Answer:

And if we solve this equation for x we got:

We can cancel
in both sides and we have this:

And then we got:


And then the length of the sides are 9+11= 20 m for the hypothenuse, 16 for the adjacent side and 9+3 = 12m for the last side.
Lenght of the smaller unknown side: 12m
Lenght of the larger unknown side: 20m
Step-by-step explanation:
For this case we have a right triangle and we can use the Pythagoras Theorem and using the info given by the triangle we can set up the following equation:

And if we solve this equation for x we got:

We can cancel
in both sides and we have this:

And then we got:


And then the length of the sides are 9+11= 20 m for the hypothenuse, 16 for the adjacent side and 9+3 = 12m for the last side side.
Lenght of the smaller unknown side: 12m
Lenght of the larger unknown side: 20m
There are as many palindromic 12-bit binary strings as there are permutations with repetition of 6-bit binary strings, which is equal to

.
There are also only two permutations with repetition of 6-bit binary strings, that when each of them is mirrored and the result is "glued" to the right side of the original string, will give a palindromic string without 10 as a substring.
Those strings are 000000 and 111111.
Therefore there are

palindromic 12-bit binary strings with 10 as substring.
Answer:
A 2√2(cos 7π/4 + i sin 7π/4)
Step-by-step explanation:
A. 2√2(cos 7π/4 + i sin 7π/4)
2 sqrt(2) ( sqrt(2)/2 - sqrt(2)/2 i)
Distribute
2-2i
This is in the fourth quadrant
B. 2√2(cos 150° + i sin 150°)
2 sqrt(2) (-sqrt(3)/2 +1/2i)
-sqrt(6) +sqrt(2) i
This is in the third quadrant (NO)
C. 2(cos 7π/4 + i sin 7π/4)
2( ( sqrt(2)/2 - sqrt(2)/2 i))
sqrt(2) - sqrt(2) i
This is the fourth quadrant
D. 2(cos 90° + i sin 90°)
2(0+i)
2i
This is on the positive y axis NO
Now we need to decide between the two in the fourth quadrant.
The point has an x coordinate of 2 and a y coordinate of -2
This aligns with point A
Answer:
i got 3617742 TELL ME IF RIGHT
Step-by-step explanation: