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sashaice [31]
2 years ago
14

While traveling to Europe, Phelan exchanged 250 US dollars for euros. He spent 150 euros on his trip. After returning to the Uni

ted States he converts his money back to US dollars. How much of the original 250 US dollars does Phelan now have? Round to the nearest cent. 1 European euro = 1.3687 US dollars
Mathematics
1 answer:
storchak [24]2 years ago
8 0

Answer: Phelan now have 44.69 US dollars. of the original 250 US dollars.

Step-by-step explanation:

Given, Phelan spent 150 euros on his trip.

1 European euro = 1.3687 US dollars

So, 150 euros = 150 × 1.3687 US dollars

= 205.305 US dollars

≈ 205.31 US dollars [ Rounded to the nearest cent. ]

Initially he had 250 US dollars, now he has left (250-205.31)US dollars that is

44.69 US dollars.

Hence, Phelan now have 44.69 US dollars. of the original 250 US dollars.

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The cosine function takes x values from all real numbers.

Therefore, the domain of the cosine function is a real numbers.
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1 year ago
The art museum is 8 1/2 miles from Alison's house. Alison has ridden her bike 2/3 of the way there so far. How far how she gone?
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Alison is 2/3 of the way to the museum, so the answer is about 2/3 of 9, or 6 miles.  Let's calculate it:

  2                              17 miles
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Robert has $50 to spend on his utility bills each month. The basic monthly charge is 23.77 electricity costs 0.1117 for each kil
Lilit [14]

Answer:

<em>The maximum number of kilowatt-hours is 235</em>

Step-by-step explanation:

<u>Inequalities</u>

Robert's monthly utility budget is represented by the inequality:

0.1116x + 23.77 < 50

Where x is the number of kilowatts of electricity used.

We are required to find the maximum number of kilowatts-hours used without going over the monthly budget. Solve the above inequality:

0.1116x + 23.77 < 50

Subtracting 23.77:

0.1116x < 50 - 23.77

0.1116x < 26.23

Dividing by 0.1116:

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1 year ago
You are traveling down a country road at a rate of 95 feet/sec when you see a large cow 300 feet in front of you and directly in
emmainna [20.7K]

Answer:

1) You can rely solely on your brakes because when doing so the car will just travel 250ft from the point you hit your brakes till the point the car stopped completely, leaving you 50ft away from the cow.

2) See attached picture.

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in ft/s^{2}

3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.

4) j(t)=\left \{ {{95t-9t^{2}; 0\le t

j'(t)=\left \{ {{95-18t; 0\leq t

(see attached picture for graph)

Step-by-step explanation:

1) In this part of the problem we need to find the time when the speed of the car is 0. Gets to a complete stop. For this we will need to take the derivative of the position function so we get:

j(t)=95t-9t^2

j'(t)=95-18t

and we set the first derivative equal to zero so we get:

95-18t=0

and solve for t

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t=\frac{95}{18}

t=5.28s

so now we calculate the position of the car after 5.28 seconds, so we get:

j(5.28)=95(5.28)-9(5.28)^{2}

j(5.28)=250.69ft

so we have that the car will stop 250.69ft after he hit the brakes, so there will be about 50ft between the car and the cow when the car stops completely, so he can rely just on the breaks.

2) For answer 2 I take the second derivative of the function so I get:

j(t)=95t-9t^{2}

j'(t)=95-18t

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and then we graph them. (See attached picture)

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

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j"(t) represents the acceleration of the car t seconds after the brakes have been hit in ft/s^{2}

3)  yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.

4) j(t)=\left \{ {{95t-9t^{2}; 0\le t

j'(t)=\left \{ {{95-18t; 0\leq t

(see attached picture for graph)

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