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Let r1 be the rate at which Tanzi uses soap, and let r2 be the rate at which Maeve uses soap. If Tanzi's bottle is completely empty when Maeve starts pouring soap, then the total amount of soap used by Tanzi is:

1200 mL

Let's say that Maeve pours x mL of soap from her bottle into Tanzi's bottle. Then the total amount of soap used by Maeve is:

(2/3) × 1200 mL - x mL

since Maeve's bottle is initially two-thirds full and she pours x mL of soap out of her bottle.

The time it takes for Tanzi's bottle to be completely emptied is:

1200 mL / r1

The time it takes for Maeve's bottle to be completely emptied is:

(2/3) × 1200 mL - x mL / r2

Since both bottles are completely emptied at the same time, these two times must be equal. Therefore, we can set the two expressions for time equal to each other and solve for x:

1200 mL / r1 = (2/3) × 1200 mL - x mL / r2

Multiplying both sides by r1 r2, we get:

1200 r2 = (2/3) × 1200 r1 r2 - x r1

Multiplying out the right-hand side and simplifying, we get:

x = (2/3) × 1200 mL - 1200 r2 / r1

Therefore, to make both bottles empty at the same time, Maeve should pour:

x = (2/3) × 1200 mL - 1200 r2 / r1

milliliters of soap from her bottle into Tanzi's bottle. 

To find the rate of change of the climber's altitude, we need to calculate the average rate of change of altitude for the time interval from 2 hours to 6 hours.

Average rate of change = (change in altitude) / (change in time)

The change in altitude from 2 hours to 6 hours is:

46,700 - 46,400 = 300 feet

The change in time from 2 hours to 6 hours is:

6 hours - 2 hours = 4 hours

Therefore, the average rate of change of altitude is:

Average rate of change = 300 feet / 4 hours = 75 feet per hour

So the rate of change of the climber's altitude from 2 hours to 6 hours is 75 feet per hour. 

Since B, A, and D are consecutive vertices of a regular n-gon, we know that angle BAD is equal to (n-2)/n × 180 degrees. Since the n-gon is regular, we know that angle ABD is equal to 180 degrees / n, since there are n total angles in the n-gon that are equal in measure.

Therefore, angle BAC is equal to angle BAD + angle ABD, which is:

(n-2)/n × 180 + 180/n

Simplifying, we get:

(2n - 2)/n × 90

Since the heptagon is regular, angle BAC is also equal to 180 degrees / 7, since there are 7 total angles in the heptagon that are equal in measure.

Setting these two expressions equal to each other, we get:

(2n - 2)/n × 90 = 180/7

Multiplying both sides by 7n and simplifying, we get:

14n - 14 = 180n/7

Multiplying both sides by 7 and simplifying, we get:

98n - 98 = 180n

Solving for n, we get:

n = 14

Therefore, the given n-gon is a regular 14-gon. Since angle BCD is an exterior angle of the heptagon, it is equal to the sum of the remote interior angles,which are angles ACB and BAC. 

Angle ACB is an interior angle of the regular heptagon, so it is equal to (7-2)/7 × 180 degrees, which simplifies to 128.57 degrees.

Angle BAC was already calculated above to be (2n - 2)/n × 90 degrees, which simplifies to 140 degrees.

Therefore, angle BCD is equal to:

angle ACB + angle BAC = 128.57 degrees + 140 degrees = 268.57 degrees

So angle BCD measures 268.57 degrees. 

We can solve this problem using the equations of motion for a projectile. Let's take the x-axis to be horizontal and the y-axis to be vertical. 

The initial velocity of the ball can be decomposed into its x- and y-components:

v₀x = v₀ cos(49°)
v₀y = v₀ sin(49°)

where v₀ is the initial speed of the ball.

Using the fact that the ball was hit 4 feet above the ground, we can set the initial height y₀ to be 4 feet.

Using the fact that the ball was caught 2.5 feet above the ground, we can set the final height y equal to 2.5 feet.

