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diamong [38]
2 years ago
9

Type the missing number that makes these fractions equal: 42/48=?/24

Mathematics
1 answer:
elixir [45]2 years ago
6 0

Answer:

21

Step-by-step explanation:

cross multiply

42 x 24 = 48x

1008 = 48x

x = 21

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Given the function f(x) = 2x, find the value of f−1(32).
Oliga [24]

Answer:

it is -30

Step-by-step explanation:

you would divide both sides by x,then f=2

using PEMDAS we come to -30

5 0
3 years ago
Part A: Abe rented a bike at $36 for 5 days. If he rents the same bike for 8 days, he has to pay a total rent of $48. Write an e
worty [1.4K]

a)

y=4 x +(1.6 X 10)


b)

y=4x +16


c)

x=0 y=0+16 (0,16)

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7 0
3 years ago
Rewrite (123 + 456) + 789 using the Associative Law of Addition
adelina 88 [10]

Answer:

123 + (456 + 789)

Step-by-step explanation:

since this is an addition problem, the parenthesis don't matter, and they can be put wherever you want.

(123 + 456) + 789

123 + 456 + 789

(123 + 456 + 789)

123 + (456 + 789)

technically, any of these would work, but I think the kind of answer the problem is looking for is;

123 + (456 + 789)

5 0
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:

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.

4 0
3 years ago
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