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DaniilM [7]
3 years ago
11

Reduce to lowest term 13/50

Mathematics
1 answer:
Rama09 [41]3 years ago
3 0

The fraction would stay the same but if you want a decimal form it would be 0.26

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Each Rubber ball is = to 2 balls btw, Find the median of the data in the pictograph below.
artcher [175]

Answer:

7

Step-by-step explanation:

The two in middle are 10 and 4

So add them both then divide by 2

So 14÷2

7

Hope this helps

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Solve the inequality 2x + 7 - 3x < 13 +2X. You must show your work and describe the graph of the solution on the number line.
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Cancel equal terms on both sides of inequality

7-3x < 13

move constant to the right side and change its sign

-3x<13-7

subtract the numbers

-3x<6

divide both sides of inequality by -3 and flip the inequality sign

answer is: x>-2

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Antonio invests $3,000, at 10% interest, compounded semiannually for 12 years. Use the compound interest formula to calculate th
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4 years ago
A ball is launched upward at 14 m/s from a platform 30 m high.Find the maximum height the ball will reach and how long it will t
BartSMP [9]

Answer:

The ball will reach a maximum height of 39.993 meters after 1.428 seconds.

Step-by-step explanation:

Let suppose that no non-conservative forces acts on the ball during its motion, then we can determine the maximum height reached by the Principle of Energy Conservation, which states that:

K_{1}+U_{g,1} = K_{2}+U_{g,2} (1)

Where:

K_{1}, K_{2} - Initial and final translational kinetic energies, measured in joules.

U_{g,1}, U_{g,2} - Initial and final gravitational potential energies, measured in joules.

By definition of translational kinetic energy and gravitational potential energy we expand and simplify the expression above:

\frac{1}{2}\cdot m\cdot v_{2}^{2}+m\cdot g\cdot y_{2}= \frac{1}{2}\cdot m\cdot v_{1}^{2}+m\cdot g\cdot y_{1} (2)

Where:

m - Mass of the ball, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

v_{1}, v_{2} - Initial and final speed of the ball, measured in meters per second.

y_{1}, y_{2} - Initial and final heights of the ball, measured in meters.

The final height of the ball is determined by the following formula:

v_{2}^{2}+2\cdot g\cdot y_{2} = v_{1}^{2}+2\cdot g\cdot y_{1}

v_{1}^{2}-v_{2}^{2}+2\cdot g \cdot y_{1}=2\cdot g\cdot y_{2}

y_{2} = y_{1}+\frac{v_{1}^{2}-v_{2}^{2}}{2\cdot g} (3)

If we know that y_{1} = 30\,m, v_{1} = 14\,\frac{m}{s}, v_{2} = 0\,\frac{m}{s} and g = 9.807\,\frac{m}{s^{2}}, the maximum height that the ball will reach is:

y_{2} = 30\,m + \frac{\left(14\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

y_{2} = 39.993\,m

The ball will reach a maximum height of 39.993 meters.

Given the absence of non-conservative forces, the ball exhibits a free fall. The time needed for the ball to reach its maximum height is computed from the following kinematic formula:

t = \frac{v_{2}-v_{1}}{-g} (4)

If we know that v_{1} = 14\,\frac{m}{s}, v_{2} = 0\,\frac{m}{s} and g = 9.807\,\frac{m}{s^{2}}, then:

t = \frac{0\,\frac{m}{s}-14\,\frac{m}{s}  }{-9.807\,\frac{m}{s^{2}} }

t = 1.428\,s

The ball will take 1.428 seconds to reach its maximum height.

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Step-by-step explanation:

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