bader

The key to this problem lies in understanding the properties of asymptotes of a hyperbola.

Asymptotes and Hyperbolas: A hyperbola has two asymptotes that intersect at the center of the hyperbola. As the hyperbola approaches its branches towards positive or negative infinity, it gets closer and closer to these asymptotes but never touches them.

Equation of Asymptotes: The equations of the asymptotes of a hyperbola with a horizontal transverse axis (like the one we're dealing with) can be expressed as:

y = ± (x * (b/a)) + k (where k is the y-coordinate of the center)

Point on Asymptote: We are given that the point (8, 2) lies on one of the asymptotes. This point will satisfy the equation of the specific asymptote it falls on.

Solve for a/b: Since the point (8, 2) is on an asymptote, we can substitute its coordinates (x = 8 and y = 2) into the equation of the asymptote (one of the ± signs will work):

2 = ± (8 * (b/a)) + k (We don't need to know the value of k for this problem)

Case 1: Positive Sign

2 = (8 * (b/a)) + k Assuming k = 0 (since we don't know its value), we have: 2 = 8 * (b/a) (b/a) = 2/8 = 1/4

Case 2: Negative Sign

2 = -(8 * (b/a)) + k Again, assuming k = 0, we have: 2 = - 8 * (b/a) (b/a) = -2/8 = -1/4 (We negate this value since it's the negative case)

Since we only care about the absolute value of a/b, both cases lead to the same answer:

|a/b| = 1/4

We are given two facts about a regular hexagon:

Perimeter (p): We know the total length of all sides of the hexagon is p units.

Area (A): We are told the area of the hexagon is A=23​ square units.

We want to find the side length of the hexagon.

Here's how we can approach this problem:

Relate Perimeter and Side Length: A regular hexagon has six sides, all with the same length. Let's call this side length "s" (units). The perimeter (p) can be expressed as the total length of all sides:
p = 6s (equation 1)

Relate Area and Side Length: There's a formula for the area of a regular hexagon based on its side length "s":
Area (A) = (3√3 * s^2) / 2 (equation 2)

Solve for Side Length: We are given A = 3/2 and want to find s. We can use equation 2 and substitute the given value of A:

(3/2) = (3√3 * s^2) / 2

Solve for s:

Multiply both sides by 2: 3 = 3√3 * s^2

Divide both sides by 3√3: s^2 = 1 / √3

Take the square root of both sides (remembering there might be positive and negative square roots): s = ± (1 / √3)^(1/2)

Since side length cannot be negative, we take the positive square root: s = 1/√3

Simplify s: Recall that the radical (√) symbol represents the square root. We can simplify 1/√3 by rationalizing the denominator, multiplying both top and bottom by √3:

s = (1/√3) * (√3/√3) s = √3 / 3

Therefore, the side length of the regular hexagon is √3 / 3 units.

The fourth root of a number is only an integer if the prime factorization of the number contains every prime factor raised to a power that is a multiple of 4.

In this case, we need to analyze the prime factorization of 1323=3×7×7.

3 is already raised to the power of 1 (which is a multiple of 4).

7 needs to be raised to a power that is a multiple of 4. Currently, it's raised to the power of 2.

Therefore, the smallest value of n that makes 1323n have a fourth root as an integer is the one that raises 7 to the fourth power. This means n=72=49​.

With n=49, 1323n=1323×49=64809, and its fourth root is 464809​=3×7=21, which is an integer.

Therefore, the answer is 49.

We can solve this problem using the Pythagorean Theorem applied to both triangles ABC and ACD since we are given a side length and looking for the hypotenuse of each right triangle.

For triangle ABC, we are given that AC=4 and solving for AB (the hypotenuse) using the Pythagorean Theorem: AB2=AC2+BC2=42+x2=42+452, so AB=42+452​.

For triangle ACD, we are given that AC=4 and solving for AD (the hypotenuse) using the Pythagorean Theorem: AD2=AC2+CD2=42+y2=42+632, so AD=42+632​.

To find the difference in length between AB and AD, we subtract the length of AC from each and find the difference: AD−AB=42+632​−42+452​.

We can evaluate these square roots on a calculator and round to the nearest hundredth to find that AD − AB ≈ 8.79 − 6.65 = 2.14 units.

Let the roots of Babette's monic quadratic be r and s. Since the leading coefficient is 1, the quadratic can be written as:

p(x)=(x−r)(x−s)=x2−(r+s)x+rs.

We are given that squaring the roots gives another monic quadratic with those squares as its roots. In other words:

(x−r2)(x−s2)=x2−(r2+s2)x+r2s2

Expanding the left side gives:

x2−(r2+2rs+s2)x+r2s2

Equating the coefficients of the corresponding terms in both quadratics, we get the system of equations:

r+s=r2+s2

r2s2=rs (notice this simplifies to r2s2−rs=0)

From the first equation, we can rearrange to get r2+s2−r−s=0. Factoring this gives (r−1)(s−1)=0. This means either r=1 or s=1.

We can consider two cases:

Case 1: r=1

Substituting r=1 into the second equation gives s2−s=0, which factors as s(s−1)=0. This means either s=0 or s=1. However, if s=0, then the leading coefficient of the quadratic wouldn't be 1, so we reject this case.

Therefore, in this case, s=1 and the quadratic is simply x2−2x+1=(x−1)2. There is only 1​ quadratic possible in this case.

Case 2: s=1

Substituting s=1 into the second equation gives r2−r=0, which factors as r(r−1)=0. This means either r=0 or r=1. However, if r=0, then the leading coefficient of the quadratic wouldn't be 1, so we reject this case.

Therefore, in this case, r=1 and the quadratic is again simply x2−2x+1=(x−1)2. There is only 1​ quadratic possible in this case.

Since both cases lead to only one possible quadratic, Babette could have been thinking of at most 2​ different monic quadratics.