if juanita and her friends made nine friendship bracelets in three hours how many could they make in 15 hours? which proportion would be used to solve this problem

Answer:

[tex]\displaystyle \int {\frac{sin(x)}{2x}} \, dx = \frac{1}{2}\sum^{\infty}_{n = 0} \frac{(-1)^nx^{2n + 1}}{(2n + 1)^2(2n)!}} + C[/tex]

General Formulas and Concepts:

Calculus

Integration

Sequences

Series

Taylor Polynomials

Power Series

Integration of Power Series:

Step-by-step explanation:

*Note:  

You could derive the Taylor Series for sin(x) using Taylor polynomials differentiation but usually you have to memorize it.

We are given the integral and are trying to find the infinite series of it:

[tex]\displaystyle \int {\frac{sin(x)}{2x}} \, dx[/tex]

We know that the power series for sin(x) is:

[tex]\displaystyle sin(x) = \sum^{\infty}_{n = 0} \frac{(-1)^nx^{2n + 1}}{(2n + 1)!}[/tex]

To find the power series for  [tex]\displaystyle \frac{sin(x)}{2x}[/tex], divide the power series by 2x:

[tex]\displaystyle \frac{sin(x)}{2x} = \sum^{\infty}_{n = 0} \bigg[ \frac{(-1)^nx^{2n + 1}}{(2n + 1)!} \cdot \frac{1}{2x} \bigg][/tex]

Simplifying it, we have:

[tex]\displaystyle \frac{sin(x)}{2x} = \sum^{\infty}_{n = 0} \frac{(-1)^nx^{2n}}{2(2n + 1)!}[/tex]

Rewrite the original integral:

[tex]\displaystyle \int {\frac{sin(x)}{2x}} \, dx = \int {\sum^{\infty}_{n = 0} \frac{(-1)^nx^{2n}}{2(2n + 1)!}} \, dx[/tex]

Integrate the power series:

[tex]\displaystyle \int {\frac{sin(x)}{2x}} \, dx = \sum^{\infty}_{n = 0} \frac{(-1)^nx^{2n + 1}}{2(2n + 1)(2n + 1)!}} + C[/tex]

Simplify the result:

[tex]\displaystyle \int {\frac{sin(x)}{2x}} \, dx = \frac{1}{2}\sum^{\infty}_{n = 0} \frac{(-1)^nx^{2n + 1}}{(2n + 1)^2(2n)!}} + C[/tex]

And we have our final answer.

Topic: AP Calculus BC (Calculus I + II)  

Unit: Power Series

If we modify the number of balls on a display it would alter the number of balls on the other displays based on an underlying mathematical rule or pattern.

The number of balls on the display could potentially represent a mathematical relationship or pattern. If we increase or decrease the number of balls on one display, it would change the number of balls on the other displays in a way that reflects this pattern.

For example, let's assume that the pattern is such that each display has twice the number of balls as the one before it. So if the first display has 1 ball, the second has 2, the third has 4, and so on. If you change the number of balls on the first display to 2, then the number of balls on the 6th display would increase according to the rule, and would become 2^6=64 instead of 1^6=32.

However, the relationship or pattern between the displays would need to be known in order to accurately predict how changing the number of balls on one display would affect the others.

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the area of the region between the curves [tex]\(y = x\) and \(y = x^2\) for \(x\) in \([-2, 1]\) is \( \frac{29}{6} \)[/tex] square units.

To find the area of the region between the curves [tex]\(y = x\) and \(y = x^2\)[/tex]for x in the interval [-2, 1], we need to set up the integral and integrate with respect to x.

First, let's graph the curves [tex]\(y = x\) and \(y = x^2\) over the interval \([-2, 1]\)[/tex] to visualize the region.

Now, let's find the points of intersection between the curves [tex]\(y = x\) and \(y = x^2\).[/tex]

Setting [tex]\(y = x\) equal to \(y = x^2\)[/tex], we get:

[tex]\[ x = x^2 \][/tex]

[tex]\[ x - x^2 = 0 \][/tex]

[tex]\[ x(1 - x) = 0 \][/tex]

This equation gives us two solutions: x = 0 and x = 1. So, the curves intersect at x = 0 and x = 1.

Now, to find the area of the region between the curves, we integrate the difference of the curves from [tex]\(x = -2\) to \(x = 0\), and from \(x = 0\) to \(x = 1\)[/tex], and then add the absolute value of these results:

[tex]\[ \text{Area} = \int_{-2}^{0} (x - x^2) \, dx + \int_{0}^{1} (x^2 - x) \, dx \][/tex]

Let's solve these integrals separately:

1. [tex]\[ \int_{-2}^{0} (x - x^2) \, dx \][/tex]

[tex]\[ = \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{-2}^{0} \][/tex]

[tex]\[ = \left[ \left(\frac{0^2}{2} - \frac{0^3}{3}\right) - \left(\frac{(-2)^2}{2} - \frac{(-2)^3}{3}\right) \right] \][/tex]

[tex]\[ = \left[ 0 - \left(\frac{4}{2} - \frac{-8}{3}\right) \right] \][/tex]

[tex]\[ = \left[ 0 - \left(2 + \frac{8}{3}\right) \right] \][/tex]

[tex]\[ = -2 - \frac{8}{3} \][/tex]

[tex]\[ = -\frac{6}{3} - \frac{8}{3} \][/tex]

[tex]\[ = -\frac{14}{3} \][/tex]

2. [tex]\[ \int_{0}^{1} (x^2 - x) \, dx \][/tex]

[tex]\[ = \left[ \frac{x^3}{3} - \frac{x^2}{2} \right]_{0}^{1} \][/tex]

[tex]\[ = \left[ \left(\frac{1^3}{3} - \frac{1^2}{2}\right) - \left(\frac{0^3}{3} - \frac{0^2}{2}\right) \right] \][/tex]

[tex]\[ = \left[ \left(\frac{1}{3} - \frac{1}{2}\right) - (0 - 0) \right] \][/tex]

[tex]\[ = \left( \frac{1}{3} - \frac{1}{2} \right) \][/tex]

[tex]\[ = \frac{1}{3} - \frac{1}{2} \][/tex]

[tex]\[ = \frac{2}{6} - \frac{3}{6} \][/tex]

[tex]\[ = -\frac{1}{6} \][/tex]

Now, we add the absolute values of these results:

[tex]\[ \text{Area} = \left| -\frac{14}{3} \right| + \left| -\frac{1}{6} \right| \]\\[/tex]

[tex]\[ \text{Area} = \frac{14}{3} + \frac{1}{6} \]\\[/tex]

[tex]\[ \text{Area} = \frac{28}{6} + \frac{1}{6} \]\\[/tex]

[tex]\[ \text{Area} = \frac{29}{6} \][/tex]

Therefore, the area of the region between the curves [tex]\(y = x\) and \(y = x^2\) for \(x\) in \([-2, 1]\) is \( \frac{29}{6} \)[/tex] square units.

The probable question maybe:

What is the area of the region between the curves [tex]\(y = x\) and \(y = x^2\)[/tex]for x in the interval [-2, 1]?

Final answer:

To find the probability of the sum of two dice being greater than 4 given the second die is a 3, first identify the sample space for the first die as {1, 2, 3, 4, 5, 6}. Then, calculate the favorable outcomes where the first die, added to 3, results in a number greater than 4, which are {2, 3, 4, 5, 6}. The probability is 5/6, rounded to approximately 0.8333.

Explanation:

Sample Space and Probability Calculation

When two dice are rolled one after another, and the second die results in a 3, we consider the outcomes of the first die only. As the first die is also a fair, six-sided die with faces numbered from 1 to 6, the sample space for the first die is S = {1, 2, 3, 4, 5, 6}.

