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SVETLANKA909090 [29]

2 years ago

11

X - 5 < - 3 Or X + 8 > 2

1 answer:

Alex Ar [27]2 years ago

7 0

Step-by-step explanation:

x-5 < -3

or, x < -3 +5

or, x < 2

Now,

x +8 >2

or, x >2 - 8

or, x >-6

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What is the ratio of 5.478 and 7.522 in mathematics

exis [7]

Answer:

Written in standard ratio form-

5.478 : 7.522

other ways of expressing it-

5.478/7.522 or

5.478 to 7.522

Simplified-

0.72826..... as an irrational number

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3 years ago

Compare the exponential functions f, g, and h which are shown below.

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3 years ago

4 1/2 ounces per 3/4 miles<br> _____ ounces per 1 mile

adelina 88 [10]

Answer: 6

Step-by-step explanation:

To get 3/4 miles to 1 mile we must take the reciprocal of the number and multiply it.

$\frac{3}{4}$ × $\frac{4}{3} =1$

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5 0

3 years ago

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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

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.

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3 years ago

12. Joe counted 5 birch trees in the park. Sara counted 235 pine trees. How many times as many pine trees as birch trees did the

Fittoniya [83]

Answer:

47

Step-by-step explanation: make sure to divide correctly

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