A 90 degree counter-clockwise rotation is the same as what kind of clockwise rotation?
a. 270 degree clockwise rotation
b. 90 degree clockwise rotation
c. 180 degree clockwise rotation
d. 360 degree clockwise rotation

The type of galaxy matched to its description.

Elliptical galaxy: B) Forms a perfect sphere or an ellipse and is flattened to some degree

Galaxy: D) Is a collection of several billion stars and interstellar matter isolated in space

Lens galaxy: A) Has a large flattened core

Spiral galaxy: C) Has a central core from which curved arms spiral outward

Elliptical galaxy: An elliptical galaxy is a type of galaxy that typically forms a perfect sphere or an ellipse shape. It is characterized by its smooth and featureless appearance, lacking the distinct spiral arms seen in spiral galaxies. Elliptical galaxies often have a flattened shape due to their rotation and gravitational interactions with other galaxies.

Galaxy: A galaxy refers to a vast collection of stars, interstellar gas, dust, and dark matter, all held together by gravity. Galaxies come in various shapes and sizes, and they can contain billions or even trillions of stars. They are the building blocks of the universe and are distributed throughout the cosmos.

Lens galaxy: A lens galaxy is a type of galaxy that has a large flattened core. It gets its name from the gravitational lensing effect it produces. Gravitational lensing occurs when the gravitational field of the lens galaxy bends and distorts the light from objects behind it, creating a lens-like effect.

Spiral galaxy: A spiral galaxy is a type of galaxy that has a central core or bulge from which curved arms spiral outward. These arms are made up of stars, gas, and dust, and they give spiral galaxies their distinct appearance. Spiral galaxies often have a flattened disk shape with a central bulge and extended arms that can stretch out across the disk.

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To add fractions with unlike denominators, we need to find the least common multiple of the denominators. In this case, the least common multiple of 3 and 4 is 12. So, we must find equivalents of each fraction with a denominator of 12.

To add fractions with unlike denominators, we need to find the least common multiple of the denominators. In this case, the least common multiple of 3 and 4 is 12. So, we must find equivalents of each fraction with a denominator of 12.

Multiplying the numerator and denominator of 2/3 by 4, we get 8/12. On the other hand, multiplying the numerator and denominator of 1/4 by 3, we get 3/12.

Therefore, we can rewrite the sum as:

8/12 + 3/12

And adding the numerators, we get:

11/12

So, 2/3 + 1/4 = 11/12.

The final answer is a proper fraction, which means that the numerator is less than the denominator, and can be further simplified if necessary.

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The distance between the tip of the ladder to the ground or the value of 'x' is 13.27 ft.

We know that the ladder is leaning on the wall and thus it makes a right-angle triangle, where:

the hypotenuse(h) is the length of the ladder,

the base(b) is the distance between the foot of the ladder and the bottom of the wall,

and the height(x) is the distance between the tip of the ladder to the bottom of the wall which we need to find.

As the question is on right angled triangle we can use the Pythagoras theorem to find the value of 'x':

[tex]Height^2 + Base^2 = Hypotenuse^2\\x^2 + b^2 = h^2[/tex]

Now we know that h= 15ft, and b=7ft.

Substituting the values in the above equation we get :

[tex]x^2 + 7^2 = 15^2\\x^2 = 225 - 49\\x = \sqrt{176}\\x= 13.27 ft.[/tex]

Therefore the distance between the tip of the ladder to the ground or the value of 'x' is 13.27 ft.

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The length of the missing segment is given as follows:

? = 4.4.

The Pythagorean Theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the lengths of the other two sides.

The theorem is expressed as follows:

c² = a² + b².

In which:

The hypotenuse length for the right triangle is given as follows:

h² = 6.6² + 8.8²

[tex]h = \sqrt{6.6^2 + 8.8^2}[/tex]

h = 11.

The hypotenuse segment is divided into a radius of 6.6 plus the missing segment of ?, thus:

6.6 + ? = 11

? = 11 - 6.6

? = 4.4.

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A: An equation that represents the problem is 1.75x - 0.45 = 4.45. B: Solving the equation from Part A gives x = 2.8. C: The solution to the equation represents the number of pounds of apple bought by Casey.

