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tankabanditka [31]

2 years ago

10

Who can help ? please

1 answer:

Assoli18 [71]2 years ago

5 0

Answer:

The apple is definitely equal to 7 each apple.

Step-by-step explanation:

Not sure about the tomato and pear.

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What the value of x 4(2x-5)=5x+4

mrs_skeptik [129]

x=8

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saveliy_v [14]

Answer:

x < 3

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

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2 years ago

Whats 30 divided by 6/7

Andre45 [30]

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3 years ago

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Find f (0) and find x when f (x) =4

Anuta_ua [19.1K]

At x = 0, y-coordinate is at -4 so that means f(0) = -4

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Hope this helps.

4 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