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3 years ago

Solve the inequality. 2(4+2x)25x+5

Mathematics

1 answer:

stealth61 [152]3 years ago

3 0

Answer:

Step-x ≤ 3

Given

2(4 + 2x) ≥ 5x + 5 ← distribute parenthesis on left side

8 + 4x ≥ 5x + 5 ( subtract 4x from both sides )

8 ≥ x + 5 ( subtract 5 from both sides )

3 ≥ x , hence

x ≤ 3by-step explanation:

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To celebrate a diners anniversary, the owners have made a 40 foot long sandwich. The sandwich is cut into three unequal pieces.

Ivahew [28]

Let x be he length of middle piece.
3x = length of longer piece.
x-5 = length of shortest piece.

x + 3x + x-5 = 40
5x - 5 = 40
5x = 45
x = 45/5
x = 9. ( Size of middle piece )

Given that the longer piece is 3 times the size of middle piece,

Longest piece = 3 x 9 = 27.

Size of longest piece = 27 feet
Size of middle piece = 9 feet
Size of smaller piece = 40 - (27+9)
= 40 - 36
= 4 feet.
Thank you.

8 0

3 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)