Part A: Write an equation that represents the problem. Define any variables.

Let x represent the number of pounds of apples Casey bought. The cost of apples is $1.75 per pound, so the total cost before using the coupon would be 1.75x. After using the $0.45 coupon, her total was $4.45. The equation representing this situation is:

1.75x - 0.45 = 4.45

Part B: Solve the equation from Part A.

Now, let's solve the equation:

1.75x - 0.45 = 4.45

Add 0.45 to both sides:

1.75x = 4.90

Now, divide both sides by 1.75:

x = 4.90 / 1.75

x = 2.8

Part C: Explain what the solution to the equation represents

The solution, x = 2.8, represents that Casey bought 2.8 pounds of apples at the grocery store.

Note: The question is incomplete. The complete question probably is: Apples are on sale at a grocery store for $1.75 per pound. Casey bought apples and used a coupon for $0.45 off her purchase. Her total was $4.45. How many pounds of apples did Casey buy? Part A: Write an equation that represents the problem. Define any variables. Part B: Solve the equation from Part A. Show all work. Part C: Explain what the solution to the equation represents.

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37.5% of the students collected between 49 and 98 kilograms of newspapers for the recycling project. Percentage method can be used here.

To answer this question, we need to determine the number of students who collected between 49 and 98 kilograms of newspapers and then calculate what percentage of the total number of students that represents.

First, we need to gather the data and sort it into categories. We can create a frequency table with intervals of 10 kilograms:
Mass Range | Number of Students
   0-9 kg               3
   10-19 kg            5
  20-29 kg           7
  30-39 kg           4
  40-49 kg           6
  50-59 kg           8
  60-69 kg           5
  70-79 kg           2
  80-89 kg           1
  90-99 kg           2

To find the number of students who collected between 49 and 98 kilograms of newspapers, we need to add up the frequencies for the 50-59, 60-69, and 70-79 kg categories. That gives us a total of 15 students.
To calculate the percentage of students who collected between 49 and 98 kilograms of newspapers, we need to divide the number of students in that range by the total number of students and then multiply by 100. In this case, we have 15 students in the range and a total of 40 students overall, so:
15/40 * 100 = 37.5%.

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Answer:

They can make 5 complete teams of 8 children even after one child goes home.

Step-by-step explanation:

If there are 43 children and they want to make teams of 8, we can find out how many complete teams they can make by dividing the total number of children by the number of children per team:

This means that they can make 5 complete teams of 8 children, with 3 children left over.

However, since one child goes home, there are only 42 children left. We can repeat the division:

This means that they can make 5 complete teams of 8 children, with 2 children left over. Therefore, they can make 5 complete teams of 8 children even after one child goes home.

It will take 6 minutes for the water hose to fill the 2-gallon bucket at a rate of 1/3 gallon per minute.

To determine the time required to fill a 2-gallon bucket using a water hose that fills at a rate of 1/3 gallon per minute, you can use a simple calculation.

First, identify the fill rate of the hose, which is 1/3 gallon per minute. Now, consider the bucket's capacity, which is 2 gallons. To find out how many minutes it takes to fill the bucket, divide the total capacity of the bucket by the fill rate:

Time (minutes) = Bucket capacity (gallons) / Fill rate (gallons per minute)

In this case:

Time (minutes) = 2 gallons / (1/3 gallons per minute)

To solve this, you can multiply the numerator and denominator by the reciprocal of the fill rate:

Time (minutes) = 2 gallons * (3 minutes per gallon)

Time (minutes) = 6 minutes

So, it will take 6 minutes for the water hose to fill the 2-gallon bucket at a rate of 1/3 gallon per minute.

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Answer: (a) To find the total number of fish that enter the lake over the 5-hour period from midnight to 5 A.M., we need to integrate the rate of fish entering the lake over this time period:

Total number of fish = ∫0^5 E(t) dt

Using the given function for E(t), we get:

Total number of fish = ∫0^5 (20 + 15 sin(πt/6)) dt

Using integration rules, we can solve this:

Total number of fish = 20t - (90/π) cos(πt/6) | from 0 to 5

Total number of fish = (100 - (90/π) cos(5π/6)) - (0 - (90/π) cos(0))

Total number of fish ≈ 121

Therefore, approximately 121 fish enter the lake over the 5-hour period.

