(3.01)

Laura lives 15 miles east of Kevin’s place. Kevin lives 8 miles south of Michelle’s place. How far does Michelle live from Laura

My name is Ann [436]

Answer:

17 miles.

Step-by-step explanation:

Let's define:

North as the positive y-axis

East as the positive x-axis.

We know that Laura lives 15 miles east of Kevin's place.

Kevin lives 8 miles south of Michelle's place.

So, if we define the origin, (0, 0) as Laura's place.

From:

"that Laura lives 15 miles east of Kevin's place."

We have that the location of Kevin's house is 15 miles west from Laura's place, then Kevin's house is at:

(0, 0) + (-15mi, 0) = (-15mi, 0)

From Kevin lives 8 miles south of Michelle's place, we know that Michelle's live 8 miles north of Kevin's place.

Then the location of Michele's house is the location of Kevin's plus (0, 8mi).

Michelle's house is located at:

(-15mi, 0) + (0, 8mi) =(-15mi, 8mi)

Now we want to find the distance between Michelle's house and Laura's house.

Michelle's house is at (-15mi, 8mi)

Laura's house is at (0mi, 0mi)

Remember that the distance between two points (a, b) and (c, d) is given by:

$D = \sqrt{(a - c)^2 + (b - d)^2}$

Then the distance between (-15mi, 8mi) and (0mi, 0mi) is:

$D = \sqrt{(-15mi - 0mi)^2 + (8mi - 0mi)^2} = 17mi$

The correct option is the first one, 17 miles.

6 0

3 years ago

A. IfA= {a, b} and B = {p, q, r}, find A x B and B x A using tree diagram.

Katarina [22]

Answer:

Step-by-step explanation:

8 0

3 years ago

A motel owner observes that when a room is priced at $60 per day, all 80 rooms of the motel are occupied. For every $3 rise in t

Zinaida [17]

Answer:

a) p(x) = 300 - 3

b) P(x) = -3 x² + 285 x

c) Price of per room per day = $ 157.5

when Number of rooms occupied , x = 47.5

Step-by-step explanation:

Given - A motel owner observes that when a room is priced at $60 per day, all 80 rooms of the motel are occupied. For every $3 rise in the charge per room per day, one more room is vacant. Each occupied room costs an additional $15 per day to maintain.

To find - a) Find the demand function, expressing p, the price charged for each room per day, as a function of x, the number of rooms occupied.

b) Find the profit function P(x).

c) Find the price of per room per day the motel should charge in order to maximize its profit.

Proof -

a)

Let

(x, y) be the point

where x represents number of rooms occupied

and y represents price of room per day.

Now,

Given that,

a room is priced at $60 per day, all 80 rooms of the motel are occupied.

So, point becomes (80, 60)

And given that For every $3 rise in the charge per room per day, one more room is vacant.

So, point becomes (79, 63)

Now, we have two points (80, 60), (79, 63)

Let us assume that,

p(x) be the price charged for each room per day

Now,

By using point - slope formula , we get

p -60 = $\frac{(63 - 60}{(79 - 80)} (x - 80)$

⇒p -60 = (-3)(x-80)

⇒p-60 = 240 -3 x

⇒p(x) = 240 + 60 -3 x

⇒p(x) = 300 - 3 x

b)

Given that,

Each occupied room costs an additional $15 per day to maintain.

Let C(x) be the cost function,

Then C(x) =15 x

now,

Revenue function,

R(x) =x*p

= x*(300 -3 x )

= 300 x - 3 x²

⇒R(x) = 300 x - 3 x²

Now,

We know

Profit function = Revenue function - Cost function

⇒P(x) = R(x)-C(x)

⇒P(x) = (300 x -3 x²) -15 x

⇒P(x) = -3 x² + 285 x

c)

P'(x) = -6 x +285

For Maximize profit , Put P'(x) = 0

⇒-6 x+ 285 =0

⇒6 x= 285

⇒x = $\frac{285}{6}$

⇒x= 47.5

∴ we get

Maximize profit is when price, p = 300 - 3x

= 300 -3(47.5)

= $157.5

⇒Price of per room per day = $ 157.5

when Number of rooms occupied , x = 47.5

5 0

3 years ago

If its -26 degree in Canada and 34 degrees in Australia what is the difference in temperature

tatyana61 [14]

Answer:

It would be 60 degrees

Step-by-step explanation:

You take the numbers and write the equation 34-(-26) to figure out the difference. You then change the signs, making it 34+26, as before you were subtracting a negative. If you complete that problem, you come to the conclusion of 60.

6 0

3 years ago

Two thirds of the families in smithville own their own homes. The total number of home owning families there is 480. How many fa

Leviafan [203]

X- the total number of families in Smithville<span>.
(2/3)x - the number of home owning families.
At the same time, </span> the number of home owning families is 480.

(2/3)x=480
x=480*3/2=720.

720 <span>families live in Smithville.</span>

8 0

3 years ago