Mathematics, 03.02.2020 07:49 Alienchild239

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.

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Answer from: Elsalvador7310

p = -10x + 200

m = -10x^2 +200x

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

p = mx + b

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

$m=\frac{y_2-y_1}{x_2-x_1}\\\\m=\frac{100-150}{10-5}=-10$

$b=p_1-mx_1\\\\b=150-(-10)(5)\\\\b=200$

The equation is:

p = -10x + 200

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

m = p * x

m=(-10x + 200)*x

m = -10x^2 +200x

Answer from: marieminnerly1628

p=-10x+200

m = -10x^2+200x

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 : $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

(x_1,y_1)=(5,150)

(x_2,y_2)=(10,100)

Substitute the values in the formula

$y-150=\frac{100-150}{10-5}(x-5)$

$y-150=\frac{-50}{5}(x-5)$

y-150=-10(x-5)

y-150=-10x+50

y=-10x+50+150

y=-10x+200

x is the no. of television sold

y is the price

Since p denotes the price

So, p=-10x+200

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

Now $Revenue = Cost \times quantity$

Since price of x telivisons = p=-10x+200

So, $Revenue = (-10x+200) \times x$

Revenue = -10x^2+200x

m denotes revenue

So, m = -10x^2+200x

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

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

Answer from: Quest

add an image

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Answer from: Quest

cant open file sorry!

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An electronics store discovers that it can sell 5 televisions per day by pricing them at $150. when...