(b) To find the average number of fish that leave the lake per hour over the 5-hour period, we need to calculate the total number of fish that leave the lake over this time period and divide by 5:

Total number of fish leaving the lake = L(0) + L(1) + L(2) + L(3) + L(4) + L(5)

Total number of fish leaving the lake = (4 + 20.1(0)^2) + (4 + 20.1(1)^2) + (4 + 20.1(2)^2) + (4 + 20.1(3)^2) + (4 + 20.1(4)^2) + (4 + 20.1(5)^2)

Total number of fish leaving the lake ≈ 257.5

Average number of fish leaving the lake per hour = Total number of fish leaving the lake / 5

Average number of fish leaving the lake per hour ≈ 51.5

Therefore, approximately 51.5 fish leave the lake per hour on average over the 5-hour period.

(c) To find the time when the greatest number of fish are in the lake, we need to find the maximum value of the function N(t) = E(t) - L(t) over the interval 0 ≤ t ≤ 8. We can do this by taking the derivative of N(t) with respect to t and setting it equal to zero:

N'(t) = E'(t) - L'(t)

N'(t) = (15π/6)cos(πt/6) - 40.2t

Setting N'(t) = 0, we get:

(15π/6)cos(πt/6) - 40.2t = 0

Simplifying and solving for t gives:

t ≈ 2.78 or t ≈ 6.22

Since 0 ≤ t ≤ 8, the time when the greatest number of fish are in the lake is t ≈ 2.78 hours after midnight (approximately 2:47 A.M.) or t ≈ 6.22 hours after midnight (approximately 6:13 A.M.).

To justify this, we can use the second derivative test. Taking the second derivative of N(t) gives:

N''(t) = -(15π2/36)sin(πt/6) - 40.2

At t ≈ 2.78, N''(t) is negative, which means that N(t) has a local maximum at this point. Similarly, at t ≈ 6.22, N''(t) is positive, which also means that N(t) has a local maximum at this point. Therefore, these are the times when the greatest number of fish are in the lake.

(d) To determine if the rate of change in the number of fish in the lake is increasing or decreasing at 5 A.M. (t = 5), we need to find the sign of the second derivative of N(t) at t = 5. Taking the second derivative of N(t) gives:

N''(t) = -(15π2/36)sin(πt/6) - 40.2

Plugging in t = 5, we get:

N''(5) = -(15π2/36)sin(5π/6) - 40.2

Simplifying, we get:

N''(5) ≈ -60.5

Since N''(5) is negative, the rate of change in the number of fish in the lake is decreasing at 5 A.M. (t = 5). This means that the number of fish entering the lake is decreasing faster than the number of fish leaving the lake, so the total number of fish in the lake is decreasing.

(a) Approximately 131 fish enter the lake over the 5-hour period from midnight to 5 A.M.

(b) The average number of fish that leave the lake per hour over the same period is approximately 14.8.

(c) The greatest number of fish in the lake occurs at time t = 2.94 hours, or approximately 2 hours and 56 minutes past midnight.

(d) The rate of change in the number of fish in the lake is increasing at 5 A.M.

(a) To find the total number of fish that enter the lake over 5 hours, we need to integrate the function E(t) from t=0 to t=5:

∫[0,5] E(t) dt = ∫[0,5] (20 + 15 sin(πt/6)) dt

This evaluates to approximately 131 fish.

(b) The average number of fish that leave the lake per hour can be found by calculating the total number of fish that leave the lake over 5 hours and dividing by 5:

∫[0,5] L(t) dt = ∫[0,5] (4 + 20.1t^2) dt

This evaluates to approximately 74 fish, so the average number of fish that leave the lake per hour is approximately 14.8.

(c) To find the time at which the greatest number of fish is in the lake, we need to find the maximum of the function N(t) = ∫[0,t] E(x) dx - ∫[0,t] L(x) dx over the interval [0,8]. We can do this by finding the critical points of N(t) and evaluating N(t) at those points. The critical point is at t = 2.94 hours, and N(t) is increasing on either side of this point, so the greatest number of fish is in the lake at time t = 2.94 hours.

