Ask question

Ask question

All categories

3 years ago

5

A relation is a special type of function. A. True B. False (30) points

1 answer:

Debora [2.8K]3 years ago

7 0

No. A relation is not a special type of function.

You might be interested in

Please help I don’t think I remember this

nikdorinn [45]

Answer:

im not sure if this is right

Step-by-step explanation:

demension 1: 9 X 10 demension 2: 10 x 4

4 0

3 years ago

Read 2 more answers

What's 18 divided by 5634❤️Pls help

White raven [17]

It's approximately 0.0032

4 0

3 years ago

Read 2 more answers

Ally is making a scale diagram of her classroom. She uses a scale factor of 3 centimeters per foot to draw the diagram. The actu

Gnoma [55]

The area of the scale drawing would be 1080 square centimeters or (1080 cm^2). I got this answer through first finding the area of the classroom in feet (which is A= lw = [18][20] = 360 ft^2) and then I proceeded to converting it to centimeters via given scale factor (3 cm = 1 foot) and had the answer of 1080 centimeters squared.

7 0

3 years ago

Read 2 more answers

An advertising company designs a campaign to introduce a new product to a metropolitan area of population 3 Million people. Let

Advocard [28]

Answer:

$P(t)=3,000,000-3,000,000e^{0.0138t}$

Step-by-step explanation:

Since P(t) increases at a rate proportional to the number of people still unaware of the product, we have

$P'(t)=K(3,000,000-P(t))$

Since no one was aware of the product at the beginning of the campaign and 50% of the people were aware of the product after 50 days of advertising

<em>P(0) = 0 and P(50) = 1,500,000 </em>

We have and ordinary differential equation of first order that we can write

$P'(t)+KP(t)= 3,000,000K$

The <em>integrating factor </em>is

$e^{Kt}$

Multiplying both sides of the equation by the integrating factor

$e^{Kt}P'(t)+e^{Kt}KP(t)= e^{Kt}3,000,000*K$

Hence

$(e^{Kt}P(t))'=3,000,000Ke^{Kt}$

Integrating both sides

$e^{Kt}P(t)=3,000,000K \int e^{Kt}dt +C$

$e^{Kt}P(t)=3,000,000K(\frac{e^{Kt}}{K})+C$

$P(t)=3,000,000+Ce^{-Kt}$

But P(0) = 0, so C = -3,000,000

and P(50) = 1,500,000

so

$e^{-50K}=\frac{1}{2}\Rightarrow K=-\frac{log(0.5)}{50}=0.0138$

And the equation that models the number of people (in millions) who become aware of the product by time t is

$P(t)=3,000,000-3,000,000e^{0.0138t}$

5 0

3 years ago

What is the slope of the line represented by the equation?<br> –2x + y = –1

Harrizon [31]

Isolate the y to one side so we get: y = 2x - 1
The slope of the line is 2.

8 0

2 years ago

Read 2 more answers