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lana66690 [7]
3 years ago
15

The school jose goes to is selling tickets to a fall musical. On the first day of ticket sales the school sold 14 adult tickets

and 4 student tickets for a total of $108. The school took in $158 on the second day by selling 7 adult tickets an 10 student tickets. What is the total price of each of the one adult ticket and one student ticket?​
Mathematics
1 answer:
galina1969 [7]3 years ago
7 0

Answer:18

Step-by-step explanation:

Adult-$14 Child-$4 ==$18

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I have 16 graded assignments. 3/4 of them were A’s and the rest were B’s. What fraction of that is B’s
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Answer:

1/4 are B's.

Step-by-step explanation:

If 3/4 of the 16 assignments are A's and the remaining are B's, all that is left is 1/4. Therefore, 1/4 of the 16 assignments are B's.

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laila [671]

(9b -7) +(7-3b) - (-3-3b) +(6+3b) = 9b -7 + 7-3b +3 + 3b + 6+3b = 12b + 9

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To find the average, we need the sum of these numbers 12b + 9, divide by 4.

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Answer is A.

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A field goal kicker lines up to kick a 44 yard (40m) field goal. He kicks it with an initial velocity of 22m/s at an angle of 55
Goryan [66]

Answer:

The ball makes the field goal.

The magnitude of the velocity of the ball is approximately 18.166 meters per second.

The direction of motion is -45.999º or 314.001º.

Step-by-step explanation:

According to the statement of the problem, we notice that ball experiments a parabolic motion, which is a combination of horizontal motion at constant velocity and vertical uniform accelerated motion, whose equations of motion are described below:

x = x_{o}+v_{o}\cdot t\cdot \cos \theta (Eq. 1)

y = y_{o} + v_{o}\cdot t \cdot \sin \theta +\frac{1}{2}\cdot g\cdot t^{2} (Eq. 2)

Where:

x_{o}, y_{o} - Coordinates of the initial position of the ball, measured in meters.

x, y - Coordinates of the final position of the ball, measured in meters.

\theta - Angle of elevation, measured in sexagesimal degrees.

v_{o} - Initial speed of the ball, measured in meters per square second.

t - Time, measured in seconds.

If we know that x_{o} = 0\,m, y_{o} = 0\,m, v_{o} = 22\,\frac{m}{s}, \theta = 55^{\circ}, g = -9.807\,\frac{m}{s} and x = 40\,m, the following system of equations is constructed:

40 = 12.618\cdot t (Eq. 1b)

y = 18.021\cdot t -4.904\cdot t^{2} (Eq. 2b)

From (Eq. 1b):

t = 3.170\,s

And from (Eq. 2b):

y = 7.847\,m

Therefore, the ball makes the field goal.

In addition, we can calculate the components of the velocity of the ball when it reaches the field goal post by means of these kinematic equations:

v_{x} = v_{o}\cdot \cos \theta (Eq. 3)

v_{y} = v_{o}\cdot \cos \theta + g\cdot t (Eq. 4)

Where:

v_{x} - Final horizontal velocity, measured in meters per second.

v_{y} - Final vertical velocity, measured in meters per second.

If we know that v_{o} = 22\,\frac{m}{s}, \theta = 55^{\circ}, g = -9.807\,\frac{m}{s} and t = 3.170\,s, then the values of the velocity components are:

v_{x} = \left(22\,\frac{m}{s} \right)\cdot \cos 55^{\circ}

v_{x} = 12.619\,\frac{m}{s}

v_{y} = \left(22\,\frac{m}{s} \right)\cdot \sin 55^{\circ} +\left(-9.807\,\frac{m}{s^{2}} \right)\cdot (3.170\,s)

v_{y} = -13.067\,\frac{m}{s}

The magnitude of the final velocity of the ball is determined by Pythagorean Theorem:

v =\sqrt{v_{x}^{2}+v_{y}^{2}} (Eq. 5)

Where v is the magnitude of the final velocity of the ball.

If we know that v_{x} = 12.619\,\frac{m}{s} and v_{y} = -13.067\,\frac{m}{s}, then:

v = \sqrt{\left(12.619\,\frac{m}{s} \right)^{2}+\left(-13.067\,\frac{m}{s}\right)^{2} }

v \approx 18.166\,\frac{m}{s}

The magnitude of the velocity of the ball is approximately 18.166 meters per second.

The direction of the final velocity is given by this trigonometrical relation:

\theta = \tan^{-1}\left(\frac{v_{y}}{v_{x}} \right) (Eq. 6)

Where \theta is the angle of the final velocity, measured in sexagesimal degrees.

If we know that v_{x} = 12.619\,\frac{m}{s} and v_{y} = -13.067\,\frac{m}{s}, the direction of the ball is:

\theta = \tan^{-1}\left(\frac{-13.067\,\frac{m}{s} }{12.619\,\frac{m}{s} } \right)

\theta = -45.999^{\circ} = 314.001^{\circ}

The direction of motion is -45.999º or 314.001º.

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