(d) The rate of change in the number of fish in the lake at 5 A.M. can be found by calculating the derivative of N(t) at t=5. The derivative is positive, so the rate of change in the number of fish is increasing at 5 A.M.

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Chris cannot build the triangular raised bed frame with the given lumber.

To determine if Chris can build her triangular raised bed frame, we need to check if the length of any one of the lumber pieces is greater than the sum of the other two. If this condition is not met, the pieces can be used to build the frame.

Let's check:

4 + 5 = 9 (no)

4 + 9 = 13 (no)

5 + 9 = 14 (yes)

Since the length of the 5-foot and 9-foot lumber pieces add up to be greater than the 4-foot piece, Chris can build her triangular raised bed frame. She can use the 4-foot and 5-foot pieces for the two shorter sides of the triangle and the 9-foot piece for the longer side.

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Answer: 16

Step-by-step explanation: to find diameter its radius x 2, so 2 x 8 is 16

a. d, days since first measurement y, acres burned since fire started

0 0, 1 4800, 2 9600, 3 14400, 4 19200, 5 24000. b. when d = -1, it tells that the area burned in the fire was 4780 acres one day before the area was first measured. When d = -3, y = 4680, this means that the area burned in the fire was 4680 acres three days before the area was first measured. c. The area burned a week before the fire measured 4800 acres was approximately 115.2 acres.

a. To complete the table, we need to plug in the values of d in the given equation and calculate the corresponding values of y.

d, days since first measurement

y, acres burned since fire started

0 0

1 4800

2 9600

3 14400

4 19200

5 24000

b. When d = -1, we have:

y = (4800)(24^(-1))^(1) = 4800 - 20 = 4780

This means that the area burned in the fire was 4780 acres one day before the area was first measured.

When d = -3, we have:

y = (4800)(24^(-3))^(1) = 4800 - 120 = 4680

This means that the area burned in the fire was 4680 acres three days before the area was first measured.

c. A week before the fire measured 4800 acres, the number of days since the fire started would be:

d = 4800 / (4800/24) = 24

Therefore, a week before the fire measured 4800 acres, the fire had been burning for 24 days. Plugging in this value in the given equation, we get:

y = (4800)(24^(-24/24))^(1) = 115.2

So, the area burned a week before the fire measured 4800 acres was approximately 115.2 acres.

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Considering the definition of probability, the probability that a randomly selected attraction at this amusement park would be a drink stand is 1/10.

Probability establishes a relationship between the number of favorable events and the total number of possible events.

The probability of any event A is defined as the ratio between the number of favorable cases (number of cases in which event A may or may not occur) and the total number of possible cases:

P(A)= number of favorable cases÷ number of possible cases

In this case, you know:

Replacing in the definition of probability:

P(A)= 2÷ 20

Solving:

P(A)= 1/10

Finally, the probability in this case is 1/10.

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Chris must work at least 36 hours to earn a minimum of $300, assuming she sells 15 books and earns the $2 bonus for each book sold.

The inequality that represents the situation is: 7.50h + 2(15) ≥ 300 where "h" represents the number of hours Chris works, and "2(15)" represents the bonus earned for selling 15 books.

The left-hand side of the inequality calculates Chris's total earnings, which is the product of her hourly wage of $7.50 and the number of hours worked, plus the bonus earned for selling 15 books.

The inequality states that the total earnings must be greater than or equal to $300, which is the minimum amount Chris wants to earn. To solve the inequality, we can simplify it by first multiplying 2 and 15 to get 30: 7.50h + 30 ≥ 300

We can isolate "h" by subtracting 30 from both sides: 7.50h ≥ 270. we can solve for "h" by dividing both sides by 7.50: h ≥ 36.

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To find the volume of a rectangle APR  prism, you need to multiply its length, width, and height. In this case, the length is 3 1/3 (or 10/3) units, the width is 4 2/3 (or 14/3) units, and the height is 25 units.

So, the volume of the rectangle can be calculated as:

Volume = length x width x height
Volume = (10/3) x (14/3) x 25
Volume = 1166.67 cubic units (rounded to two decimal places)

Therefore, the volume of the rectangle with a length of 3 1/3, a width of 4 2/3, and a height of 25 is approximately 1166.67 cubic units.

