Answer:
[tex]-3/13[/tex]
Step-by-step explanation:
So we have the expression:
[tex]\frac{y^3+\sqrt{x-2}}{|4x-y|}[/tex]
And we want to evaluate it for x=6 and y=-2.
Thus, substitute:
[tex]=\frac{(-2)^3+\sqrt{(6)-2}}{|4(6)-(-2)|}[/tex]
Let's do the numerator first:
[tex](-2)^3+\sqrt{6-2}[/tex]
Cube the first term and subtract under the radical:
[tex]=-8+\sqrt4[/tex]
Simplify:
[tex]=-8+2[/tex]
Add:
[tex]=-6[/tex]
Now, do the denominator:
[tex]|4(6)-(-2)|[/tex]
Multiply:
[tex]=|24-(-2)|[/tex]
Simplify and add:
[tex]=|24+2|\\=|26|[/tex]
Remove the absolute value bars:
[tex]=26[/tex]
So, together:
[tex]=\frac{(-2)^3+\sqrt{(6)-2}}{|4(6)-(-2)|}\\=-6/26[/tex]
Reduce:
[tex]=-3/13[/tex]
And that's our answer: :)
