An isosceles trapezoid abcd with height 2 units has all its vertices on the parabola y=a(x+1)(x−5). what is the value of a, if points a and d belong to the x−axis and m∠bad=60 degrees?

Step-by-step explanation:

Here, we assume that point A is (-1, 0) and point B is (z, 2). Since angle BAD is 60°, we have ...

  2/(z+1) = tan(60°) = √3

  z = (2/√3) -1

Putting the value of z into the equation for the parabola lets us find "a".

  2 = a(z +1)(z -5) = a(2/√3)(2/√3 -6)

Then ...

  a = 2/(4/3 -12/√3) = (6√3)/(4√3 -36)

  a = -(3+9√3)/52 ≈ −0.357470332079

c.

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,