What is the equation of the line that passes through the point (4,7) and has a slope of 0?

The change in the scale factor from the old scale to the new scale is option A; 3 to 1.

The ratio used to depict the relationship between the dimensions of a model or scaled figure and the corresponding dimensions of the real figure or object is called the scale. On the other hand, a scale factor is a value that is used to multiply all of an object's parts in order to produce an expanded or decreased figure.

Given, A scale drawing of a house addition shows a scale factor of 1 in. = 3.3 ft.

Since Josh decides to make the house addition smaller,

he changes the scale of the drawing to 1 in. = 1.1 ft.

Old: 1 inch = 3.3 feet.

New: 1 inch = 1.1 feet.

Old; 3.3 / 1.1 = 3.

New; 1.1 / 1.1 =  1

The change in the scale factor from the old scale to the new scale is option A; 3 to 1.

Learn more about scale here:

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The original purchase price of the property, after accounting for 4% annual appreciation over two years with a final selling price of $620,000, is approximately $573,000 when rounded to the nearest $1,000.

The question involves finding the original price of a property given its final selling price after a period of annual appreciation. To calculate the original purchase price, we need to use the formula for compound appreciation in reverse, known as discounting. Since the appreciation is 4% per year for two years and the final selling price is $620,000, the original purchase price (P) can be found using the formula:

P = Final Price / (1 + rate of appreciation)^number of years

So, substituting the given values, we get:

P = $620,000 / (1 + 0.04)^2

Now, calculate the denominator:

(1 + 0.04)^2 = 1.0816

And then divide by the final price:

P = $620,000 / 1.0816

P = $573,029.94

When rounded to the nearest $1,000, the original purchase price is approximately $573,000.

"The dimensions of the ice surface are approximately 25 yd in length and 29 yd in width.

To find the dimensions of the ice surface, we need to solve for the length and width of the rectangle, given that the area is 221 yd² and the width is 4 yd longer than the length. Let's denote the length of the rink as l and the width as w. We can then set up the following equations based on the given information:

1. The area of a rectangle is given by the product of its length and width, so we have:

[tex]\[ l \times w = 221 \][/tex]

 2. The width is 4 yd longer than the length, so we have:

[tex]\[ w = l + 4 \][/tex]

Now we can substitute the second equation into the first equation to express the area solely in terms of the length I:

[tex]\[ l \times (l + 4) = 221 \][/tex]

Expanding the equation, we get:

[tex]\[ l^2 + 4l = 221 \][/tex]

Rearranging the terms to set the equation to zero, we have a quadratic equation:

[tex]\[ l^2 + 4l - 221 = 0 \][/tex]

To solve this quadratic equation, we can factor it or use the quadratic formula. Factoring, we look for two numbers that multiply to -221 and add up to 4. These numbers are 13 and -17. So we can rewrite the equation as:

[tex]\[ (l + 17)(l - 13) = 0 \][/tex]

Setting each factor equal to zero gives us two possible solutions for l:

[tex]\[ l + 17 = 0 \quad \text{or} \quad l - 13 = 0 \] \[ l = -17 \quad \text{or} \quad l = 13 \][/tex]

Since a negative length does not make sense in this context, we discard l = -17 and take l = 13 yd as the length of the rink.

Now we can find the width w by adding 4 yd to the length:

[tex]\[ w = l + 4 \] \[ w = 13 + 4 \] \[ w = 17 \][/tex]

However, we made a mistake in the factoring process. The correct factors of 221 that add up to 4 are 17 and 13, not -17 and 13. The correct length should be 13 yd, and the width should be:

[tex]\[ w = l + 4 \] \[ w = 13 + 4 \] \[ w = 29 \][/tex]

Therefore, the correct dimensions of the ice surface are 13 yd in length and 29 yd in width.

Using the Triangle Inequality Theorem and the Pythagorean theorem, we can determine if a given set of measures can form a triangle, and if so, what type of triangle (acute, obtuse, or right) it is. We find that measures a) 4,7,9; b) 10,13,16; c) 8,8,11; d) 9,12,15 and f) 4.5,6,10.2 can all form triangles, but e) 5,14,20 cannot. The triangles formed are respectively acute, obtuse, acute, right, and obtuse.

In the discipline of Mathematics, specifically Geometry, we use the Triangle Inequality Theorem and the Pythagorean theorem to determine if given measures can constitute the sides of a triangle and also classify the triangle. The Triangle Inequality Theorem states that the sum of the lengths of any two sides of a triangle must be larger than the length of the third side. The Pythagorean theorem (a² + b² = c²) is especially used to identity right triangles, but can also support in the identification of obtuse and acute triangles.

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Answer:

The correct answer is

D

The Pythagorean theorem, usually applied in right-angled triangles, states the square of the hypotenuse equals the sum of the squares of the other two sides. This can be summarized by the equation: a² + b² = c². It is a fundamental principle in geometry.