Using the fact that the ball traveled 166 feet horizontally, we can set the horizontal displacement x equal to 166 feet.

Using the fact that there is no air resistance, we can set the acceleration due to gravity to be -32.2 feet per second squared.

Using the equations of motion for a projectile, we can write:

x = v₀x t
y = y₀ + v₀y t + (1/2)gt²

where t is the time of flight of the ball.

The time of flight can be found by setting y equal to 2.5 feet and solving for t:

2.5 = 4 + v₀ sin(49°) t + (1/2)(-32.2) t²

Solvingfor t, we get:

t = (v₀ sin(49°) + sqrt((v₀ sin(49°))² + 2(32.2)(4-2.5)))/32.2

Simplifying, we get:

t = (v₀ sin(49°) + sqrt((v₀ sin(49°))² + 25.84))/32.2

Using the fact that x = 166, we can write:

166 = v₀ cos(49°) t

Substituting the expression for t we derived earlier, we get:

166 = v₀ cos(49°) [(v₀ sin(49°) + sqrt((v₀ sin(49°))² + 25.84))/32.2]

Solving for v₀, we get:

v₀ = sqrt(166² (32.2)² / (cos²(49°) [(sin(49°))^2 + 25.84 / (32.2)^2]))

Simplifying, we get:

v₀ ≈ 125.4 feet per second

Therefore, the initial speed of the ball was approximately 125.4 feet per second. 

We can solve this system of equations using Gaussian elimination. Writing the augmented matrix, we have:

$$\left[\begin{array}{ccccc|c} 1 & 2 & 3 & 4 & 5 & 41 \\ 2 & 3 & 4 & 5 & 1 & 15 \\ 3 & 4 & 5 & 1 & 2 & 34 \\ 4 & 5 & 1 & 2 & 3 & 68 \\ 5 & 1 & 2 & 3 & 4 & 57 \end{array}\right]$$

Subtracting twice the first row from the second row, three times the first row from the third row, four times the first row from the fourth row, and five times the first row from the fifth row, we get:

$$\left[\begin{array}{ccccc|c} 1 & 2 & 3 & 4 & 5 & 41 \\ 0 & -1 & -2 & -3 & -9 & -67 \\ 0 & -2 & -4 & -11 & -13 & -83 \\ 0 & -3 & -11 & -14 & -13 & -108 \\ 0 & -9 & -13 & -17 & -11 & -148 \end{array}\right]$$

Adding twice the second row to the third row, three times thesecond row to the fourth row, and nine times the second row to the fifth row, we get:

$$\left[\begin{array}{ccccc|c} 1 & 2 & 3 & 4 & 5 & 41 \\ 0 & -1 & -2 & -3 & -9 & -67 \\ 0 & 0 & 0 & -17 & -31 & -213 \\ 0 & 0 & -17 & -5 & 10 & -289 \\ 0 & 0 & -31 & 8 & 67 & -691 \end{array}\right]$$

Finally, adding 17 times the third row to the fourth row, and 31 times the third row to the fifth row, we get:

$$\left[\begin{array}{ccccc|c} 1 & 2 & 3 & 4 & 5 & 41 \\ 0 & -1 & -2 & -3 & -9 & -67 \\ 0 & 0 & 0 & -17 & -31 & -213 \\ 0 & 0 & 0 & -74 & -472 & -3111 \\ 0 & 0 & 0 & 441 & 1044 & -1353 \end{array}\right]$$

We can now solve for the variables by back-substitution. From the last row, we have:

441d + 1044e = -1353

Solving for d and e, we get:

d = (-472(441) - 74(1044))/(-74(-17) - 472(0)) = 14
e = (-1353 - 441(14))/1044 = -2

Substituting d and e back into the fourth row, we get:

-74c - 472 = 3111 - 441(14)
-74c = 3897
c = -51

Substituting c, d, and e back into the third row, we get:

-17b - 31(-2) = 213 + 51 + 14
-17b = 311
b = -\frac{311}{17}

Substituting b, c, d, and e back into the second row, we get:

-a - 2(-\frac{311}{17}) - 3(-51) - 4(14) - 5(-2) = -67
-a = -\frac{1847}{17}

Substituting a, b, c, d, and e back into the first row, we get:

\frac{-1847}{17} + 2(-\frac{311}{17}) + 3(-51) + 4(14) + 5(-2) = 41

Therefore, the ordered quintuplet (a, b, c, d, e) that satisfies the system of equations is:

(a, b, c, d, e) = (-\frac{1847}{17}, -\frac{311}{17}, -51, 14, -2) 

The berries are 99% water, which means that 1% of their weight is not water. Therefore, the weight of the non-water content of the berries is:

0.01 * 200 kg = 2 kg

This means that the weight of the water in the berries is:

200 kg - 2 kg = 198 kg

After two days in the sun, the water content of the berries is 90%, which means that 10% of their weight is not water. Therefore, the weight of the non-water content of the berries is:

0.1 * x = x/10

where x is the total weight of the berries after two days.

This means that the weight of the water in the berries is:

x - x/10 = 9x/10

We know that the weight of the water in the berries has decreased from 198 kg to 9x/10 kg. We can set up an equation to solve for x:

198 kg = (90/99)(9x/10) kg

Simplifying, we get:

19800/90 = (9/10)x

Multiplying both sides by 10/9, we get:

220 = x

Therefore, the total weight of the berries after two days is 220 kg. 

Let A be the amount of money Alice has and B be the amount of money Bob has.

From the first condition, we have:

A + n = 7(B - n)

Simplifying, we get:

A + n = 7B - 7n
A + 8n = 7B

From the second condition, we have:

A - n = k(B + n)

Simplifying, we get:

A - n = kB + kn
A - kn = kB + n
A = (k + 1)B + (k + 1)n

We want to find the ratio of A to B when neither gives the other any money, which means n = 0. Substituting n = 0 into the equations we derived earlier, we get:

A + 8n = 7B  becomes  A = 7B
A = (k + 1)B + (k + 1)n  becomes  A = (k + 1)B

Since both equations give us the value of A in terms of B, we can equate them to get:

7B = (k + 1)B

Solving for k, we get:

k = 6

Therefore, the ratio of the amount of money Alice has to the amount Bob has is:

A/B = (k + 1) = 7/1 = 7:1

1. Using Vieta's formulas, we know that A + B = 5 and AB = 3. We want to find a quadratic with roots A^2 and B^2. 

We know that (x - A^2)(x - B^2) = 0 has roots A^2 and B^2. Expanding this equation, we get:

x^2 - (A^2 + B^2) x + A^2B^2 = 0

Now we need to express A^2 + B^2 and A^2B^2 in terms of A and B. We know that:

(A + B)^2 = A^2 + 2AB + B^2

(A - B)^2 = A^2 - 2AB + B^2

Adding these two equations gives:

2A^2 + 2B^2 = 2(A^2 + B^2) + 2AB

Simplifying, we get:

A^2 + B^2 = (A + B)^2 - 2AB
A^2 + B^2 = 25 - 2(3)
A^2 + B^2 = 19

Using the fact that AB = 3, we can substitute into the quadratic we derived earlier to get:

x^2 - 19x + 9 = 0

Therefore, the quadratic whose roots are A^2 and B^2 is x^2 - 19x + 9.

2. We have A - B = 4 and A^3 - B^3 = 52. 

(a) To find all possible values of AB, we can use the identity:

A^3 - B^3 = (A - B)(A^2 + AB + B^2)

Substituting in A - B = 4 and simplifying, we get:

52 = 4(A^2 + AB + B^2)
13 = A^2 + AB + B^2

Using the fact that AB = (A - B)^2 - A^2 - B^2, we can substitute into the equation above to get:

13 = 2A^2 + 2B^2 - 8A
13 = 2(A - 2)^2 + 2B^2 - 17

Rearranging and simplifying, we get:

2B^2 = -2(A - 2)^2 + 30
B^2 = -(A - 2)^2 + 15

Since B^2 is non-negative for all real numbers, we know that -(A - 2)^2 + 15 ≥ 0. Solving for A, we get:

(A - 2)^2 ≤ 15

Therefore, -√15 + 2 ≤ A ≤ √15 + 2.

