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3 years ago

Solve for x: 1 04>x>7 0 4 0-2 >x> 1 -2

Mathematics

1 answer:

Luda [366]3 years ago

8 0

Answer:

Your answer is in Step by Step

Step-by-step explanation:

=104x^7040-2x^1-2

We move all terms to the left:

-(104x^7040-2x^1-2)=0

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What is two plus two

Strike441 [17]

well .... 2 + 2 = 4

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3 years ago

Read 2 more answers

Whats the answer two -x+4=1

blagie [28]

Its suposed to be-5+4=-1

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Before beginning voice lessons, Chad already knew how to sing 805 pieces, and he expects

NeTakaya

Let's create an equation to find the # of weeks until Chad learns 937 pieces.

x = # of weeks

937 = 2x + 805

Subtract 805 from both sides.

132 = 2x

Divide both sides by 2

66 = x

Chad will need 66 weeks of lessons to learn 937 pieces.

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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:

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

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3 years ago

Bob currently has $1,000 in a saving account he saves $25 per month he saves the same amount each month and does not take any mo

erastova [34]

In 18 months he will have 1450

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3 years ago