Answered

An electronics store discovers that it can sell 5 televisions per day by pricing them at $150. When the televisions are on sale for $100 the store sells 10 of them every day. Write a linear equation to compare the price of a television, p, to the number sold, x. Then write a quadratic equation to compare the revenue, m, from selling televisions to the number sold, x.

Answer :

luisejr77

Answer:

[tex]p = -10x + 200[/tex]

[tex]m = -10x^2 +200x[/tex]

Step-by-step explanation:

We know that at a price of $ 150, 5 televisions are sold and at a price of $ 100, 10 televisions are sold.

We must write a linear equation for this situation.

The equation of the line will have the following form

[tex]p = mx + b[/tex]

Where m is the slope of the line and b is the intercept with the p axis

[tex]m=\frac{y_2-y_1}{x_2-x_1}\\\\m=\frac{100-150}{10-5}=-10[/tex]

[tex]b=p_1-mx_1\\\\b=150-(-10)(5)\\\\b=200[/tex]

The equation is:

[tex]p = -10x + 200[/tex]

Now we know that the revenue m is the product of the price p for the quantity sold x.

[tex]m = p * x[/tex]

[tex]m=(-10x + 200)*x[/tex]

[tex]m = -10x^2 +200x[/tex]

Answer:

[tex]p=-10x+200[/tex]

[tex]m = -10x^2+200x[/tex]

Step-by-step explanation:

Given : An electronics store discovers that it can sell 5 televisions per day by pricing them at $150. When the televisions are on sale for $100 the store sells 10 of them every day.

To Find: Write a linear equation to compare the price of a television, p, to the number sold, x. Then write a quadratic equation to compare the revenue, m, from selling televisions to the number sold, x.

Solution:

Let p be the price and x be the no. of television sold

We are given that  it can sell 5 televisions per day by pricing them at $150. When the televisions are on sale for $100 the store sells 10 of them every day. i.e(5,150) and (10,100)

Now to find the linear equation to compare the price of a television, p, to the number sold, x.

We will use two point slope form

Formula : [tex]y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]

[tex](x_1,y_1)=(5,150)[/tex]

[tex](x_2,y_2)=(10,100)[/tex]

Substitute the values in the formula

[tex]y-150=\frac{100-150}{10-5}(x-5)[/tex]

[tex]y-150=\frac{-50}{5}(x-5)[/tex]

[tex]y-150=-10(x-5)[/tex]

[tex]y-150=-10x+50[/tex]

[tex]y=-10x+50+150[/tex]

[tex]y=-10x+200[/tex]

x is the no. of television sold

y is the price

Since p denotes the price

So,  [tex]p=-10x+200[/tex]

Thus a linear equation to compare the price of a television, p, to the number sold, x is   [tex]p=-10x+200[/tex]

Now [tex]Revenue = Cost \times quantity[/tex]

Since price of x telivisons = [tex]p=-10x+200[/tex]

So,  [tex]Revenue = (-10x+200) \times x[/tex]

[tex]Revenue = -10x^2+200x[/tex]

m denotes revenue

So,[tex]m = -10x^2+200x[/tex]

Thus a quadratic equation to compare the revenue, m, from selling televisions to the number sold, x. is [tex]m = -10x^2+200x[/tex]

Hence a linear equation to compare the price of a television, p, to the number sold, x is   [tex]p=-10x+200[/tex] and a quadratic equation to compare the revenue, m, from selling televisions to the number sold, x. is [tex]m = -10x^2+200x[/tex]

Other Questions