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3 years ago

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Berto has $12 to put gas in his car. If gas costs $3.75 per gallon, which ordered pair relating number of gallons of gas, x, to

the total cost of the gas, y, includes the greatest amount of gas Berto can buy?

2 answers:

Nastasia [14]3 years ago

6 0

Berto can purchase 12/3.75 gallons, this is 3.2 gallons of gas. So the coordinate on this line would be (3.2, 12).
Good Luck! :D

Ber [7]3 years ago

5 0

Answer:

The required ordered pair is (3.2,12).

Step-by-step explanation:

Let the x be the relating number of gallons of gas and y be the total cost of the gas.

It is given that the cost of gas is $3.75 per gallon. Total cost of x gallons is

$y=3.75x$

Berto has $12 to put gas in his car.

$y\leq 12$

$3.75x \leq 12$

Divide both sides by 3.75.

$x \leq 3.2$

The maximum number of gallons of gas is 3.2 and the total cost of the gas is 12.

Therefore the required ordered pair is (3.2,12).

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Goryan [66]

Answer:

Generally the barrier width is $a = 1.9322 *10^{-9} \ m$

Step-by-step explanation:

From the question we are told that

The tunneling probability required is $T = 1 * 10^{-5}$

The barrier height is $V_o = 0.4 eV$

The electron energy is $E = 0.08eV$

Generally the wave number is mathematically represented as

$k = \sqrt{ \frac{2 * m [V_o - E]}{\= h^2} }$

Here m is the mass of the electron with the value $m = 9.11 *10^{-31} \ kg$

h is is know as h-bar and the value is $\= h = 1.054*10^{-34} \ J \cdot s$

So

$k = \sqrt{ \frac{2 * 9.11 *10^{-31 } [0.4 - 0.04] * 1.6*10^{-19}}{[1.054*10^{-34}^2]} }$

=> $k = 3.073582 *10^{9} \ m^{-1}$

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$T = 16 * \frac{E}{V_o } * [1 - \frac{E}{V_o} ] * e^{-2 * k * a}$

So

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=> $6.944*10^{-6}= e^{-2 * 3.0736 *10^{9} * a}$

Taking natural log of both sides

$ln[6.944*10^{-6}] = -2 * 3.0736 *10^{9} * a}$

=> $-11.8776 = -2 * 3.0736 *10^{9} * a}$

=> $a = 1.9322 *10^{-9} \ m$

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