The Pythagorean theorem is a mathematical principle that applies specifically to right-angled triangles. The theorem, credited to the ancient Greek philosopher Pythagoras, stipulates that the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This relationship can be represented by the equation: a² + b² = c², where c is the length of the hypotenuse, and a and b are the lengths of the other two sides.

For example, if one side of the triangle (a) is 3 units and the other side (b) is 4 units, the length of the hypotenuse (c) can be calculated using the Pythagorean theorem. The calculation would be set as follows: 3² + 4² = c². When solved, it results in c = √(3² + 4²) = √(9 + 16) = √25 = 5 units. Therefore, the length of the hypotenuse in this case is 5 units.

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To simplify the expression (4+2k)⋅6+3k, distribute 6 to both terms inside the parentheses and combine like terms.

To simplify the expression (4+2k)⋅6+3k, we need to distribute the 6 to both terms inside the parentheses:

(4⋅6) + (2k⋅6) + 3k

Simplifying further, we have:

24 + 12k + 3k

Combining like terms, we get:

24 + 15k

So, the simplified expression is 24 + 15k.

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For rolling the dice 200 times, the same logic applies. Regardless of the number of rolls, the game with the higher probability of winning, which is the second game, is preferred.

To determine which game is more favorable for rolling the dice, we need to calculate the expected value for each game. The expected value represents the average outcome of rolling the dice.

For the first game, the possible outcomes are integers between 1 and 6. The arithmetic mean should be between 3.25 and 3.75 to win a prize. So, we need to find the probability of rolling numbers that satisfy this condition.

Let's denote [tex]\( p_1 \)[/tex] as the probability of winning the first game. To find [tex]\( p_1 \)[/tex], we calculate the probability of rolling numbers between 13 and 18 (inclusive) since the sum of these numbers falls within the range of 3.25 to 3.75.

[tex]\[ p_1 = \frac{6}{6^2} = \frac{1}{6} \][/tex]

For the second game, we need the arithmetic mean to be more than 4.5. This means the sum of the numbers should be more than [tex]\( 4.5 \times 6 = 27 \)[/tex]. Since the maximum sum of rolling six dice is 36, all outcomes satisfy this condition.

Let's denote [tex]\( p_2 \)[/tex] as the probability of winning the second game. Since all outcomes are favorable, [tex]\( p_2 = 1 \)[/tex].

Now, we calculate the expected values for each game:

[tex]\[ E_1 = p_1 \times \text{Prize amount for game 1} \][/tex]

[tex]\[ E_2 = p_2 \times \text{Prize amount for game 2} \][/tex]

Given that the prize amount is the same for both games, we can compare the expected values directly.

For rolling the dice 20 times:

[tex]\[ E_1 = \frac{1}{6} \times \text{Prize amount} \][/tex]

[tex]\[ E_2 = 1 \times \text{Prize amount} \][/tex]

Since [tex]\( \frac{1}{6} \)[/tex] is less than 1, it's better to choose the game with the higher probability, which is the second game.

For rolling the dice 200 times, the same logic applies. Regardless of the number of rolls, the game with the higher probability of winning, which is the second game, is preferred.

The complete question is:

In a six-sided dice game, a prize is won if the arithmetic mean of number rolled is 3.25-3.75. In a second game, a prize is won if the arithmetic mean is more than 4.5. In which game would you rather roll the dice 20 times or 200 times

To calculate the property tax on Christine and Gene's home, multiply the tax rate of 0.0175 (converted from $17.50 per $1,000) by the assessed value of $230,000. The annual property tax on their home would be $4,025.

The student is asking how to calculate the property tax on Christine and Gene's home. To calculate this, you need to use the assessed value of the property and the property tax rate. Here is the step-by-step calculation:

The calculation is:

Property tax = 0.0175 * $230,000 = $4,025

Therefore, the annual property tax on their home is $4,025.

Answer:550(N)

Step-by-step explanation:

Answer:

550 N

Step-by-step explanation:

I got the answer right.

Answer:

[tex]x^5+2x^3+6x+\frac{1}{5}[/tex]

Step-by-step explanation:

We have been given an equation [tex]x^5+2x^3+6x+\frac{1}{5}[/tex]. We are asked to write our given equation in standard form.

We know that to write an equation in standard form, we need to write the degree terms in descending order.

Upon looking at our given equation, we can see that all terms are in descending order of degree, therefore, our given equation is already writen in standard form.

Final answer:

To find the combined capacity of the two water towers, you add their capacities together by summing the significant figures and keeping the exponent the same. The result is a combined water capacity of 17.13×105 gallons.

Explanation:

The question involves adding two large numbers that are expressed in scientific notation. To find the combined water capacity of the two water towers, we simply add the quantities together. The first tower holds 7.35×105 gallons, and the second tower holds 9.78×105 gallons.

To combine these:

Add the significant figures (the numbers before the exponent): 7.35 + 9.78.

Calculate this sum: 7.35 + 9.78 = 17.13.

Since both powers of ten are the same (105), you can keep the exponent as is.

Combine the significant figures with the exponent to get the final answer: 17.13×105 gallons.

This is the total combined water capacity of the two towers.