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A square pyramid with the maximum volume that can fit inside a cube has a same base as a cube ( 30 cm x 30 cm ) . The height of the pyramid is also same as a side length of a cube ( h = 30 cm ).

    The volume of the pyramid:

    V = 1/3 · 30² · 30 = 1/3 · 900 · 30 = 9,000 cm³

    The maximum volume of the pyramid is 9,000 cm³.

The y-values that a function approaches when the x-values are extremely large or extremely small is called the function's asymptotic behavior.

When we talk about the asymptotic behavior of a function, we are referring to what happens to the values of the function as the input (x-values) either tends to positive infinity or negative infinity.

In other words, we are interested in how the function behaves when the input values become extremely large or extremely small.

To understand asymptotic behavior, let's consider two types of asymptotes: horizontal and vertical asymptotes.

Horizontal Asymptotes:

A horizontal asymptote is a horizontal line that a function approaches as the x-values become extremely large or extremely small. We usually denote horizontal asymptotes as y = c, where c is a constant.

For example, let's consider the function f(x) = (2x^2 + 3) / (x^2 - 1). As x approaches positive or negative infinity, we can observe the following behavior:

As x becomes extremely large or extremely small, the function becomes closer and closer to the line y = 2. Therefore, we say that y = 2 is a horizontal asymptote for this function.

Vertical Asymptotes:

A vertical asymptote is a vertical line that the function approaches as the x-values approach a particular value. It typically occurs when there is a division by zero or when the function tends to infinity at a specific point.

For example, consider the function g(x) = 1 / (x - 2). As x approaches 2 from either side (but never equal to 2), we can observe the following behavior:

As x approaches 2 from the left (x < 2), the function g(x) becomes increasingly negative, tending towards negative infinity.

As x approaches 2 from the right (x > 2), the function g(x) becomes increasingly positive, tending towards positive infinity.

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In this context, the slope of the line of best fit, represented by the equation y = 1.25x + 7, represents the relationship between the age of an infant (in months) and its weight (in pounds) and the independent variable, x, represents the age of the infant in months, and the dependent variable, y, represents the weight of the infant in pounds.

The slope of the line, 1.25, indicates the rate at which the infant's weight changes with respect to its age. Specifically, it shows that for each additional month of age, the infant's weight is expected to increase by 1.25 pounds. This means that, on average, an infant gains 1.25 pounds per month.

In conclusion, the slope (1.25) in this context represents the average weight gain per month for an infant, based on the data collected by the doctor. It helps to understand the general association between an infant's age and its weight, and can be useful in predicting an infant's weight at a given age. However, it's important to remember that this is an average value and individual infants may have different weight gain patterns.

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The exponential term (87^t)^(1.75) increases, leading to an exponential growth in the number of people who receive the email.

The relationship between the elapsed time t, in days, since Bharat sent the letter and the number of people P(t) who receive the email is modeled by the following function:

P(t) = 2401 * (87^t)^(1.75)

In this function, t represents the number of days that have passed since Bharat sent the letter, and P(t) represents the number of people who receive the email at that time.

The function is an exponential growth model where the base is 87, and the exponent is t raised to the power of 1.75. The constant 2401 is a scaling factor that determines the initial number of people who receive the email at t=0.

As time passes, the exponential term (87^t)^(1.75) increases, leading to an exponential growth in the number of people who receive the email.

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[tex]\sf y =\dfrac{2}{5}x-4[/tex]

Step-by-step explanation:

        To find the equation of the required line, first we need to find the slope of the given line in the graph.

          Choose two points from the graph.

         (0 ,4)    x₁ = 0 & y₁ = 4

         (2,-1)     x₂ = 2 & y₂ = -1

[tex]\sf \boxed{\sf \bf Slope=\dfrac{y_2-y_1}{x_2-x_1}}[/tex]

           [tex]\sf = \dfrac{-1-4}{2-0}\\\\=\dfrac{-5}{2}[/tex]

       [tex]\sf m_1=\dfrac{-5}{2}[/tex]

[tex]\sf \text{Slope of the perpendicular line m = $\dfrac{-1}{m_1}$}[/tex]

                                                    [tex]\sf = -1 \div \dfrac{-5}{2}\\\\=-1 * \dfrac{-2}{5}\\\\=\dfrac{2}{5}[/tex]

[tex]\boxed{\sf slope \ intercept \ form \ : \ y = mx + b}[/tex]

Here, m is slope and b is y-intercept.

