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3 years ago

10

A taxi service has two rates, as follows: Rate A charges a flat $2.50 for every mile; rate B charges $15 plus $1.50 per mile. At

least how many miles (expressed as a whole number) must one travel to make Rate B cheaper?

1 answer:

Vilka [71]3 years ago

5 0

Rate A: 2.50x
Rate B: 1.50x + 15

2.50x > 1.50x + 15
-1.50x -1.50x

1.00x > 15
x > 15

You must drive at least 15 miles before Rate B is cheaper

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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

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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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Answer:

Exact answer not possible: See below.

Step-by-step explanation:

The total that Tammy would pay is the sum of 2A and 5B, where A and B are the number of candies costing $2 and $5, respectively. Therefore:

2A + 2B = Total Spent

We are told Tammy spent <u>at least</u> $75, which can be written as "Total Spent > $75."

The equation becomes 2A + 5B > 75

Rearranging:

2A + 5B > 75

2A > 75 - 5B

A > (75 - 5B)/2

To find the maximum of the candy A bought, one can try different values of A and B that result in a total of at least $75. If the amount spent were $75 exactly, a solution would be 35 A (for $70), leaving $5 for 1 candy B. But we don't know the exact amount. The problems states "at least $75." As far as we know, Tammy may have spent $105, $405, $1,005, or even $4,005 (200 A and 1 B). One cannot pick a maximum simply since the maximum spent is not defined. The next possible value above $75 would be $77, which represents 36 A and 1 B candies.

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