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Ratling [72]
3 years ago
8

In a volcano, erupting lava flows continuously through a tube system about 20 kilometers to the sea. Assume a lava flow of 0.5 k

ilometer per hour and calculate how long its takes to reach the sea.
Mathematics
1 answer:
butalik [34]3 years ago
3 0

Answer:

40 hours

Step-by-step explanation:

Lets take "x" as the # of hours it takes the lava to reach the sea

since there is a lava flow of 0.5 kilos per hour

we will multiply "x" with 0.5

0.5x

now the tube system is 20 kilometers long

this means that at a rate of 0.5 kilos per hour the lava will reach the sea once it is 20 kilometers deep

so

0.5x=20

divide "0.5" by both sides to eliminate the coefficient

\frac{0.5x}{0.5} =\frac{20}{0.5}

x=40

it will take 40 hours for the lava to reach the sea

Hope this helps!

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Answer:

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Step-by-step explanation:

6 0
3 years ago
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Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. f(x,y)= e^xy; X
kotykmax [81]

Answer:

f(x,y) = e^{xy} is maximum at x = 2 and y = 2 and f(2,2) = e^{4}

Step-by-step explanation:

Since f(x,y) = e^{xy} and x³ + y³ = 16, Ф(x,y) = x³ + y³ - 16

df/dx = ye^{xy}, df/dy = xe^{xy}, dФ/dx = 3x² and dФ/dy = 3y²

From the method of Lagrange multipliers,

df/dx = λdΦ/dx and df/dy = λdΦ/dy

ye^{xy} = 3λx² (1) and  xe^{xy} = 3λy² (2)

multiplying (1) by x and (2) by y, we have

xye^{xy} = 3λx³ (4) and  xye^{xy} = 3λy³ (5)

So,  3λx³ = 3λy³

⇒ x = y

Substituting x = y into the constraint equation, we have

x³ + y³ = 16

x³ + x³ = 16

2x³ = 16

x³ = 16/2

x³ = 8

x = ∛8

x = 2 ⇒ y = 2, since x = y

So, f(x,y) = f(2,2) = e^{2 X2} = e^{4}

We need to determine if this is a maximum or minimum point by considering other points that fit into the constraint equation.

Since x³ + y³ = 16 when x = 0, y is maximum when y = 0, x = maximum

So, 0³ + y³ = 16

y³ = 16

y = ∛16

Also, when y = 0, x = maximum

So, x³ + 0³ = 16

x³ = 16

x = ∛16

and f(0,∛16) = e^{0X\sqrt[3]{16} } = e^{0} = 1.

Also, f(∛16, 0) = e^{\sqrt[3]{16}X0 } = e^{0} = 1.

Since f(0,∛16) = f(∛16, 0) = 1 < f(2,2) = e^{4}

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8 0
3 years ago
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OLga [1]

Answer:

1.) Exponential Growth

2.) Exponential Decay

3.) Exponential Growth

4.) Exponential Decay

Step-by-step explanation:

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            ↓

always increasing

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<em>             </em>↓

always decreasing

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            ↓

always increasing

<u>4.) </u><u><em>f (x) </em></u><u>= 320 (1/6)^</u><u><em>x</em></u>

<em>          </em>   ↓

always decreasing

<u><em>EXPLANATION:</em></u>

It's exponential growth when the base of our exponential is bigger than 1, which means those numbers get bigger. It's exponential decay when the base of our exponential is in between 1 and 0 and those numbers get smaller.

7 0
3 years ago
Assistance please!!!
Talja [164]

2:1 is the perimeter

3 0
3 years ago
The length of a rectangle is 4 cm less than it’s width. If the area of the rectangle is 165 cm^2. What are the dimensions of the
Monica [59]

Answer:

The dimensions are 11 cm by 15 cm

Step-by-step explanation:

Area of a rectangle = (length)(width).

Let L represent the length and W the width.

Since L = W - 4, we have

Area of rectangle = (W - 4)W = 165 cm^2.

Thus, w^2 - 4W - 165 = 0.  We can solve for W using the quadratic formula:

a = 1, b = -4 and c = -165.  Thus,

        -(-4) ± √ [ (-4)² - 4(1)(-165) ]

W = ---------------------------------------

                        2

         4 ± √ [16 + 660 ]

W = ------------------------------

                       2

         4 ± √ [ 676 ]                    4 ± 26

W = -----------------------  =   W = -------------  =  W  = 2 ± 13

                  2                                  2

Thus, W is 2 + 13 = 15.  It cannot be negative, so we discard 2 - 13 = -11.

If the width, W, is 15, then the length is 4 less, or L = 11.

The dimensions are 11 cm by 15 cm

6 0
3 years ago
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