Please help me! I really need help on this ASAP

the vertex of the parabola is at the point; (5, -1)

which agrees with answer "B" in the list of options

Step-by-step explanation:

Notice that this is the equation of a parabola with branches that open horizontally (not vertically), since the variable the goes squared is the y-variable instead of "x".

By analyzing it we can then write it by isolating the term in "x" on one side of the equation, and use at the same time the fact that it is being written in "vertex" form:

Therefore, the "y-value" of the vertex must be that which renders zero in the expression squared, that is y = -1. On the other hand, the x-value of the vertex is that which renders zero for the variable "x":  x=5.

Then, the vertex of the parabola is at the point; (5, -1)

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2