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2 years ago

Common denominator for 5 and 2?

Mathematics

2 answers:

Ghella [55]2 years ago

7 0

Answer:

Common denominator of 2 and 5 if you want to add or subtract two fractions with 2 and 5 as denominators.

Step-by-step explanation:

Step 1) Take the LCD and divide each denominator by it as follows:

10/2 = 5

10/5 = 2

Step 2) Multiply each nominator with the respective answers from Step 1:

3 x 5 = 15

2 x 2 = 4

Step 3) Put it all together to solve the problem:

15/10 + 4/10 = 19/10

= 3/2 + 2/5 = 19/10

=10

frutty [35]2 years ago

3 0

Answer:

It's 10

Step-by-step explanation:

hope it will help u .....

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PLEASE ANSWER ASAP CORRECT ANSWER GETS BRAINLIST

geniusboy [140]

I guess the number 2 is to indicate the question...

Then the equation would be 1/3 - 1/2x = -2/3

1/3 - 1/2x - 1/3 = -2/3 - 1/3

-1/2x = -3/3

-1/2x = -1

-1/2x *2 = - 1*2

-x = -2

-x * -1 = -2 * -1

x = 2

Good luck

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At the beach this weekend, the ratio of sandcastles to kites was 3 to 5. If there were 15 kites spotted, how many items were see

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Answer: 24

Step-by-step explanation:

The ratio of sandcastles to kites was 3 to 5 so if 15 kites were spotted that meant that the 5 was multiplied by 3 so then we multiply the 3 by 3. Now there are 9 sandcastles and 15 kites so now we add 9 to 15 to figure out how many items were seen in all. Nine added to 15 equals 24 which means that 24 items were seen in total.

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Soloha48 [4]

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

You operate a gaming Web site, www.mudbeast.net, where users must pay a small fee to log on. When you charged $3 the demand was

Doss [256]

Answer:

A) The linear relation between price and demand is:

$d=-550x+2750$

The revenue R is:

$R=-550x^2+2750x$

B) The profit functionP is:

$P=-550x^2+2750x-30$

C) The largest monthly profit is obtained with a log-on fee of $2.5 per month. This corresponds to a profit of $3407.5.

Step-by-step explanation:

We have a site where the number of log-ons depends on our monthly fee. A linear relation is established between the price (log-on fee) and the number of log-ons.

We have two points for this linear relationship:

At price x=3, the demand is d=1100.
At price x=2.5, the demand is d=1375.

We will model the relation:

$d=mx+b$

We can calculate the slope m as:

$m=\dfrac{\Delta d}{\Delta x}=\dfrac{d_2-d_1}{x_2-x_1}=\dfrac{1375-1100}{2.5-3}\\\\\\m=\dfrac{275}{-0.5}=-550$

Then, replacing one point in the linear equation, we can calculate the intercept b:

$d_1=mx_1+b\\\\1100=(-550)\cdot 3+b\\\\1100=-1650+b\\\\b=1100+1650=2750$

Then, the linear relation between demand and price is:

$d=-550x+2750$

The revenue R can be expressed as the multiplication of the price and the demand:

$R=x\cdot d=x(-550x+2750)=-550x^2+2750x$

If we have a fixed cost of $30 per month, the profit P is:

$P=R-FC=-550x^2+2750x-30$

We can maximize the profit by deriving the profit function and making it equal to zero.

$\dfrac{dP}{dx}=0\\\\\\\dfrac{dP}{dx}=-550(2x)+2750(1)=0\\\\\\-1100x+2750=0\\\\x=\dfrac{2750}{1100}=2.5$

This corresponds to a profit of:

$P(2.5)=-550(2.5)^2+2750(2.5)-30\\\\P(2.5)=-550\cdot 6.25+6875-30\\\\P(2.5)=-3437.5+6875-30\\\\P(2.5)=3407.5$

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2 years ago