\left[x _{2}\right] = \left[ \frac{-1+i \,\sqrt{3}+2\,by+\left( -2\,i \right) \,\sqrt{3}\,by}{2^{\frac{2}{3}}\,\sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}+\frac{\frac{ - \sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}{24}+\left( \frac{-1}{24}\,i \right) \,\sqrt{3}\,\sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}{\sqrt[3]{2}}\right][x2]=⎣⎢⎢⎢⎢⎡2323√(432by+√(−6912+41472by+103680by2+55296by3))−1+i√3+2by+(−2i)√3by+3√224−3√(432by+√(−6912+41472by+103680by2+55296by3))+(24−1i)√33√(432by+√(−6912+41472by+103680by2+55296by3))⎦⎥⎥⎥⎥⎤
totally answer.
The answer to that would be many
Multiply the second fraction by three.
4/9 + 3/9 = 7/9
Answer:
A hole at x=-4.
Step-by-step explanation:
This is a fraction so we have to worry about division by zero.
The only time we will be dividing by 0 is when x+4 is 0.
Solving the equation
x+4=0 for x:
Subtract 4 on both sides:
x=-4
So there is either a vertical asymptote or a hole at x=-4.
These are the kinds of discontinuities we can have for a rational function.
If there is a hole at x=-4, then x=-4 will make the top zero and can be cancelled out after simplification.
If is is a vertical asymptote, x=-4 will make the top NOT zero.
Let's see what -4 for x in x^2+6x+8 gives us:
(-4)^2+6(-4)+8
16+-24+8
-8+8
0
Top and bottom are 0 when x=-4.
Let's see what happens after simplication.
We are going to factor a^2+bx+c if factorable by finding two numbers that multiply to be c and add up to be b.
So what 2 numbers together multiply to be 8 and add up to be 6.
I hoped you said 4 and 2 because (4)(2)=8 where 4+2=6.

We we able to cancel out that factor that was giving us x=-4 is a zero.
Therefore there is a hole at x=-4.
The answer is B. y = 0.75 + 2.25. because i've done this assement.