Substitute the m value in the above equation,

                    [tex]\sf y =\dfrac{2}{5}x + b[/tex]

    The line is passing through (5 , -2),

                     [tex]\sf -2 = \dfrac{2}{5}*5+b[/tex]

                     -2 =  2 + b

                -2 - 2 = b

                      b = -4

Slope-intercept form:

                [tex]\sf y = \dfrac{2}{5}x-4[/tex]

Answer:the two sides still equal.

Step-by-step explanation:To solve 2x + 3 = 5, Sylvia first subtracted 5 from both sides. What did she do wrong? Because Sylvia did the same operation to both sides, the equation is still correct: the two sides still equal.

Answer:

1

Step-by-step explanation:

or, 2x=5-3

or, 2x=2

or, x=2/2

x=1

Answer: Specifically, the pickup truck has lost about 56% of its value compared to the initial value, while the sedan has lost about 58% of its value.

Step-by-step explanation:

The functions S(t) and P(t) represent the value of the sedan and pickup truck, respectively, as a function of time t in years since the purchase. Let's analyze each function:

For S(t)=24,400(0.82)^t, the coefficient 24,400 represents the initial value or starting point of the function. This means that the value of the sedan at the time of purchase was $24,400.

The base 0.82 represents the rate of depreciation or decrease in value of the sedan over time. Specifically, the sedan's value decreases by 18% per year (100% - 82%).

For P(t)=35,900(0.71)^t/2, the coefficient 35,900 represents the initial value or starting point of the function.

This means that the value of the pickup truck at the time of purchase was $35,900. The base 0.71 represents the rate of depreciation or decrease in value of the pickup truck over time.

Specifically, the pickup truck's value decreases by approximately 29% every two years, since the exponent is divided by 2.

Comparing the two functions, we can see that the initial value of the pickup truck was higher than the initial value of the sedan.

However, the rate of depreciation of the pickup truck is greater than that of the sedan. This means that the pickup truck will lose its value at a faster rate than the sedan.

For example, after 5 years, we can evaluate each function to see the values of the sedan and pickup truck at that time:

S(5) = 24,400(0.82)^5 ≈ $10,373.67

P(5) = 35,900(0.71)^(5/2) ≈ $15,864.48

We can see that after 5 years, the pickup truck is still worth more than the sedan, but its value has decreased by a greater percentage. Specifically, the pickup truck has lost about 56% of its value compared to the initial value, while the sedan has lost about 58% of its value.

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Answer:

(-13)^3

Step-by-step explanation:

Exponents can be used for repeated multiplication.

In this case, the number "negative 13" is repeated several times, all connected with multiplication.

There are a total of three "negative 13"s being multiplied together ("negative 13" appears three times on the page).

To rewrite using exponents, we would write one of the following:

(-13)^3

[tex](-13)^3[/tex]

Since polygon A prime B prime C prime D prime above is above the image of polygon ABCD, the type of transformation shown is: D. vertical translation.

In Mathematics and Geometry, a transformation can be defined as the movement of a point from its initial position to a new location. This ultimately implies that, when a function or object is transformed, all of its points would also be transformed.

In Mathematics and Geometry, a horizontal translation to the right is modeled by this mathematical equation g(x) = f(x - N) while a vertical translation to the positive y-direction (upward) is modeled by this mathematical equation g(x) = f(x) + N.

Where:

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Answer:

Vertical translation

Step-by-step explanation:

I am in the middle of taking the quiz and belive this is the correct answer!

