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2 years ago

Can u tell what is 4/5 minus 3/10

Mathematics

2 answers:

algol [13]2 years ago

8 0

Answer:

4/5 - 3/10 is 0.5.

Step-by-step explanation:

You want to divide top and bottom (Whatever you do to one side do to the other) So start off dividing 5 by 10 which wil give you 0.5, next you want to divide 4 by 3 which gives a repeating decimal so youll use the whole decimal (0.5) as an answer

balandron [24]2 years ago

3 0

Answer: 1/2

Explanation: To subtract unlike fractions such as 4/5 - 3/10, first find a common denominator.

The common denominator of 5 and 10 is the

least common multiple of 5 and 10 which is 10.

To get a 10 in the denominator of 4/5, we multiply both the numerator and the denominator of 4/5 by 2 to get the equivalent fraction 8/10.

Notice that 3/10 already has a 10 in the

denominator so it stays the same.

So we have 8/10 - 3/10 which is 5/10.

5/10 can be simplified to 1/2.

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1 year ago

M(x)=−(x−5) 2 +25m, left parenthesis, x, right parenthesis, equals, minus, left parenthesis, x, minus, 5, right parenthesis, squ

Aleksandr [31]

Answer:

Step-by-step explanation:

Given the function that represents the amount of mosquitoes as;

M(x) = -(x-5)² + 25

To get the maximum mosquitoes possible, we need to first find x;

At the maximum dM/dx = 0

dM//dx = -2(x-5)

0 = -2(x-5)

-2(x-5) = 0

-2x + 10 = 0

-2x = -10

x = 10/2

x = 5

Substitute x = 5 into the function;

M(5) = −(5−5)² +25

M(5) = 0+25

M(5) = 25

Hence the maximum number of mosquitoes is 25

8 0

2 years ago

Either enter an exact answer in terms of T or use 3.14 for 7 and enter your answer as a decimal.

muminat

Answer:

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Step-by-step explanation:

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2 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

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Answer:

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