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konstantin123 [22]
3 years ago
6

Plato Question!!!!

Mathematics
2 answers:
Katyanochek1 [597]3 years ago
7 0

Answer:

(2, 120° )

Step-by-step explanation:

To convert from rectangular to polar form, that is

(x, y) → (r, Θ ), use

r = \sqrt{x^2+y^2}

Θ = tan^{-1}( \frac{y}{x})

here (x, y ) = (- 1, \sqrt{3})

r = \sqrt{(-1)^2+(\sqrt{3} } )^2

  = \sqrt{1+3} = \sqrt{4} = 2

Θ = tan^{-1}(\sqrt{3}) = 60° ← related acute angle

Note (- 1, \sqrt{3}) is in the second quadrant so Θ must be in the second quadrant.

Θ = 180° - 60° = 120°

(- 1, \sqrt{3}) → (2, 120°)

Keith_Richards [23]3 years ago
5 0

Answer:

Step-by-step explanation:

Answer:

Step-by-step explanation:

The polar form of a Cartesian coordinates is given by (r, @) where r = √x^2 + y^2

@ = tan^-1 y/x

r = √-1^2 + √3^2

r = √4

r= 2

@ = tan ^-1 √3/-1 -tan √3/1 = -60°

@ = 180-60 = 120°

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Sauron [17]

It is not a function with domain as { 9 / 2 , 0} and range as {1 / 4, -4, 6}.

We are given a relation:

{(9 / 2, 1 / 4),(4 1 / 2, -4),(0, 5 / 6)}

We can also write it as:

{(9 / 2, 1 / 4),(9 / 2, -4),(0, 5 / 6)}

The domain of the relation will be:

Domain = { 9 / 2 , 0}

The range of the relation will be:

Range = {1 / 4, -4, 6}

We know that every function can be a relation but every relation cannot be a function.

A relation means the connection between the input and the output.

A function means that for every input there should only be one output.

Here, this relation is not a function because:

The input 9 / 2 has 2 outputs as 1 / 4 and -4 which is not possible in a function.

Therefore, we get that it is not a function with domain as { 9 / 2 , 0} and range as {1 / 4, -4, 6}.

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Answer: x is less than or equal to -4.25


Step-by-step explanation:

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Topic: The Quadratic Formula
Finger [1]

Answer:

Step-by-step explanation:

The quadratic formula for a equation of form

ax²+bx + c = 0 is

x= \frac{-b +- \sqrt{b^2-4ac} }{2a}

For the first equation,

x²+3x-4=0,

we can match that up with the form

ax²+bx + c = 0

to get that

ax² =  x²

divide both sides by x²

a=1

3x = bx

divide both sides by x

3 = b

-4 = c

. We can match this up because no constant multiplied by x could equal x² and no constant multiplied by another constant could equal x, so corresponding terms must match up.

Plugging our values into the equation, we get

x= \frac{-3 +- \sqrt{3^2-4(1)(-4)} }{2(1)} \\= \frac{-3+-\sqrt{25} }{2} \\ = \frac{-3+-5}{2} \\= -8/2 or 2/2\\=  -4 or 1

as our possible solutions

Plugging our values back into the equation, x²+3x-4=0, we see that both f(-4) and f(1) are equal to 0. Therefore, this has 2 real solutions.

Next, we have

x²+3x+4=0

Matching coefficients up, we can see that a = 1, b=3, and c=4. The quadratic equation is thus

x= \frac{-3 +- \sqrt{3^2-4(1)(4)} }{2(1)}\\= \frac{-3 +- \sqrt{9-16} }{2}\\= \frac{-3 +- \sqrt{-7} }{2}\\

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4x² + 1= 4x,

we can start by subtracting 4x from both sides to maintain the desired form, resulting in

4x²-4x+1=0

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x=\frac{-(-4) +- \sqrt{(-4)^2-4(4)(1)} }{2(4)} \\= \frac{4+-\sqrt{16-16} }{8} \\= \frac{4+-0}{8} \\= 1/2

Plugging 1/2 into 4x²+1=4x, this works as the only solution. This equation has one real solution

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