The question asks for the probability of the sum being greater than 4 given that the second die is a 3. This means we are looking for the sum to be 5 or more. We can calculate the possible outcomes where the first die, when added to 3, results in a total greater than 4.

Therefore, the outcomes in the sample space that result in a sum greater than 4 are {2, 3, 4, 5, 6}. The probability of this event, given the second die is a 3, is the number of favorable outcomes divided by the total number of possible outcomes of the first die. There are 5 favorable outcomes and 6 possible outcomes, so the probability is 5/6 or approximately 0.8333 when rounded to four decimal places.

To construct the sample space, we need to consider all possible outcomes of rolling two dice. The probability of the sum of the dots on the dice being greater than 4 given that the second die is a 3 is 5/36.

To construct the sample space, we need to consider all possible outcomes of rolling two dice. Since each die has six sides numbered 1 to 6, the sample space will consist of 36 outcomes. We can represent the outcomes as pairs of numbers, where the first number represents the result of the first die and the second number represents the result of the second die. For example, (1, 1) represents both dice landing on 1, (1, 2) represents the first die landing on 1 and the second die landing on 2, and so on.

To determine the probability of the sum of the dots on the dice being greater than 4 given that the second die is a 3, we need to identify the outcomes where the second die is 3 and the sum is greater than 4. These outcomes are (2, 3), (3, 3), (4, 3), (5, 3), and (6, 3). There are a total of 5 outcomes that satisfy these conditions. Since the sample space has 36 outcomes, the probability is 5/36. To find the probability that the sum of the dots on two dice is greater than 4 given the second die is a 3, we list the possible outcomes for the first die as {1, 2, 3, 4, 5, 6}. The favorable outcomes are those that, when added to 3, result in a number greater than 4: {2, 3, 4, 5, 6}. This results in a probability of 5/6.

Answer:

152.4 cm

Step-by-step explanation:

Wesly is 5 feet and 3 inches tall. Since there are 12 inches in one foot, 5 feet has 60 inches. This means Wesly is 63 inches tall. If one inch is 2.54 cm then 60(2.54) is Wesly's height in centimeters. 60(2.54) = 152.4

Answer:

497.2 miles

Step-by-step explanation:

Great question, it is always good to ask away and get rid of any doubts that you may be having.

To begin solving this problem we first need to calculate how much gas Tyler has in his jeep after stopping at the gas station. We calculate this by multiplying the total bill by the price per gallon of gas, and then we add the amount that was left in the tank.

[tex](22/1.25)+5 = 22.6gallons[/tex]

After stopping at the gas station Tyler has 22.6 gallons of gas in his jeep. Since he gets 22 miles per gallon we multiply this by the amount of gallons in his car to calculate the distance they can travel.

[tex]\frac{22.miles}{gallon} * 22.6gallons = 497.2miles[/tex]

Tyler and his friends can travel 497.2 miles with the amount of gas they have.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

When selecting three batteries at random with replacement, there are 125 points in the sample space. When selecting three batteries at random without replacement, there are 60 points in the sample space.

To determine the number of points in the sample space when selecting three batteries at random with replacement, we use the counting principle. Since there are 5 batteries, each with an equal likelihood of being selected, and we are selecting three batteries, we can calculate the number of points in the sample space as follows:

Number of points = Number of possible outcomes for the first battery * Number of possible outcomes for the second battery * Number of possible outcomes for the third battery

Since there are 5 batteries, the number of possible outcomes for each battery is 5. Therefore,

Number of points = 5 * 5 * 5 = 125

So, there are 125 points in the sample space when selecting three batteries at random with replacement.

When selecting three batteries at random without replacement, the number of points in the sample space will be different. We can calculate it using a similar approach, but taking into account that once a battery is selected, it is removed from the bag and cannot be selected again. Therefore, for the first battery, there are 5 possible outcomes. For the second battery, there will be 4 possible outcomes since one battery has already been selected. For the third battery, there will be 3 possible outcomes. Therefore,

Number of points = 5 * 4 * 3 = 60

So, there are 60 points in the sample space when selecting three batteries at random without replacement.