[tex]8 \frac{1}{3} \pi[/tex]

Step-by-step explanation:

volume of a sphere = 4/3 pi r²

r = 2.5

4/3× pi× 2.5² = 25/3pi

25/3 as a mixed number is 8 and 1/3

therefore rhe answer is 8 and 1/3 pi

Answer:

16.7 units

Step-by-step explanation:

its a 45°-45°-90° right triangle, so n1=4.9

r=4.9[tex]\sqrt{2}[/tex] =6.9

perimeter = 4.9+4.9+6.9=16.7 units

The probability that the outcome is all heads if at least one coin shows a head is 8/49.

To solve these problems, we'll use the basic principles of probability.

The probability of an event K (at least one head) can be calculated by subtracting the probability of the complement of K (no heads) from 1.

Since the coins can either show all heads or not, the complement of K is the event of no heads, which is denoted as T (tails for all coins). Therefore, we have:

P(K) = 1 - P(T)

Each coin toss is independent, and the probability of getting tails on a single toss is 1/2. Since there are three coins tossed independently, we multiply the probabilities together:

P(T) = ([tex]\frac{1}{2}[/tex]) * ([tex]\frac{1}{2}[/tex]) * ([tex]\frac{1}{2}[/tex]) = [tex]\frac{1}{8}[/tex]

Substituting this into the equation for P(K):

P(K) = 1 - P(T) = 1 - [tex]\frac{1}{8}[/tex] = [tex]\frac{7}{8}[/tex]

So, the probability of event K (at least one head) is [tex]\frac{7}{8}[/tex].

The probability that the outcome is all heads if at least one coin shows a head can be calculated using conditional probability. We want to find P(H | K), which represents the probability of event H (all heads) given event K (at least one head).

The formula for conditional probability is:

P(H | K) = [tex]\frac{P(H \∩ K) }{ P(K)}[/tex]

To find P(H∩K), we need to determine the probability of the intersection of events H and K (i.e., the probability of getting all heads and at least one head).

Since H is a subset of K (if all coins show heads, then at least one head is shown), we have:

P(H∩K) = P(H)

Therefore, P(H∩K) is the same as P(H). According to the problem, P(H) = [tex]\frac{1}{7}[/tex].

Now, substituting P(H∩K) = P(H) and P(K) = [tex]\frac{7}{8}[/tex] into the conditional probability formula:

P(H | K) = [tex]\frac{P(H\∩K) }{ P(K)}[/tex] = ([tex]\frac{1}{7}[/tex]) / ([tex]\frac{7}{8}[/tex]) = ([tex]\frac{1}{7}[/tex]) * ([tex]\frac{8}{7}[/tex]) = [tex]\frac{8}{49}[/tex]

So, the probability that the outcome is all heads if at least one coin shows a head is [tex]\frac{8}{49}[/tex].

To summarize:

P(K) = [tex]\frac{7}{8}[/tex]

P(H | K) = [tex]\frac{8}{49}[/tex]

P(H∩K) = [tex]\frac{1}{7}[/tex]

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The solution of the given system of equations is x = 3.2 and y = 1.5.

To solve this system of equations, we can use the method of elimination, where we eliminate one of the variables by adding or subtracting the equations.

First, let's eliminate y by multiplying the first equation by 2 and the second equation by 15:

12x/5 + 2y/15 = 4.6 (multiply the first equation by 2)

3x/2 - 10y = 18 (multiply the second equation by 15)

Now we can eliminate y by multiplying the first equation by 5 and adding it to the second equation:

12x + y/5 = 23 (multiply the first equation by 5 and simplify)

12x - y = 54 (subtract the second equation from the previous equation)

Adding the two equations, we get:

24x = 77

Therefore, x = 77/24.

Substituting x = 77/24 into the first equation, we get:

6(77/24)/5 + y/15 = 2.3

Simplifying this equation, we get:

y = 1.5

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The number that is equal to the place values, 7 hundred thousands 4 thousands 3 tens and 6 ones, is 704,036

From the question, we are to determine the number that is equal to the given place values

From the given information, the given place value is

7 hundred thousands 4 thousands 3 tens and 6 ones

Now, we will write each of the values in figures

7 hundred thousands = 700,000

4 thousands = 4,000

3 tens = 30

6 ones = 6

To determine the number that is equal to the place values, we will sum all the digits

700,000 + 4,000 + 30 + 6

704,036

Hence,

The number that is equal to the place value is 704,036

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