       // Fill the array with an assortment of Circle and Square objects
       shapes[0] = new Circle();
       shapes[1] = new Square();
       shapes[2] = new Circle();
       shapes[3] = new Square();
       shapes[4] = new Circle();
       shapes[5] = new Square();
   }
}
```
This code creates an array of 6 shape references named `shapes` and fills it with an assortment of `Circle` and `Square` objects.

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1. There are a fixed number of trials

2. There are only two possible outcomes

3. The probability of success (being fully vaccinated) remains constant for each trial

4. The trials are independent

To verify that these samples satisfy the conditions of the binomial experiment, let's check each requirement in context:

1) There are a fixed number of trials: In this case, there are 14 trials for each sample, as the researcher took a random sample of 14 individuals from both Fairfax County and Lee County.

2) There are only two possible outcomes: In this study, the possible outcomes for each individual are either being fully vaccinated or not being fully vaccinated.

3) The probability of success (being fully vaccinated) remains constant for each trial: The probability of success in Fairfax County is 74%, and in Lee County, it is 40%. These probabilities remain constant for each individual sampled in their respective county.

4) The trials are independent: Each individual's vaccination status is independent of the others, as one person's vaccination status does not affect the probability of another person being vaccinated in the sample.

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The volume of the given solid over the indicated region of integration, f(x,y) = 2x + 2y - 5 and R = {(x,y): -4 ≤ x ≤ 1, 3 ≤ y ≤ 6}, is 63 units³.

To find the volume, we will integrate the function f(x,y) over the region R. Follow these steps:

1. Set up the double integral: ∬R (2x + 2y - 5) dA

2. Set up the limits of integration: ∫(from -4 to 1) ∫(from 3 to 6) (2x + 2y - 5) dy dx
3. Integrate with respect to y: ∫(from -4 to 1) [(2xy + 2y²/2 - 5y)] (evaluated from 3 to 6) dx
4. Simplify and evaluate: ∫(from -4 to 1) [6x + 18 - 5(6-3)] dx
5. Integrate with respect to x: [3x² + 18x - 15x] (evaluated from -4 to 1)
6. Simplify and evaluate: (3 + 18 - 15) - (-12 + 72 + 60)
7. Calculate the final result: 6 - (-30) = 36

The volume of the region is 63 units³.

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The solution is : The value is, f(x)·g(x) = x³ +x² -2x -2

Given that,

Let f(x)=x+1

and g(x)= x^2 – 2.

Substitute the function definitions and simplify.

 f(x)·g(x)

 = (x +1)(x² -2)

 = x(x² -2) +1(x² -2)

 = x³ -2x +x² -2

so, we get,

 f(x)·g(x) = x³ +x² -2x -2

The value is, f(x)·g(x) = x³ +x² -2x -2

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complete question:

Let f(x)=x+1 and g(x)= x^2 – 2. Find f (x)· g(x)

Answer:

(0;7)

Step-by-step explanation:

The graph will cross the y-axis when x is zero

Replace x with 0 in the given function:

[tex]7 \times {0}^{2} + 2 \times 0 + 7[/tex]

[tex]0 + 0 + 7 = 7[/tex]

x = 0,

y = 7

The series ₙ=₁Σ∞ (n²+n+6) / (n⁴ + n²) is convergent.

Comparison test:

Let Σuₙ and Σvₙ be the two series of positive real numbers and there is a natural number in such that uₙ ≤ vₙ for all n ≥ m, k being a fixed positive number. Then,

i) Σuₙ is convergent if Σvₙ is convergent.

ii) Σvₙ is divergent if Σuₙ is divergent.

Given, ₙ=₁Σ∞ ( 6ⁿ⁺¹) / (5ⁿ -6 )

Let  ₙ=₁Σ∞ vₙ  be the given series where vₙ = ( 6ⁿ⁺¹) / (5ⁿ -6 )

Now, we have  ( 6ⁿ⁺¹) / (5ⁿ -6 ) > (6/5)ⁿ , ∀ n ≥ 2.

uₙ < vₙ  , where uₙ = (6/5)ⁿ.

By geometric series Σ(6/5)ⁿ is divergent.

Therefor, the series diverges by the comparison test. Each term is greater than that of a divergent geometric series.

Comparison test(limit form)

Let Σuₙ and Σvₙ be two series of positive real numbers and [tex]\lim_{n \to \infty} \frac{u}{v} = l\\[/tex], where l is a non-zero finite number. Then the two series  Σuₙ and Σvₙ converge and diverge together.

Given, ₙ₋₁Σ∞ (n²+n+6) / (n⁴ + n²)

Let ₙ₋₁Σ∞ uₙ be the given series

Then, uₙ = (n²+n+6) / (n⁴ + n²)

Let, vₙ = 1/(n²)

Then, [tex]\lim_{n \to \infty} \frac{u}{v\\}[/tex] = (n²+n+6). n² / (n⁴ + n²)

= [tex]\lim_{n \to \infty} \\[/tex] [n⁴ (1 + 1/n + 6/n²)] / n⁴ (1 + 1/n²)

= 1 (non-zero finite number)

Since, Σ (1/n²) convergent by p-series[p = 2>1]

Therefore, By comparison test

Σuₙ is convergent.

Hence, the series ₙ=₁Σ∞ (n²+n+6) / (n⁴ + n²) is convergent.

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Complete question:

Determine whether the series converges or diverges.

