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3 years ago

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Subtract. Susan began a trip with 18 1/2 gallons of gas in the gas tank of her car. If she used 17 3/4 gallons on the trip, how

many gallons did she have left when she finished the trip?
a 1 1/4 gallons
b 1 3/4 gallons
c 1/2 gallons
d 3/4 gallons

2 answers:

Gemiola [76]3 years ago

8 0

D. 18.5 minus 17.75 is .75

cestrela7 [59]3 years ago

6 0

Answer:

Option (d) $\frac{3}{4}$ gallon is correct.

Step-by-step explanation:

Amount of gas present in the gas tank when she began the trip is $18\frac{1}{2} =\frac{37}{2}$ gallons.

Amount of gas she used in the trip is $17\frac{3}{4} =\frac{71}{4}$ gallons

Now, in order to find the amount gas she has left when she finished the trip, we need to subtract the amount of gas used during the trip from the gas present in the gas tank at the begining.

So, the amount of gas she have left is $\frac{37}{2} -\frac{71}{4}$

$=\frac{74-71}{4}$

$=\frac{3}{4}$

Hence, she left with $\frac{3}{4}$ gallon of gas.

So, option D is correct.

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Irina18 [472]

Sh(2x) = (e^2x + e^-2x)/2

<span>Thus the integral becomes </span>

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2 years ago

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

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Dmitry [639]

Answer:

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Step-by-step explanation:

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40 - 52.5

= -12.5 feet

= 12.5 feet below the water

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2 years ago

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JulsSmile [24]

Answer:

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2 years ago

drek231 [11]

Answer:

y=-3x-5

Step-by-step explanation:

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divide them (-6/2 and -10/2) to get

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Hope i helped

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2 years ago