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The axis of symmetry of the graph of y=ax2+bx+c is x=−2ab​. In general, the axis of symmetry is a vertical line, so we have −2ab​=−1. Solving for b, we find b=2a​=16.

First, we combine the terms in the equation so that it is in the standard form of a quadratic equation: y=−5x2+15x−10. To find the vertex of a parabola in vertex form, we complete the square.

To complete the square, we take half of the square coefficient, square it, and add it to both sides of the equation:

y = -5x^2 + 15x - 10 y + \frac{25}{4} = -5x^2 + 15x + \frac{25}{4} - 10

The left side of the equation is now a perfect square, so we can take the square root of both sides:

y + \frac{25}{4} = \sqrt{(-5x^2 + 15x + \frac{25}{4})^2} y + \frac{25}{4} = \sqrt{(-5x + \frac{15}{2})^2}

To get the equation in vertex form, we move the constant term to the right side of the equation:

y = -\sqrt{(-5x + \frac{15}{2})^2} - \frac{25}{4}

The vertex of a parabola in vertex form is always at the point where the x -coordinate is equal to half of the coefficient of the x2 term. In this case, the coefficient of the x2 term is −5, so the vertex is at the point (1015​,−425​)=(23​,−425​)​.

We can rewrite the equation as \begin{align*} x -y^2 + 8y &= 13 + 2y + y^2 - 7y - 14\ 2y^2 + 9y + x &= 0\ 2(y^2 + \tfrac{9}{2}y + \tfrac{x}{2}) &= 0 \ 0 &= (y+\tfrac{9}{4})^2 + x - \tfrac{81}{16} \end{align*}This is a parabola, opening upwards, with vertex at (1681​,4−9​).

Enter the letters of the points that lie in quadrant IV.

Quadrant IV is the bottom left quadrant, so it contains points with negative x-coordinates and negative y-coordinates.

B, F all have negative x-coordinates and negative y-coordinates, so the answer is B, F.

Since ∠ACB=90∘, we have ∠ACE=30∘. Also, since △DEF is equilateral, ∠DEF=60∘. Then ∠AED=60∘−30∘=30∘. Furthermore, △AED is isosceles since AD=AE, so ∠ADE=30∘. Since the sum of the angles in a triangle is 180∘, we have \begin{align*} \angle AED + \angle ADE + \angle EAD &= 180^\circ \ 30^\circ + 30^\circ + \angle EAD &= 180^\circ \ \angle EAD &= 120^\circ. \end{align*}By the Law of Sines in △ADE, we have \begin{align*} \frac{AD}{\sin \angle EAD} &= \frac{AE}{\sin \angle ADE} \ \frac{AE}{5} &= \frac{AE}{\sin 30^\circ} \ 5 &= \sin 30^\circ \ AE &= \frac{5}{\sin 30^\circ} \ AE &= \boxed{\frac{10}{\sqrt{3}}}. \end{align*}

So the answer is AE = 10/sqrt(3).

Squaring the first equation gives [(x+y)^2=x^2+2xy+y^2,]so 102=32+2xy or 2xy=26. Dividing both sides of x+y=10 by 2, we get x/2+y/2=5, so x+y=10. Thus, (x,y) lies on the line y=10−x. Also, since 2xy=26, xy=13, so (x,y) lies on the hyperbola xy=13. The solutions are (x,y)=(2,8), (−2,12), (4,6), (−4,14).

Therefore, the ordered pairs (x,y) are (2,8),(4,6),(8,2),(12,−2),(14,−4)​

The quadratic has nonreal roots if and only if its discriminant is negative.

The discriminant of the quadratic ax2+bx+c is b2−4ac.

The discriminant of the quadratic 12x2+kx+27 is k2−4⋅12⋅27=k2−1296.

Therefore, the quadratic has nonreal roots if and only if k2−1296<0.

This inequality is equivalent to (k−36)(k+36)<0.

Therefore, the quadratic has nonreal roots if and only if k<−36 or k>36.

In interval notation, the solution is:

k \in (-\infty,-36) \cup (36,\infty)