Answer:
You can proceed as follows:
Step-by-step explanation:
First solve the quadratic inequality 
. To do that, factorize, then we have that 
. This implies that
or
In the first case the solution is the empty set 
. In the second case the solution is the interval 
. Now we have that
![A=[1,4]](https://tex.z-dn.net/?f=A%3D%5B1%2C4%5D)
![C=(3,4]](https://tex.z-dn.net/?f=C%3D%283%2C4%5D)
.
To show that 
consider 
. Then 
, this implies that 
, then 
. Now, to show that 
consider , then , then 
, then ![x\in [1,4] \, \text{and}\, x\in (3,5)](https://tex.z-dn.net/?f=x%5Cin%20%5B1%2C4%5D%20%5C%2C%20%5Ctext%7Band%7D%5C%2C%20x%5Cin%20%283%2C5%29)
, this implies that .
Observe the image below.
Answer:
Charges what?
Step-by-step explanation:
Unfinished...
-12x/-4= -27
(-4)-12x/-4 it will cancel the 4 but you gotta multiply on the other side by 4 too
now have -12x =108
-12x=108
divide 108 by -12x which x=-9
Interesting question. Let the 2 unknown sides be x and x+1.
Then (11 m)^2 + x^2 = sum of the squares of the 2 shortest sides
= (x+1)^2
121 + x^2 = x^2 + 2x + 1. Then 121 = 2x + 1, and 2x = 120, or x = 60.
Then the hyp. has length 60+1= 61.
We must check these results. Using the Pyth. Thm. (a^2 + b^2 = c^2),
11^2 + 60^2 = 61^2
121 + 3600 = 3721 This is true, so our answers are correct.
The longer side is 60 and the hyp. has length 61.
