Answer:
yields 2 real roots
Step-by-step explanation:
To find the number of real roots for a quadratic, we apply the discriminate. The discriminate is the inside portion of the square root from the quadratic formula.
•
yields 2 real roots
•
yields 1 real root
• [tx]b^2-4ac<0[/tex] yields no real roots
Example:
where a=4, b=-7, and c=2
.
It has 2 real roots
Answer:

Step-by-step explanation:
For this case we need to find the following integral:

And for this case we can use the substitution
from here we see that
, and if we solve for x we got
, so then we can rewrite the integral like this:

And if we distribute the exponents we have this:

Now we can do the integrals one by one:

And reordering the terms we have"

And rewriting in terms of x we got:

And that would be our final answer.
Answer:
An imaginary line would need to be drawn at any angle to the center of the side opposite the angle to reflect across an axis of symmetry so that a regular pentagon can carry onto itself. Hope this answer helps.