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sukhopar [10]
3 years ago
8

The distance between two cities is 145 miles. A truck can cover this distance in 2.5 hours. The speed of a car is 1.5 times fast

er than the speed of the truck. How long does it take them to meet, if they start moving towards each other simultaneously?
Mathematics
1 answer:
nadezda [96]3 years ago
4 0

The speed at which the distance is covered by the two vehicles is the sum of their speeds:

... (truck speed) + 1.5×(truck speed) = 2.5×(truck speed)


The time required to cover the distance is inversely proportional to speed, so is

... (2.5 hours)/2.5 = 1.0 hours


It takes 1.0 hours for the vehicles to meet.


_____

The speed of the truck is (145 mi)/(2.5 h) = 58 mi/h

The speed of the car is 1.5 × 58 mi/h = 87 mi/h.

The speed at which they move toward each other is

... 58 mi/h + 87 mi/h = 145 mi/h

So, the time it takes to cover 145 miles is

... (145 mi)/(145 mi/h) = 1.0 h

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What is the total amount that Matthew's bank will receive after lending him $8,000 for four years at an interest rate of 6 perce
bija089 [108]

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Total amount that Matthew's bank will receive is $10099.81.

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We need to find the amount Matthew's bank will receive after lending him $8,000 for four years at an interest rate of 6 percent, compounded annually.

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How many pounds of a 15% copper alloy must be mixed with 700lb of a 30% copper alloy to maybe a 25.5% copper alloy
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Let x represent the mass (in pounds) of that 15\% copper alloy required, such that the final mixture would contain 25.5\% copper by mass.

Consider: if x pounds of that 15\% copper alloy is mixed with 700 pounds that 30\% copper alloy, what would be the mass of copper in the mixture?

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Therefore, the mixture would contain (210 + 0.15\, x) \; \rm lb of copper.

The mass of that mixture would be (700 + x)\; \rm lb. The mass fraction of copper in that mixture would be:

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This ratio is supposed to be equal to 25.5\%. These two pieces of equations combine to give an equation about x:

\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\% = 25.5\%.

\displaystyle \frac{210 + 0.15\, x}{700 + x} = 0.255.

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210 + 0.15\, x= 0.255\, (700 + x).

(0.255 - 0.15)\, x= 210 - 0.255 \times 700.

\displaystyle x = \frac{210 - 0.255 \times 700}{0.255 - 0.15} = 300.

Therefore, 300\; \rm lb of that 15\% alloy would be required.

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