9514 1404 393
Answer:
not a right triangle
Step-by-step explanation:
These lengths will form a triangle iff ...
10 + 20 > 30 . . . . . false
These side lengths do not form a triangle. It is NOT a right triangle.
_____
<em>Additional comment</em>
We notice the differences between these numbers are the same: 10. The only right triangle with side lengths having the same difference is some multiple of a 3-4-5 right triangle. For a right triangle with side lengths having a difference of 10, the sides would have to be 30, 40, 50--not 10, 20, 30.
For this case we have that the distance between two points is given by:
We have the points:
Substituting:
Thus, the distance between the points is 13
Answer:
Answer:
Boat level variation = 23'
Step-by-step explanation:
The graph shows that the variation between Amsterdam and Randall is referred to the Randall level
Answer:
1. -8x² - 16x - 6 a = -8, b = -16, c = -6
1. 4x² - 1 a = 4, b = 0, c = -1
Step-by-step explanation:
use the 'foil' method:
(4x + 2)(-2x - 3)
F: -8x² O: -12x I: -4x L: -6
-8x² - 16x - 6
(2x + 1)(2x - 1)
F: 4x² O: -2x I: 2x L: -1
4x² - 1
Do you mean no repeating numbers within the two sets? Because if so, I don't think it's possible.
I started by trying to figure out what the numbers in the tenths place should be. I used subtraction: 71 - ___ = ____. If you try it out, you can't subtract anything with a 6, 7, 8, or 9 in the tenths place because it will leave you with 11 (a repeating digit number), 10 (has a 0), or less (1-digit numbers). Also, a 3 cannot go into the tenths place because when you do 71 - 3_ , your answer will always begin with a 3 (problem because the 3 repeats), or it will contain a 0.
Therefore, the numbers left for the tenths place are: 1, 2, 4, and 5. 1 and 5 pair up, leaving 4 and 2.
71 - 5_ = 1_ and 71 - 4_ = 2_.
Then, I tried to figure out what numbers go in the ones place. That lead me to realize they act in pairs. The pairs possible in the ones place are 2 and 9, 3 and 8, 4 and 7, 5 and 6. These numbers always go together to result with the final "1" in the "71". Using this information, I looked at the numbers I already used: 1, 2, 4, and 5. Now, looking at the pairs, I eliminated the pairs containing a number already used. This leaves me with only one pair: 3 and 8. Obviously, you need two more pairs to solve the problem, which leads me to my point of saying: This problem is impossible to solve.
I really hope someone can prove me wrong! But this is the solution I have reached for now. :)
