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Zanzabum
3 years ago
12

Can you work out this math puzzle?

Mathematics
1 answer:
alexandr402 [8]3 years ago
5 0
Triangle=10
Circle=5
Square:3.5
Half square:1.75
Answer to ?:26.25
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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
to recover a large chair in your home, you purchase 9 1/2 yards of upholstery fabric at $11.00 per yard. If there is 7% sales ta
KATRIN_1 [288]
11.00*9.5=104.5
104.5*0.07=7.315
7.315+104.5=111.815
6 0
3 years ago
Write an equivalent expression for 4.5x(2 - 6)
ki77a [65]

4.5x(2 - 6) \\  \\ 1. \: 4.5x \times  - 4 \\ 2. \:  =  - 18x

5 0
3 years ago
Read 2 more answers
You estimate that a baby pig weighs 20 pounds. The actual weight of the baby pig is 16 pounds. Find the percent error
nlexa [21]
Divide 16 by 20 to find the percent.

16/20 = 0.8 = 8%

To check, multiply 20 by 0.08

20 x 0.08 = 16

Brainliest answer please? 3 more away from Expert!
7 0
4 years ago
Find the area of the figure. Round to the nearest tenth if necessary.
Anestetic [448]
So I went ahead and divided up the figure so that you would understand what I'm going to tell you. Let's start with the red rectangle.

The distance from point R to point I is 6 units and the distance from Point R to point S is 5 units. 6*5 = 30 so that red rectangle has an area of 30 units

Now let's calculate triangle AVI. Area of a triangle is Base * Height. The base is 1 and the height is 5 which means the area of triangle AVI is 5.

Next, triangle ALT. From Point A to point L is 4 unit. From Point L to point T is 3 units. 4*3 is 12. So triangle ALT has an area of 12.

Triangle TSI. From Point T to point L is 3 units. From Point L to X is 2. Triangle TSI has an area of 6.

Now we add all of them up. 30+5+12+6 = 53. In total, RSTUV has an area of 53

8 0
4 years ago
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