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3 years ago

What is 17/147 in lowest terms

Mathematics

2 answers:

katovenus [111]3 years ago

5 0

Hi there!

To find your answer, let us see what number can go evenly into 17 and 147.

$\frac{17}{147}$

Since it looks like nothing can go into each number evenly, $\frac{17}{147}$ is already simplified down to its lowest terms and it can't go any lower than that.

Hope this helps! :D

Message me if you need anymore help on anything!

Feliz [49]3 years ago

4 0

17 is prime so it is in lowest terms already

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3 years ago

Find all polar coordinates of point P where P = ordered pair 3 comma negative pi divided by 3 .

jek_recluse [69]

Answer:

The all polar coordinates of P are:

(3 , -π/3) , (3 , 5π/3) , (-3 , 2π/3) , (-3 , -4π/3)

Step-by-step explanation:

* Lets study the polar coordinates of a point

- In polar coordinates there is an infinite number of coordinates

for a given point.

- The polar coordinates of a point (x , y) is (r , θ), where

r = √ ( x2 + y2 )

θ = tan-1 ( y / x )

# Ex: the following four points are all coordinates for the same point.

* (5 , π/3) = (5 , −5π/3) = (−5 , 4π/3) =(−5 , −2π/3)

- These four points only represent the coordinates of the point without

rotating more than once

- So the point (r,θ) can be represented by any of the following

coordinate pairs (r , θ + 2π n) and (−r , θ + (2n + 1) π), where n is

any integer.

* Now lets solve the problem

∵ P = (3 , -π/3)

∵ (r , θ + 2πn)

∴ r = 3 an d Ф = -π/3

- let n = 1

∴ P = (3 , -π/3 + 2π)

∴ P = (3 , 5π/3)

∵ P =(3 , -π/3)

∵ P = (-r , θ + (2n + 1) π)

Let n = 0

∴ P = (-3 , -π/3 + (2×0 + 1) π)

∴ P = (-3 , -π/3 + (0 + 1) π)

∴ P = (-3 , -π/3 + π)

∴ P = (-3 , 2π/3)

∵ P =(3 , -π/3)

∵ P = (-r , θ + (2n + 1) π)

Let n = -1

∴ P = (-3 , -π/3 + (2(-1) + 1) π)

∴ P = (-3 , -π/3 + (-2 + 1) π)

∴ P = (-3 , -π/3 + -π) = (-3 , -4π/3)

∴ P = (-3 , -4π/3)

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3 years ago

Simplify the exponential expression. (−8) 2 =

sweet-ann [11.9K]

Answer:

-16

Step-by-step explanation:

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3 years ago

Graph the hyperbola using the transverse axis, vertices, and co-vertices:

Reptile [31]

See attachment for the graph of the hyperbola 12x^2 - 3y^2 - 108 = 0

<h3>How to graph the hyperbola?</h3>

The equation of the hyperbola is given as:

12x^2 - 3y^2 - 108 = 0

Start by calculating the transverse axis

So, we have:

Transverse axis

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

Where

a = 3 and b = 6

The transverse axis is calculated as:

y = ±b/a(x - h) + k

So, we have:

y = ±6/3(x - 0) + 0

Evaluate the difference and sum

y = ±6/3x

Evaluate the quotient

y = ±2x

This means that the transverse axes are y = 2x and y =-2x

The vertices

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

The co-vertices

In the above section, we have:

The vertices of the given hyperbola are (0, 0)

This means that

(h, k) = 0

And

a = 3 and b = 6

The co-vertices are

(h - a, k) and (h + a, k)

So, we have:

(0 - 3, 0) and (0 + 3, 0)

Evaluate

(-3,0) and (3, 0)

See attachment for the graph of the hyperbola