Let n be a nonzero integer. find a domain on which f (x) = (1 − xn)1/n coincides with its inverse. hint: the answer depends on whether n is even or odd.

A regular polygon has the same number of lines of symmetry as it has sides. Therefore, a regular polygon with 40 sides will have 40 lines of symmetry.

In Mathematics, symmetry is a significant aspect when dealing with geometric figures, such as polygons. In a regular polygon, the number of lines of symmetry is equal to the number of sides. Therefore, a regular polygon with 40 sides will have 40 lines of symmetry.

This is because, in a regular polygon, all sides and angles are equal, meaning it can be reflected or folded over each of its sides and still remain the same shape. So, you can draw a line from each vertex to the opposite side (or vertex in some cases) and get a perfect fold or reflection.

Remember, this is only true for regular polygons, as irregular polygons won't have the same number of lines of symmetry because their sides and angles may not be equal.

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Answer:

[tex]2\sqrt{1+tan(x)} +C[/tex]

Step-by-step explanation:

To start solving this you need to use substitution. I let u = 1+tan(x). Next you need to find du/dx, which is sec^2(x) using trigonometric properties. Solve for dx and get dx = du / sec^2(x). Next put the new dx back in. This gives you integral sec^2(x) / sqrt u * du / sec^2(x). The sec^2(x) cancels and the new expression is integral 1/sqrt u * du, which can be simplified to integral u^-1/2 * du. You then take the integral and get 2u^1/2. Lastly, substitute the original u back in and get 2 sqrt 1+tan(x) + C.

Answer:

10 1/3 = 3 /31.      2 2/3 = 3/8.          5 1/3 = 3/16.

Step-by-step explanation:

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Answer:

wow

Step-by-step explanation:

3 angles inside a triangle = 180 degrees

 since the 2 sides are equal

 angle B  and C are the same

180 - 50 = 130 degrees

130 / 2 =65

 Angle B is 65 degrees

We can use the points (-4, -1) and (0, -2) to solve.

Slope formula: y2-y1/x2-x1

= -2-(-1)/0-(-4)

= -1/4

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Final answer:

To find the equation for f–1(x), swap the roles of x and f(x) in the equation for f(x), solve for f(x), and replace it with x.

Explanation:

To find the equation for f–1(x), we need to swap the roles of x and f(x) in the equation for f(x) and solve for f(x). So, given that f(x) = 1/4x + 3, we can start by replacing f(x) with x in the equation:

x = 1/4f–1(x) + 3

Now we can solve for f–1(x). First, subtract 3 from both sides of the equation:

x - 3 = 1/4f–1(x)

Then, multiply both sides of the equation by 4:

4(x - 3) = f–1(x)

So, the equation for f–1(x) is f–1(x) = 4(x - 3).

Final answer:

To find the inverse of the function f(x) = 1/4x + 3, swap x and y and solve for y to get the inverse function f⁻¹(x) = 1/(4(x - 3)).

Explanation:

To find the equation for the inverse function f⁻¹(x) of the given function f(x) = 1/4x + 3, we need to swap the variables x and y and then solve for y.

Therefore, the inverse function is f⁻¹(x) = 1/(4(x - 3)), showcasing the step-by-step process to derive the inverse relationship.

Answer:

Option B is right.

Step-by-step explanation:

Given that there are totally 9 books unread by Mrs. Reid.  OUt of these she wants to take only 3 books on the trip.

For books order does not count.

Hence combinations would be appropriate to calculate.

3 books out of 9 can be selected in

9C 3 ways

No of ways of selecting 3 books for the trip out of 9 = [tex]\frac{9(8)(7)}{1(2)(3)} =84[/tex]

There are just 84 ways of choosing 3 books to bring on the trip out of 9 unread books.

Option B is right.

The correct answer is 84.

To solve this problem, we can use the concept of combinations from combinatorics. A combination is a way of selecting items from a collection, such that the order of selection does not matter. In this case, Mrs. Reid wants to select 3 books out of 9 without regard to the order in which they are chosen.

The number of ways to choose 3 books from 9 is given by the combination formula:

[tex]\[ C(n, k) = \frac{n!}{k!(n-k)!} \][/tex]

where[tex]\( n \)[/tex] is the total number of items to choose from, [tex]\( k \)[/tex]is the number of items  to choose, and [tex]\( ! \)[/tex] denotes factorial, which is the product of all positive integers up to that number.

For Mrs. Reid's situation:

[tex]\[ n = 9 \] \[ k = 3 \][/tex]

So we plug these values into the formula:

[tex]\[ C(9, 3) = \frac{9!}{3!(9-3)!} \][/tex]

Calculating the factorials:

[tex]\[ 9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \] \[ 3! = 3 \times 2 \times 1 \] \[ (9-3)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 \][/tex]

Now we simplify the expression by canceling out the common terms in the numerator and the denominator:

[tex]\[ C(9, 3) = \frac{9 \times 8 \times 7 \times 6!}{3! \times 6!} \] \[ C(9, 3) = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \][/tex]

Divide out the common factors:

[tex]\[ C(9, 3) = \frac{9}{3} \times \frac{8}{2} \times \frac{7}{1} \] \[ C(9, 3) = 3 \times 4 \times 7 \][/tex]

Now we multiply the remaining numbers [tex]\[ C(9, 3) = 3 \times 4 \times 7 = 84 \][/tex]

Therefore, Mrs. Reid can choose 3 books to bring on the trip in 84 different ways.

Given that the certain type of spore colony having 50 spores and a rate of growth of 3 times in 24 hour or one day period, it would grow to (3 times 50) 150 spores at the end of the first day, (3 times 150) 450 in the 2nd day, (3 times 450) 1350 in the third day, and (3 times 1350) 4050 spores at the end of the fourth day.

The spore colony, which triples in size each day, will contain 4050 spores at the end of the 4th day starting from an initial count of 50 spores.

This question involves exponential growth, where the size of the spore colony triples every day.

To solve this, we will calculate the number of spores at the end of each day:

Initial spores= 50

End of Day 1,
50 * 3 = 150

End of Day 2,
150 * 3 = 450

End of Day 3,
450 * 3 = 1350

End of Day 4,
1350 * 3 = 4050

Therefore, at the end of the 4th day, the spore colony would contain 4050 spores.

To determine if a probability of contamination more than twice the mean of 14.4 is unusual, the standard deviation must be considered. In a normal distribution, occurrences more than two standard deviations from the mean are typically unusual.

Whether a probability of contamination more than twice the mean of 14.4 is unusual or not depends on the standard deviation of the data. In a typical normal distribution, values more than two standard deviations away from the mean could be considered unusual. Considering a mean of 14.4, if the standard deviation is small, then a value twice as large would likely be an outlier. However, if the standard deviation is large, such a value could fall within the realm of normal variation.

To ascertain if a value is unusual, one would compare it to the range defined by a certain number of standard deviations from the mean. A common rule of thumb is the empirical rule, which states that for a normally distributed dataset, roughly 68% of data falls within one standard deviation of the mean, 95% within two standard deviations, and 99.7% within three standard deviations.

In this context, if a probability of contamination is more than twice the mean (greater than 28.8), and assuming a normal distribution, it would be considered unusual if it falls outside of two standard deviations from the mean.

To solve the linear equation 6x + 3y = 15 for y, isolate y, which yields y = (15 - 6x) / 3. Substituting x as 2, we find y = 1, and when x is 5, y equals -5.

In order to solve the equation 6x + 3y = 15 for y, we must first isolate y on one side of the equation. This can be done by first subtracting 6x from both sides of the equation, resulting in 3y = 15 - 6x. To solve for y, we then divide both sides by 3, obtaining y = (15 - 6x) / 3.

When x = 2, we substitute it into the equation to find y:
y = (15 - 6(2)) / 3y = (15 - 12) / 3y = 3 / 3y = 1.

Similarly, when x = 5, we find y by substitution:
y = (15 - 6(5)) / 3y = (15 - 30) / 3y = -15 / 3y = -5.

C—34. Subscriptions

Answer:

B

Step-by-step explanation:

Our three consecutive integers are 37, 38, and 39.

Work is provided in the image attached.