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marusya05 [52]
2 years ago
8

Answer 4. Pls pls ASAP ASAP pls ASAP please ASAp

Mathematics
1 answer:
vlabodo [156]2 years ago
8 0

Answer:

56 in2

Step-by-step explanation:

14x8x1/2 (1/2 because a trapezoid formula divides by 2)

112x1/2

=56

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The answer is C that is X=7 and X=-1

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3 years ago
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
lisov135 [29]

Answer:

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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?

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  • 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.

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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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