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Gnoma [55]
3 years ago
6

A square has a perimeter of 20 in what is the length of each side

Mathematics
1 answer:
Dvinal [7]3 years ago
5 0
It is a 4 by 5 because 4 times 5 eguels 20
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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

  Polar Coordinates (r , θ)

# 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  

  center (0 , 0) and transverse axis parallel to the y-axis is

  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)

∴ a = ± 9 ⇒ a² = 81

∵ 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

3 0
3 years ago
Geoffery went on a trip across South America. He took 1,750 US dollars (USD) with him. In Chile, he traded his money for Chilean
musickatia [10]
Given:
$1,750
Rate = $1 = CLP 496.50
Spent : 217,513 pesos
Rate = CLP 1 = VEF 0.1283
Spent : 38,422.19 bolivars
Rate = VEF 1 = PGY1063.95 
Spent : 12,857,441.39 guarani
Rate = PYG 1 = $0.00002186

$1750 * CLP 496.50 = CLP 868,875
CLP 868,875 - 217,513 = CLP 651,362

CLP 651,362 * VEF 0.1283 = VEF 83,569.7446
VEF 83,569.7446 - 38,422.19 = VEF 45,147.5546

VEF 45,147.5546 * PYG 1063.95 = PYG 48,034,740.70
PYG 48,034,740.70 - 12,857,441.39 = PYG 35,177,299.30

PYG 35,177,299.30 * $0.00002186 = $768.976763 ⇒ $768.98


4 0
4 years ago
Read 2 more answers
Find the distance between (-1,-5) and (4,3)
yawa3891 [41]

Answer:

\sqrt{89} = 9.83

Step-by-step explanation:

\sqrt{89}

\sqrt{25 + 64}  \\\sqrt{5^2 + 8^2}

7 0
3 years ago
Square of the complex number (i - 2)?
Elena L [17]
(i-2)^2=\\--------------------\\use:(a-b)^2=a^2-2ab+b^2\\--------------------\\=i^2-2\times i\times2+2^2=-1-4i+4=3-4i\\\\\\\\i=\sqrt{-1}\Rightarrow i^2=-1
7 0
3 years ago
Read 2 more answers
Which graph shows the sequence –7, –4, –1, 2, ...?
cricket20 [7]

The first graph is the answer
8 0
3 years ago
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