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Reika [66]
4 years ago
5

A cone - shaped building has a height of 6 meters and a base with a diameter of 10 meters. The building will be filled with road

salt that costs $25 per cubic meter. How much will it cost to fill the building with road salt? Use 3.14 for π Round to the nearest hundredth HELP ME ASAP!!!!

Mathematics
2 answers:
AveGali [126]4 years ago
7 0
Formula
V= (1/3) pi r^2 h
r = d/2 = 10/2 = 5
h = 6
pi = 3.14
V = (1/3) * 3.14 * 5^2 * 6
V = (1/3) * 471
V = 157 cubic meters.

1 cubic foot cost $25
157 cubic feet cost x Cross multiply.

x = 25 * 157
x = 3925 dollars.



MArishka [77]4 years ago
3 0
The formula for the volume of a cone is
.. V = (1/3)*π*r^2*h
The radius (r) is half the diameter, so is 5 m. Fill in the numbers and do the arithmetic.
.. V = (1/3)*3.14*(5 m)^2*(6 m) = 157 m^3

The cost will be $25 for each of those cubic meters, so is
$25*157 = $3925.00.

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Step-by-step explanation:

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3 years ago
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Determine two pairs of polar coordinates for the point (5, 5) with 0° ≤ θ < 360°.
inna [77]

Answer:

First part

The answer is (5 square root 2, 45°), (-5 square root 2, 225°) ⇒ answer (d)

Second part

The equation in standard form for the hyperbola is y²/81 - x²/19 = 1 ⇒ answer(b)

Step-by-step explanation:

First part:

* Lets study the Polar form and the Cartesian form

- The important difference between Cartesian coordinates and

  polar coordinates:

# In Cartesian coordinates there is exactly one set of coordinates

  for any given point.

# In polar coordinates there is an infinite number of coordinates

   for a given point. For instance, the following four points are all

   coordinates for the same point.

# In the polar the coordinates the origin is called the pole, and

  the x axis is called the polar axis.

# The angle measurement θ can be expressed in radians

   or degrees.

- To convert from Cartesian Coordinates (x , y) to

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# r = ± √(x² + y²)

# θ = tan^-1 (y / x)

* Lets solve the problem

- The point in the Cartesian coordinates is (5 , 5)

∵ x = 5 and y = 5

∴ r = ± √(5² + 5²) = ± √50 = ± 5√2

∴ tanФ = (5/5) = 1

∵ tanФ is positive

∴ Angle Ф could be in the first or third quadrant

∵ Ф = tan^-1 (1) = 45°

∴ Ф in the first quadrant is 45°

∴ Ф in the third quadrant is 180 + 45 = 225°

* The answer is (5√2 , 45°) , (-5√2 , 225°)

Second part:

* Lets study the standard form of the hyperbola equation

- The standard form of the equation of a hyperbola with  

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  y²/a² - x²/b² = 1, where

• the length of the transverse axis is 2a

• the coordinates of the vertices are (0 , ±a)

• the length of the conjugate axis is 2b

• the coordinates of the co-vertices are (±b , 0)

•      the coordinates of the foci are (0 , ± c),  

• the distance between the foci is 2c, where c² = a² + b²

* Lets solve our problem

∵ The vertices are (0 , 9) and (0 , -9)

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∵ The foci at (0 , 10) , (0 , -10)

∴ c = ± 10

∵ c² = a² + b²

∴ (10)² = (9)² + b² ⇒ 100 = 81 + b² ⇒ subtract 81 from both sides

∴ b² = 19

∵ The equation is  y²/a² - x²/b² = 1

∴  y²/81 - x²/19 = 1

* The equation in standard form for the hyperbola is y²/81 - x²/19 = 1

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Step-by-step explanation:

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