Answer:
12 is to the right of -15 on a number line.
Step-by-step explanation:
Hope this helps! - CJ
We can rewrite the expression under the radical as
then taking the fourth root, we get
![\sqrt[4]{\left(\dfrac32a^2b^3c^4\right)^4}=\left|\dfrac32a^2b^3c^4\right|](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cleft%28%5Cdfrac32a%5E2b%5E3c%5E4%5Cright%29%5E4%7D%3D%5Cleft%7C%5Cdfrac32a%5E2b%5E3c%5E4%5Cright%7C)
Why the absolute value? It's for the same reason that
since both 
and 
return the same number , and 
captures both possibilities. From here, we have
The absolute values disappear on all but the 
term because all of 
, 
and 
are positive, while 
could potentially be negative. So we end up with
Checking the <span>discontinuity at point -4
from the left f(-4) = 4
from the right f(-4) = (-4+2)² = (-2)² = 4
∴ The function is continues at -4
</span>
<span>Checking the <span>discontinuity at point -2
from the left f(-2) = </span></span><span><span>(-2+2)² = 0
</span>from the right f(-2) = -(1/2)*(-2)+1 = 2
∴ The function is jump discontinues at -2
</span>
<span>Checking the <span>discontinuity at point 4
from the left f(4) = </span></span><span><span>-(1/2)*4+1 = -1
</span>from the right f(4) = -1
but there no equality in the equation so,
</span><span>∴ The function is discontinues at 4
The correct choice is the second
point </span>discontinuity at x = 4 and jump <span>discontinuity at x = -2</span>
Answer:
0 = 0
Step-by-step explanation:
