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Benutzernamemacabresubwoofer
Punkte1982
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 #1
avatar+1982 
0

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. 

 #1
avatar+1982 
0

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) 

 #1
avatar+1982 
+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.

 #1
avatar+1982 
0

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.