Now we canuse this range of values for A to find the corresponding range of values for AB. We know that AB = 3 + B(A - B), so substituting in A - B = 4, we get:

AB = 3 + 4B

Since B can take any real value, the range of possible values for AB is (-∞, ∞).

(b) To find all possible values of A + B, we can add A - B = 4 to A^3 - B^3 = 52 to get:

A^3 + 3A^2B + 3AB^2 + B^3 = 56

Using the fact that A^3 - B^3 = 52 and A - B = 4, we can substitute and simplify to get:

4A^2 + 12AB + 4B^2 = 56
A^2 + 3AB + B^2 = 14

Using the fact that A + B = (A - B) + 2B = 4 + 2B, we can substitute and simplify to get:

A^2 + 3AB + B^2 = (A + B)^2 - AB = 14
(A + B)^2 = 14 + AB

Since AB can take any real value, the range of possible values for (A + B)^2 is [14, ∞). Taking the square root of both sidesand considering both positive and negative values, we get:

A + B = ±√(14 + AB)

Therefore, the possible values of A + B are A + B = √(14 + AB) and A + B = -√(14 + AB).

(c) Using the expression we derived in part (a), we know that:

B^2 = -(A - 2)^2 + 15

Since B^2 is non-negative for all real numbers, we know that -(A - 2)^2 + 15 ≥ 0. Solving for A, we get:

(A - 2)^2 ≤ 15

Therefore, -√15 + 2 ≤ A ≤ √15 + 2. 

Using the fact that A - B = 4, we can solve for B in terms of A to get:

B = A - 4

Substituting the range of values for A, we get:

-√15 - 2 ≤ B ≤ √15 - 2

Therefore, the possible values of A and B are:

-√15 + 2 ≤ A ≤ √15 + 2, and -√15 - 2 ≤ B ≤ √15 - 2.

The two correct answers to choose are B) tether 1.8m, flag 4.08m and D) tether 0.6m, flag 1.36m.

To find the values of k and m that make the system have infinitely many solutions, we need to find conditions under which the two equations are equivalent. In other words, we want to find values of k and m such that the second equation can be obtained from the first equation by multiplying both sides by a constant and adding a constant.

Let's subtract the first equation from the second equation:

(6a + 2b) - (3a + 2b) = k + 3a + mb - 2

Simplifying, we get:

3a = k + 3a + mb - 2

Rearranging, we get:

2a - mb = k - 2

Now we can use the fact that the system has infinitely many solutions to find values of k and m. Since the system has infinitely many solutions, the equation 2a - mb = k - 2 must be equivalent to one of the given equations. In other words, there must be values of k and m such that:

2a - mb = 2 - (3a + 2b)

Multiplying both sides by -1, we get:

3a + 2b - 2a + mb = -2

Simplifying, we get:

a + 2b + mb = -2

Now we have two equations:

2a - mb = k - 2
a + 2b + mb = -2

We can solve for k and m by eliminating one of the variables, say a or b, from the equations. Let's eliminate a by multiplying the second equation by 2 and subtracting it from the first equation:

2(2a - mb = k - 2) - (a + 2b + mb = -2)

Simplifying, we get:

3a - 5b = 2k - 2

Now we want this equation to be true for all values of a and b, which means the coefficients of a and b on the left-hand side must be equal to 0:

3a - 5b = 0

Equating the coefficients with the previous equation, we get:

2k - 2 = 0

Solving for k, we get:

k = 1

Now substituting k = 1 into the equation 2a - mb = k - 2, we get:

2a - mb = -1

Rearranging, we get:

mb - 2a = 1

We can solve for m by setting a = b = 1:

m - 2 = 1

Solving for m, we get:

m = 3

Therefore, the values of k and m that make the system have infinitely many solutions are k = 1 and m = 3.