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Dahasolnce [82]
3 years ago
9

If you multiply 1m times unlimedted what will u get

Mathematics
2 answers:
erastovalidia [21]3 years ago
6 0
You will get unlimited/ infinite
hjlf3 years ago
3 0

Answer:

∞

simple and correct answer, infinite!

✌️:)

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1. The manager of FrozonAir Refrigerator factory notices that on Monday it cost the company a total of $25,000 to build 30 refri
Vinvika [58]

Answer:

The fixed costs per day for the factory is $10,000 and marginal cost of one refrigerator for $500

Step-by-step explanation:

Let

Cost per refrigerator (marginal cost) = r

Fixed costs per day for the factory = f

Total cost = variable cost + fixed cost

Monday

25,000 = 30r + f

Tuesday

30,000 = 40r + f

Using Monday,

f = 25000 -30r

Substitute f = 25000 - 30r into Tuesday

30,000 = 40r + f

30,000 = 40r + (25,000 -30r)

30,000 = 40r + 25,000 - 30r

Collect like terms

30,000 - 25,000 = 40r - 30r

5,000 = 10r

Divide both sides by 10

r = 5,000 / 10

= 500

r = $500

Substitute r= 500 into Monday equation

f = 25,000 -30r

= 25,000 - 30(500)

= 25,000 - 15,000

= 10,000

f = $10,000

Therefore, the fixed costs per day for the factory is $10,000 and marginal cost of one refrigerator for $500

7 0
3 years ago
How many boards he can paint
katrin2010 [14]
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4 0
3 years ago
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BARSIC [14]
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5 0
3 years ago
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The sum of the angle measures of any triangle is
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3 0
3 years ago
an inverted conical water tank with a height of 20 ft and a radius of 8 ft is drained through a hole in the vertex (bottom) at a
viktelen [127]

Answer:

the rate of change of the water depth when the water depth is 10 ft is;  \mathbf{\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi} \  \ ft/s}

Step-by-step explanation:

Given that:

the inverted conical water tank with a height of 20 ft and a radius of 8 ft  is drained through a hole in the vertex (bottom) at a rate of 4 ft^3/sec.

We are meant to find the  rate of change of the water depth when the water depth is 10 ft.

The diagrammatic expression below clearly interprets the question.

From the image below, assuming h = the depth of the tank at  a time t and r = radius of the cone shaped at a time t

Then the similar triangles  ΔOCD and ΔOAB is as follows:

\dfrac{h}{r}= \dfrac{20}{8}    ( similar triangle property)

\dfrac{h}{r}= \dfrac{5}{2}

\dfrac{h}{r}= 2.5

h = 2.5r

r = \dfrac{h}{2.5}

The volume of the water in the tank is represented by the equation:

V = \dfrac{1}{3} \pi r^2 h

V = \dfrac{1}{3} \pi (\dfrac{h^2}{6.25}) h

V = \dfrac{1}{18.75} \pi \ h^3

The rate of change of the water depth  is :

\dfrac{dv}{dt}= \dfrac{\pi r^2}{6.25}\  \dfrac{dh}{dt}

Since the water is drained  through a hole in the vertex (bottom) at a rate of 4 ft^3/sec

Then,

\dfrac{dv}{dt}= - 4  \ ft^3/sec

Therefore,

-4 = \dfrac{\pi r^2}{6.25}\  \dfrac{dh}{dt}

the rate of change of the water at depth h = 10 ft is:

-4 = \dfrac{ 100 \ \pi }{6.25}\  \dfrac{dh}{dt}

100 \pi \dfrac{dh}{dt}  = -4 \times 6.25

100  \pi \dfrac{dh}{dt}  = -25

\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi}

Thus, the rate of change of the water depth when the water depth is 10 ft is;  \mathtt{\dfrac{dh}{dt}  = \dfrac{-25}{100  \pi} \  \ ft/s}

4 0
4 years ago
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