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nekit [7.7K]
3 years ago
14

Lorena is driving a truck that is painting the lines on a new road. The lines she paints are 0.152 meters wide, and she will pai

nt a total length of 5.45 × 105 meters of lines. The shape created by the lines will be a rectangle, and the area will be the product of the length and width.
Approximately what area of the road will Lorena be painting?

5.602 × 10 (5) m2
8.284 × 10 (4) m2
8.698 × 10 (1) m2
8.284 × 10 (5) m2
5.602 × 10 (4) m2

Mathematics
1 answer:
Alina [70]3 years ago
5 0

Answer:

  8.284 × 10^4 m^2

Step-by-step explanation:

The area is the product of length and width:

  (5.45·10^5 m)·(0.152 m) = 0.8284·10^5 m^2 = 8.284·10^4 m^2

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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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300\; \rm lb.

Step-by-step explanation:

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?

  • Mass of copper in x pounds of that 15\% copper alloy: (0.15\, x)\; \rm lb.
  • Mass of copper in 700 pounds of that 30\% copper alloy: 700 \times 0.30 = 210\; \rm lb.

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:

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

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.

Simplify and solve for x:

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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