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2 years ago

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Kathleen lost a tooth today. Now she has lost 4 more than her sister Cara lost. Write an expression to represent the number of t

eeth Cara has lost. Let represent the number of teeth Kathleen lost. Expression:

1 answer:

Alex Ar [27]2 years ago

7 0

Answer:

k-4=c

Step-by-step explanation:

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Answer:

I guess the correct answer is 116 cm

Step-by-step explanation:

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3 years ago

Solve the equation. <br><br><br> 12=2+z/-6

Bogdan [553]

Answer:

$\boxed{\bold{z=-60}}$

Step-by-step explanation:

Switch Sides

$\bold{2+\frac{z}{-6}=12}$

Subtract 2 From Both Sides

$\bold{2+\frac{z}{-6}-2=12-2}$

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$\bold{\frac{z}{-6}=10}$

Multiply Both Sides By -6

$\bold{\frac{z\left(-6\right)}{-6}=10\left(-6\right)}$

Simplify

$\bold{z=-60}$

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3 years ago

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Answer: y = x - 2

Step-by-step explanation:

Subtract 5 from both sides.

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3 years ago

{6 3/6 + 10 4/6) + 9 2/6

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Answer:

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Step-by-step explanation:

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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:

$300\; \rm lb$ .

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

2 years ago