Calculate the resistance of a piece of aluminum wire with a diameter of 100 mils and a length of two miles, at 68°F. Hint: Be sure to first convert mils to cmils and use the K value for aluminum found in the reference. (Round the FINAL answer to two decimal places.)

Answer:

Step-by-step explanation:

Let's model the cost by the following exponential function: c (t) = (7.50) * (1.035) ^ t Where, c (t): cost of the movie after t years. 7.50: initial cost of the movie in $ 1,035: annual percentage increase due to inflation. t: time in years. for t = 0 We have: c (t) = (7.50) * (1.035) ^ 0 c (t) = (7.50) * (1) c (t) = 7.50 Answer: The graph that best models the function is: GRAPH 2 (from left to right).

The difference between the outside perimeter of the walkway and the perimeter of the pool in algebraic expression is (2x + 14).

Pool:

Length = 2x - 1

Width = 12

The perimeter of the pool = 2(length + width)

= 2{(2x - 1) + 12}

= 2(2x - 1 + 12)

= 2(2x + 11)

= 4x + 22

Outside perimeter of the walkway:

Length = (2x - 1) + x

= 3x - 1

Width = 12 + 7

= 19

The perimeter of the walkway = 2(length + width)

= 2{(3x - 1) + 19}

= 2(3x - 1 + 19

= 2(3x + 18)

= 6x + 36

Difference between the outside perimeter of the walkway and the perimeter of the pool = The perimeter of the walkway - The perimeter of the pool

= (6x + 36) - (4x + 22)

= 6x + 36 - 4x - 22

= 2x + 14

Hence, (2x + 14) is the expression which represents the difference between the perimeter of the walkway and the perimeter of the pool.

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Answer:

It is 8

Step-by-step explanation:

The length of the base of the isosceles triangle is 18 inches.

Let the length of the base be 'b' inches. According to the given information, each of the congruent sides of the isosceles triangle is 8 inches less than twice the base. So, the length of each congruent side is (2b - 8) inches.

The perimeter of the triangle is the sum of all three sides, which is given as 74 inches. So, we can write the equation:

(2b - 8) + (2b - 8) + b = 74

Simplifying the equation, we get:

5b - 16 = 74

5b = 90

b = 18

Therefore, the length of the base is 18 inches.

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Answer:

Step-by-step explanation:

The answer is fifteen

Explanation:  6-3/0.4-0.2=3 over 0.2 which is equivalent to 15

Answer B: 15

Step-by-step explanation:

Answer:

The model that is most appropriate for the set is:

                Quadratic.

Step-by-step explanation:

Clearly after looking at the graph of the function which is made from the given set of points ; we observe that the graph is similar to a graph of the quadratic function whose vertex is at (0,-2) and also the line x=0 is the axis of symmetry of the graph.

Since, the graph is symmetric to this line.

Hence, the model is:

Quadratic.

Answer:

The answer is a. the real answer

Step-by-step explanation:

Solution: (1) The expression [tex](5\sqrt{3} )^x[/tex]  [tex]\text{ is written as }[/tex] [tex]5^x3^{\frac{x}{2}[/tex].

[tex](5\sqrt{3} )^x=5^x(\sqrt{3} )^x\\(5\sqrt{3} )^x=5^x(3^\frac{1}{2} )^x\\(5\sqrt{3} )^x=5^x3^\frac{x}{2}[/tex]

The value of u is [tex]\frac{x}{2}[/tex].

(2) The expression [tex](\frac{1}{2})^x-3[/tex] is written as [tex]2^{-x}-3[/tex].

[tex](\frac{1}{2})^x-3=(2^{-1})^x-3\\(\frac{1}{2})^x-3=2^{-x}-3[/tex]

The value of u is -x.

(3) The expression [tex]\frac{9}{3\sqrt{3} }[/tex] is written as [tex]\sqrt{3} (2^0)[/tex].

[tex]\frac{9}{3\sqrt{3} }=\frac{9}{3\sqrt{3} }(\frac{\sqrt{3}}{\sqrt{3}} )\\\frac{9}{3\sqrt{3} }=\frac{9(\sqrt{3}) }{3(3) }\\\frac{9}{3\sqrt{3} }=\sqrt{3} \\\frac{9}{3\sqrt{3} }=\sqrt{3}(2^0)[/tex]

THe value of u is 0.

(4)The expression [tex]\frac{16}{3(\sqrt{2^{x}} )}[/tex] is written as [tex]\frac{1}{3}(2^{4-\frac{x}{2}})[/tex].

[tex]\frac{16}{3(\sqrt{2^{x}} )}=\frac{2^4}{3(2^x)^{1/2}}\\\frac{16}{3(\sqrt{2^{x}} )}=\frac{1}{3}(2^{4-\frac{x}{2}})[/tex]

The value of u is [tex]4-\frac{x}{2}[/tex].

