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anyanavicka [17]
1 year ago
10

Order six and three tenths repeating, negative six and four ninths, 630%, and −6.1 from least to greatest.

Mathematics
1 answer:
NNADVOKAT [17]1 year ago
4 0

The order of six and three tenths repeating, negative six and four ninths, 630%, and −6.1 from least to greatest is negative six and four ninths, −6.1, 630%, six and three tenths repeating. option D

<h3>Ascending order</h3>

This is the arrangements of numbers from the least to the greatest, that is, from lower value to higher value.

six and three tenths repeating

= 6.333

negative six and four ninths

= -6 4/9

= -6.444

630%

= 630/100

= 6.3

−6.1

From least to greatest:

  • negative six and four ninths
  • −6.1
  • 630%
  • six and three tenths repeating

Therefore, the order six and three tenths repeating, negative six and four ninths, 630%, and −6.1 from least to greatest is negative six and four ninths, −6.1, 630%, six and three tenths repeating. option D

Learn more about ascending order:

brainly.com/question/12783355

#SPJ1

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3. Determine the slope of the line that has the following coordinates: (-3, 6) (4,-2)
poizon [28]

Answer:

-8/7

Step-by-step explanation:

(-3, 6) & (4,-2)

To find the slope of the line, we use the slope formula: (y₂ - y₁) / (x₂ - x₁)

Plug in these values:

(-2 - 6) / (4 - (-3))

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Hope this helps!

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2 years ago
Given the function f(x)=1/2x+8 find x so that, f(x)=10<br><br> please help im stumped
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Step-by-step explanation:

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2 years ago
Find two unit vectors orthogonal to both 8, 5, 1 and −1, 1, 0 .
Elina [12.6K]
In order to do this, you must first find the "cross product" of these vectors. To do that, we can use several methods. To simplify this first, I suggest you compute:

‹1, -1, 1› × ‹0, 1, 1›

You are interested in vectors orthogonal to the originals, which don't change when you scale them. Using 0,-1,1 is much easier than 6s and 7s.

So what methods are there to compute this? You can review them here (or presumably in your class notes or textbook):
http://en.wikipedia.org/wiki/Cross_produ...

In addition to these methods, sometimes I like to set up:
‹1, -1, 1› • ‹a, b, c› = 0
‹0, 1, 1› • ‹a, b, c› = 0

That is the dot product, and having these dot products equal zero guarantees orthogonality. You can convert that to:

a - b + c = 0
b + c = 0

This is two equations, three unknowns, so you can solve it with one free parameter:

b = -c
a = c - b = -2c

The computation, regardless of method, yields:
‹1, -1, 1› × ‹0, 1, 1› = ‹-2, -1, 1›

The above method, solving equations, works because you'd just plug in c=1 to obtain this solution. However, it is not a unit vector. There will always be two unit vectors (if you find one, then its negative will be the other of course). To find the unit vector, we need to find the magnitude of our vector:

|| ‹-2, -1, 1› || = √( (-2)² + (-1)² + (1)² ) = √( 4 + 1 + 1 ) = √6

Then we divide that vector by its magnitude to yield one solution:

‹ -2/√6 , -1/√6 , 1/√6 ›

And take the negative for the other:

‹ 2/√6 , 1/√6 , -1/√6 ›
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3 years ago
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