Not easy:
- 4 < x^4 + 4x^2 ---> 0 < x^4+4x^2+4 = (x^2+2)^2, so it is always positive. Always satisfied. Curious ...
Now the other one:
x^4+4x^2<21 --> x^4+4x^2-21 < 0 --> (x^2+2)^2 - 25 < 0
(x^2+2)^2<25 ->>> x^2+2 < 5 ---> x^2 <3,
So -sqrt(3)<x<sqrt(3) is the answer! Brainliest?
Answer:
No solutions.
Step-by-step explanation:
5x = 8x^-1/3
Divide 8 into both sides.
5/8x = x^-1/3
Divide both sides by x.
5/8 = x^-4/3
Multiply both sides by the exponent -3/4.
5/8^-3/4 = x
1.422624 = x
Plug in 1.422624 for x to check.
It does not work. There are no real solutions.
Step-by-step explanation:
You must first add 20 to all inqualies, leaving u with 0 is greater than or equal to 2x which is also greater than or equal to 40.
Divde everything by 2 to get rid of the 2 from the x
0 divided by 2 is 0, 2x divded by 2 is x, 40 divided by 2 is 20.
Thats the answers
Answer:
-7/3
Step-by-step explanation:
-7 2/3=-23/3
5 1/3=16/3
-----------------
16/3+(-23/3)
16/3-23/3
-7/3
You just need to try with all of the pair f.ex
0,-2 -> put 0 in the x-> y=8*0-2=-2 so ok it is q possible pair of the function..
You try every pair till you find set where all the pairs fit
So set 1
1 and 6: 8*1-2=6 ok
2 and 14 : 8*2-2=14 ok
Hope I could help
