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3 years ago

Does this image have a set of inputs to a set of possible outputs where each input is related to exactly one outputs Yes or No?

Mathematics

1 answer:

borishaifa [10]3 years ago

3 0

Answer: Yes

This graph passes the vertical line test. This is a test where we try to draw a single vertical line through more than one point on the curve. In this case, such a thing is not possible. Any input x leads to exactly one output y. This graph is a function.

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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

-11 2/3 x (-4 1/5)=?

qaws [65]

Answer:

Step-by-step explanation:

$(-11\frac{2}{3} )$ * $(-4\frac{1}{5} )$ $= (-\frac{35}{3} )(-\frac{21}{5})$

$(-\frac{35}{3} )(-\frac{21}{5})$ $=\frac{735}{15}$

$\frac{735}{15}=49$

6 0

3 years ago

What is the best approximation for the input value when f(x)=g(x)?

Lostsunrise [7]

Answer:

$x=0$ and $x=1$ .

Step-by-step explanation:

If we have to different functions like the ones attached, one is a parabolic function and the other is a radical function. To know where $f(x)=g(x)$ , we just have to equalize them and find the solution for that equation: