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Marina CMI [18]
3 years ago
6

Three boxes are stacked one on top of the other. One box is 5 feet 10 inches tall , one is 4 feet 2 inches tall and one is 2 fee

t 3 inches tall. How high is the stack ?
Mathematics
1 answer:
g100num [7]3 years ago
7 0

Answer:

136 inches

Step-by-step explanation:


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Write the trig ratios based on E
lidiya [134]

Answer:

\

Step-by-step explanation:

8 0
3 years ago
A 1,600.00 principal earns 7% interest compounded semiannually twice per year' after 33 years what is the balance in the account
Dmitrij [34]
A=1600(1+0.07/2)^(2*33)=15,494.7
7 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
Solve for b:<br><br> 6b+14 = -7-b
vladimir2022 [97]

Answer:

b = -3

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

6b+14=−7−b

6b+14=−7+−b

6b+14=−b−7

Step 2: Add b to both sides.

6b+14+b=−b−7+b

7b+14=−7

Step 3: Subtract 14 from both sides.

7b+14−14=−7−14

7b=−21

Step 4: Divide both sides by 7.

\frac{7b}{7} = \frac{-21}{7}

b= -3

3 0
3 years ago
Read 2 more answers
A rectangular area adjacent to a river is fenced​ in; no fence is needed on the river side. The enclosed area is 1500 square fee
ZanzabumX [31]

Answer:

a) C(x) = 15000/x + 6x +80

b) Domain of C(x)  { R x>0 }

Step-by-step explanation:

We have:  

Enclosed area = 1500 ft²   = x*y      from which     y  =  1500 / x    (a) where x is perpendicular to the river

Cost = cost of sides of fenced area perpendicular to the river  + cost of side parallel to river + cost of 4 post then

Cost = 10*y + 2*3*x + 4*20 and accoding to (a)  y = 1500/x

Then

C(x)  = 10* ( 1500/x ) + 6*x + 80

C(x) = 15000/x + 6x +80

Domain of C(x)  { R x>0 }

5 0
3 years ago
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