X^2-y^3=6
x^2-6=y^3
y= (x^2-6)^1 by 3
Answer:
2 real solutions
Step-by-step explanation:
Remember this messy thing?

The <em>quadratic formula</em>, as it's called, gives us the roots to any quadratic equation in standard form (ax² + bx + c = 0). The information on the <em>type</em> of roots is contained entirely in that bit under the square root symbol (b² - 4ac), called the <em>discriminant</em>. If it's non-negative, we'll have <em>real</em> roots, if it's negative, we'll have <em>complex roots</em>.
For our equation, we have a discrimant of (-3)² - 4(6)(-4) = 9 + 96 = 105, which is non-negative, so we'll have real solutions, and since quadratics are degree 2, we'll have exactly 2 real solutions.
Quick answer I don't think this has an answer.
If you take the cos-1(2 sqrt(2)) your calculator should have a fit. Let's check that out. Mine certainly does. So there is something wrong with the question. If there is something to add in please do it and I will it least put an answer in the comments. As it stands, nothing will work.
If you put your calculator in radians, you will get an answer but it will not be anything resembling the choices you've listed.
If you meant sqrt(2) / 2 that would give 45o. Put it in your calculator like this 2 ^ 0.5 divided by 2 = 0.707
Cos - 1 (0.707) = 45
<span>Proportional Relationship should be the correct answer.</span>