The number of possible outcomes for selecting three batteries with replacement is 125. For selecting three batteries without replacement, the total is 60. The counting principle multiplies the number of options for each choice to calculate the sample space.

When selecting three batteries from a bag containing five different brands at random, the number of points in the sample space can be determined using the counting principle. The sample space represents all possible outcomes of an experiment.

a. Sampling with replacement

When batteries are selected with replacement, after each selection, the chosen battery is put back into the bag, making it available for the next selection. Thus, for each pick of the three batteries, there are five choices. The total number of points in the sample space for selecting three batteries with replacement is therefore 5 × 5 × 5 = 125.

b. Sampling without replacement

When batteries are selected without replacement, the chosen battery is not put back in the bag, which means the options decrease with each selection. The first pick has five choices, the second pick will have four (since one battery is already picked), and the third pick will have three. The total number of points in the sample space for selecting three batteries without replacement is therefore 5 × 4 × 3 = 60.

The equivalent value of the expression k = ( 2F/p )

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.

It demonstrates the equality of the relationship between the expressions printed on the left and right sides.

Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.

Given data ,

Let the equation be represented as A

Now , the value of A is

Substituting the values in the equation , we get

F = ( 1/2 ) kp

On simplifying , we get

Multiply by 2 on both sides , we get

2F = kp

Divide by p on both sides , we get

k = 2F/p

Hence , the expression is k = 2F/p

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To find tan x/2, given that tan x=3 and x terminates in π < x < (3π/2), we can use the half-angle formula for tangent. The value of tan (x/2) is ±1/√2.

To find tan x/2, given that tan x=3 and x terminates in π < x < (3π/2), we can use the half-angle formula for tangent. The half-angle formula for tangent is tan(x/2) = ±√((1-cosx) / (1+cosx)). Since tan x=3, we need to find the value of cos x first.

Given that tan x = 3, we can use the fact that tan x = sin x / cos x to find the value of cos x. Rearranging the equation, we have cos x = sin x / tan x = 1 / 3. Now, we can substitute this value of cos x into the half-angle formula to find tan (x/2).

tan (x/2) = ±√((1-cos x) / (1+cos x))
tan (x/2) = ±√((1-1/3) / (1+1/3))
tan (x/2) = ±√((2/3) / (4/3))
tan (x/2) = ±√(2/4)
tan (x/2) = ±√(1/2)
tan (x/2) = ±1/√2

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By setting up and solving a system of two equations based on the problem, we find that the larger country is 86,673 square kilometers in area and the smaller is 86,300 square kilometers.

This problem can be solved by using a system of two equations in two variables. Let's denote areas of two countries as x (for the bigger country) and y (for the smaller country).

The information from the problem provides us with two equations:
1) x + y = 172,973 (since combined area of two countries is 172,973 square kilometers),
2) x - y = 373 (since Country A is larger by 373 square kilometers).

Solving these equations will give the areas of the individual countries. Countries' areas can be found by adding the two equations together to eliminate y, you get 2x = 173,346, so x = 86,673 square kilometers.

Then, you can substitute x into the first equation: 86,673 + y = 172,973, from which y = 172,973 - 86,673 = 86,300 square kilometers.

So, the larger country's area is 86,673 square kilometers and the smaller country's area is 86,300 square kilometers.

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a. There could be 5 different choices and b. There could be 6 different choices.

Multiplication is the mathematical operation that is used to determine the product of two or more numbers. If an event can occur in m different ways and if following it, a second event can occur in n different ways, then the two events in succession can occur in m × n different ways.

Given tha the number of different kinds of soups = 2

The number of different kinds of salad = 3

(a) If you are having either soup or salad, then the total number of choices you have;

2 + 3 = 5

Hence, the total number of choices you have 5.

(b)  If you are having both soup and salad, then the number of choices you have;

2 x 3 = 6

Hence, the total number of choices you have 6.

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