Final Answer:

1. [tex]\( (5\sqrt{3})^x \) in the form \( 3^u \): \( 3^{x\log_3(5\sqrt{3})} \)[/tex]

2. [tex]\( (1/2)^{x-3} \) in the form \( 2^u \): \( 2^{-(x-3)} \)[/tex]

3. [tex]\( \frac{9}{3\sqrt{3}} \) in the form \( 2^u \): \( 2^{2/3} \)[/tex]

4. [tex]\( \frac{16}{3\sqrt{2^x}} \) in the form \( 2^u \): \( 2^{(2x-4)/3} \)[/tex]

Explanation:

1. [tex]\( (5\sqrt{3})^x \) in the form \( 3^u \): \( 3^{x\log_3(5\sqrt{3})} \)[/tex]

To express [tex]\( (5\sqrt{3})^x \)[/tex] in the form [tex]\( 3^u \)[/tex], we use the logarithmic property [tex]\( a^{\frac{1}{n}} = n\sqrt[n]{a} \)[/tex]. Applying this, we get [tex]\( 5\sqrt{3} = 3^{\frac{\log_3(5\sqrt{3})}{\log_3(3)}} \)[/tex], and raising it to the power ( x ) yields the desired form.

2. [tex]\( (1/2)^{x-3} \) in the form \( 2^u \): \( 2^{-(x-3)} \)[/tex]

To rewrite [tex]\( (1/2)^{x-3} \)[/tex] in the form [tex]\( 2^u \)[/tex], we use the property [tex]\( a^{-b} = \frac{1}{a^b} \)[/tex]. Applying this, we transform the expression to [tex]\( 2^{-(x-3)} \)[/tex], achieving the desired form.

3. [tex]\( \frac{9}{3\sqrt{3}} \) in the form \( 2^u \): \( 2^{2/3} \)[/tex]

By canceling out the common factor of 3 in the numerator and denominator, we simplify [tex]\( \frac{9}{3\sqrt{3}} \)[/tex]to [tex]\( 2^{2/3} \), as \( \sqrt{3} = 3^{1/2} \)[/tex].

4. [tex]\( \frac{16}{3\sqrt{2^x}} \)[/tex] in the form [tex]\( 2^u \): \( 2^{(2x-4)/3} \)[/tex]

Applying the property[tex]\( a^{-b} = \frac{1}{a^b} \)[/tex] to the denominator, we simplify [tex]\( \frac{16}{3\sqrt{2^x}}[/tex] to [tex]\( 2^{(2x-4)/3} \)[/tex], achieving the desired form.

Final answer:

To convert a whole number to a percentage, multiply the number by 100 and add the percent symbol. For instance, converting the whole number 3 to a percentage results in 300%.

Explanation:

How to Convert a Whole Number to a Percentage

To convert a whole number to a percentage, you simply multiply the number by 100 and add the percent (%) symbol to the result. This is because a percentage represents a part per hundred. For example, to convert the whole number 3 to a percentage, you multiply it by 100, which equals 300%. Here's the step-by-step process:

Start with the whole number (e.g., 3).

Multiply the whole number by 100 (3 x 100 = 300).

Add the percent symbol to express the result as a percentage (300%).

This method is a straightforward percentage calculation, turning a whole number into its equivalent percentage form.

The original number of spokes on the wheel was 36.

The question seeks to determine the original number of spokes on a bicycle wheel, given that the current number is 32 after reducing by 4. To find the original number of spokes, we need to use a simple algebraic equation that adds the removed spokes back to the current number.

Let x represent the original number of spokes. The manufacturer has reduced the number of spokes by 4, so we subtract 4 from the original number to get the current number of spokes. The equation that represents this scenario is:

x - 4 = 32

To find the original number of spokes, we need to solve for x:

x = 32 + 4

x = 36

Therefore, the original number of spokes on the wheel was 36.

Final answer:

The question is related to the mathematics subject area and addresses the high school grade level. It involves understanding the normal distribution, mean, standard deviation, and the application of these concepts in quality control and filling volumes for beverage containers.

Explanation:

The subject of the question relates to the concept of normal distribution, which is a fundamental concept in statistics, a branch of mathematics. The student is provided with the information that beer bottles are filled so that they contain an average of 475 ml of beer, with the amount of beer being normally distributed and having a standard deviation of 8 ml. This scenario involves understanding concepts like mean, standard deviation, and the properties of a normal distribution.

To visualize similar quantities, consider that an eyedropper holds about 1 milliliter, a juice box holds about 25 centiliters (or 250 milliliters), and a soda bottle holds about 1 liter. The provided beer bottle contains roughly half the amount of a typical soda bottle. Knowledge of standard units such as liters and milliliters is essential in this context.

Furthermore, the question can be connected with practical scenarios such as the filling volumes for beverage containers, where manufacturers need to ensure that the bottles and cans contain the labeled amount of liquid. Quality control relies on statistics to test whether the actual amounts meet the desired standards, factoring in variations from measurement processes or filling mechanisms.