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GarryVolchara [31]
3 years ago
13

What equation has been incorrectly solved. -4(6-b)=4

Mathematics
1 answer:
Ivenika [448]3 years ago
7 0
The distribution of the -4 was incorrectly put in as -4b when it should be 4b

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Which statement is TRUE about the graph?
Lunna [17]

Answer:

c

Step-by-step explanation:

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In which quadrant is the number –14 – 5i located on the complex plane?
loris [4]

Answer:

The complex number r = -14 - i\,5 belongs to the third quadrant of the complex plane.

Step-by-step explanation:

Let be r = -14 - i\,5. In the complex plane, if \Re (r) < 0 (real component) and \Im (r) < 0 (imaginary component), the number belongs to the third quadrant of the complex plane. The complex number r = -14 - i\,5 belongs to the third quadrant of the complex plane.

5 0
3 years ago
Jacinta is working on a jigsaw puzzle that has 403 pieces. She has completed 1/4 of the puzzle. About how many pieces has she pl
slava [35]
403 / 4 = 100.75 
to get any fraction of a number, as long as the numerator is 1 in the fraction, just divide by the denominator. In this case it's 4. Since you can't place .75 of a puzzle piece, she's placed 100 pieces in the puzzle.
4 0
3 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 ›
7 0
3 years ago
HELP ASAP! A college campus has a lawn shaped like a circle with a radius, r, of 60 meters. Which is the best approximation of t
astra-53 [7]

Answer:

he answer is 2800, hope this helps!

Step-by-step explanation: :))

4 0
3 years ago
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