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3 years ago

15

A mechanic charges $55 an hour for an oil change and spends $45 a day on supplies. Which expression represents the amount the me

chanic earns in one day?

1 answer:

san4es73 [151]3 years ago

5 0

Answer:

55x-45=y

Step-by-step explanation:

in terms of algeabra 55 dollars per hour x amount of hours 45 spent so subtract 45; y is the amount per day. The variables can be swapped

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4 years ago

The equation y=123.1(1.065)* models the number of college students (in thousands) who studied abroad each year from 1998 through

xenn [34]

The estimate of the number of students studying abroad in 2003 is 169 and the estimate of the number of students studying abroad in 2018 is 433

<h3>a. Estimate the number of students studying abroad in 2003.</h3>

The function is given as:

y = 123(1.065)^x

Where x represents years from 1998 to 2013

2003 is 5 years from 1998.

This means that

x = 5

Substitute the known values in the above equation

y = 123(1.065)^5

Evaluate the exponent

y = 123 * 1.37008666342

Evaluate the product

y = 168.520659601

Approximate

y = 169

Hence, the estimate of the number of students studying abroad in 2003 is 169

<h3>b. Assuming this equation continues to be valid in the future, use this equation to predict the number of students studying abroad in 2018.</h3>

2018 is 20 years from 1998.

This means that

x = 20

Substitute the known values in the above equation

y = 123(1.065)^20

Evaluate the exponent

y = 123 * 3.52364506352

Evaluate the product

y = 433.408342813

Approximate

y = 433

Hence, the estimate of the number of students studying abroad in 2018 is 433

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2 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

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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

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AnnyKZ [126]

The degree is highest power in the monomials.

4 = 1

2z = 1

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4 years ago

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