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3 years ago

10 lbs ≈

Mathematics

2 answers:

castortr0y [4]3 years ago

8 0

Answer:

The answer is D. 4.5

Step-by-step explanation:

Brainliest will be very well appreceated

sergejj [24]3 years ago

7 0

Answer:

D. 4.5 kg

Step-by-step explanation:

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Answer this question to get marked as brainliest plus 15 points!!!!

Alchen [17]

Answer:

Step-by-step explanation:

complementary means it equals 90

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3 years ago

This is a scale drawing of a field. The scale is 1 in.: 10 ft. What is the actual length of the

creativ13 [48]

Answer:7ft

Step-by-step explanation:

i got it right

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3 years ago

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What does x equal in 5x+y=-24

qaws [65]

Answer

24/5

Step-by-step explanation:

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4 years ago

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The total land area of the world is about fifty-seven million, two hundred eighty thousand square miles. Write that number

expeople1 [14]

The total land area of the world is about fifty-seven million, two hundred eighty thousand square miles is 57, 280, 000 square miles

<em><u>Solution:</u></em>

Given that, we have to write the given words in number

The total land area of the world is about fifty-seven million, two hundred eighty thousand square miles

The number of zeros in million is 6

Here given that 57 million

So, 57 million means 57, followed by 6 digits

two hundred eighty thousand means 280 followed by 3 zeros

two hundred eighty thousand = 280 thousand

Thus the number is: 57, 280, 000

5 0

3 years ago

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:

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

5 0

3 years ago