ₙ=₁Σ^∞ ( 6ⁿ⁺¹) / (5ⁿ -6 )

Although part of your question is missing, you might be referring to this full question: " Suppose A,B,C and D are different points on the same line. How many segments have endpoints at these points?"

The number of segments that have end points at these points are 6.

The number of ways of selecting x objects from n unlike objects is given by combinations

ⁿCₓ = n! / ( ( n - x )! x! ), where

n = total number of objects

x = number of chosen objects from the set

A, B, C, and D are four different points on the same line.

By counting the number of ways to choose 2 points from the 4 points, we can calculate the number of line segments with endpoints at these points.

So, the combination is ⁴C₂.

⁴C₂ = (4!)/(2!(4-2)!)

      = (4 x 3)/(2 x 1) = 6

So, there are total of 6 segments with endpoints at A, B, C, and D.

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Answer: 1/3

Step-by-step explanation: The line goes two points up and six points right. 2/6 simplified is 1/3

Answer:

slope = [tex]\frac{1}{3}[/tex]

Step-by-step explanation:

calculate the slope m using the slope formula

m = [tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1} }[/tex]

with (x₁, y₁ ) = (- 3, 1) and (x₂, y₂ ) = (3, 3) ← 2 points on the line

m = [tex]\frac{3-1}{3-(-3)}[/tex] = [tex]\frac{2}{3+3}[/tex] = [tex]\frac{2}{6}[/tex] = [tex]\frac{1}{3}[/tex]

Answer:

Step-by-step explanation:

yeap, the midpoint is the center of it and half the distance between those two points is the radius.

[tex]~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-2}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{7}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 4 -2}{2}~~~ ,~~~ \cfrac{ 7 +5}{2} \right) \implies \left(\cfrac{ 2 }{2}~~~ ,~~~ \cfrac{ 12 }{2} \right)\implies \stackrel{ center }{(1~~,~~6)} \\\\[-0.35em] ~\dotfill[/tex]

[tex]~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-2}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{7})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{ diameter }{d}=\sqrt{(~~4 - (-2)~~)^2 + (~~7 - 5~~)^2} \implies d=\sqrt{(4 +2)^2 + (7 -5)^2} \\\\\\ d=\sqrt{( 6 )^2 + ( 2 )^2} \implies d=\sqrt{ 36 + 4 } \implies d=\sqrt{ 40 }~\hfill \stackrel{radius}{\cfrac{\sqrt{40}}{2}} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{1}{h}~~,~~\underset{6}{k})}\qquad \stackrel{radius}{\underset{\frac{\sqrt{40}}{2}}{r}} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ x - 1 ~~ )^2 ~~ + ~~ ( ~~ y-6 ~~ )^2~~ = ~~\left( \frac{\sqrt{40}}{2} \right)^2\implies \boxed{(x-1)^2+(y-6)^2=10}[/tex]

option D (4, 7, 10, 13, ...) is the number pattern that uses the rule "add 3."

What is arithmetic sequence?

An arithmetic sequence is a sequence of numbers in which each term after the first is found by adding a fixed constant number, called the common difference, to the preceding term.

A number pattern is a sequence of numbers that follows a certain rule or pattern. In this case, we are given four number patterns (A, B, C, and D) and asked to identify which one uses the rule "add 3."

Let's take a closer look at each of the patterns:

A: 2, 6, 18, 48, ...

To determine the rule for this sequence, we need to look at the relationship between each pair of consecutive numbers. We can see that each number in the sequence is obtained by multiplying the previous number by 3.

B: 3, 7, 11, 15,...

Similarly, to determine the rule for this sequence, we need to look at the relationship between each pair of consecutive numbers. We can see that each number in the sequence is obtained by adding 4 to the previous number.

C: 3, 9, 27, 54,...

To determine the rule for this sequence, we need to look at the relationship between each pair of consecutive numbers. We can see that each number in the sequence is obtained by multiplying the previous number by 3. However, this pattern starts with 3 instead of 2, so it is a variation of pattern A.

D: 4, 7, 10, 13,...

Finally, we come to pattern D. We can see that each number in the sequence is obtained by adding 3 to the previous number.

Therefore, option D (4, 7, 10, 13, ...) is the number pattern that uses the rule "add 3."

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The expression that describes the relationship is as follows:

D. The number of sheets of paper is 50 times the number of packages

The number of sheet of paper in x packages is described by the rule y = 50x.

Therefore, the verbal description that describes the relationship can be represented as follows:

y = 50x

where

Therefore, the number of sheets of paper is 50 times the number of packages describes the relationship.

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Answer:

To find a basis for W, we need to find a set of linearly independent matrices in W that span W.

Let's first rewrite the condition for A to be in W:

A = [a a - c

b 2a + c]

We can rewrite A in terms of a linear combination of matrices as follows:

A = a [1 1

0 2] + b [0 -1

1 0] + c [0 1

0 1]

Therefore, any matrix A in W can be written as a linear combination of the three matrices:

B1 = [1 1

0 2],

B2 = [0 -1

1 0],

B3 = [0 1

0 1]

We just need to check that these three matrices are linearly independent. To do this, we set up the equation

c1 B1 + c2 B2 + c3 B3 = 0

where c1, c2, c3 are scalars. This gives the system of linear equations

c1 = 0

c2 - c3 = 0

c1 + c2 + c3 = 0

The solution to this system is c1 = 0, c2 = c3, and any value for c2. This means that the three matrices are linearly independent, and hence they form a basis for W.

give thanks for more! welcome <3

Step-by-step explanation: