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Here is a link to the graph of g(x) = -2x^2 - 12x - 24: https://www.desmos.com/calculator/wlsjngcwe9

The ratio is 1:4. If you want an explanation on how I got it let me know.

This question has already been answered. Here is the link: https://web2.0calc.com/questions/math_13723

The equation for circumference is πd and the so the circumference of the circle formed by the second hand is 12π. In 30 minutes, the second hand will travel around the clock 30 times, 1 per minute. This gives us the equation 12π * 30 = 360π cm.

The number of ways to order Shakespeare's and Dickens' book is 6!. There will be 7 spaces (in front of all the books, in between the first and second book, in between the second and third book...all the way until at the end of all the books) that Conrad's books can go with them since they can't be next to each other. The number of ways of doing this is binom{7}{3}⋅3! since once they are in their spots they can be ordered 3! ways.

Thus, the number of ways is 6!⋅binom{7}{3}⋅3!=151200.

Using a little bit of case work, we find that P can have 19 different values.

This can be found by using a right triangle formed using the radius of the smaller circle, which is 5 cm, and half of one of the chords of the larger circle, which equals 12. The radius of the larger circle is 13 cm.

A rhombus is a parallelogram with opposite equal acute angles, opposite equal obtuse angles, and four equal sides.  A rhombus inscribed in a rectangle touches the sides of the rectangle so by this we can infer that the diagonals of the largest inscribed rhombus are equal to the length and breadth of the rectangle. If we have the length(l) and breadth(b) of the rectangle, the length of the diagonal of the largest rhombus inscribed inside it is d1 = l and d2 = b. The area of a rhombus is given by the formula,

Area = (d1*d2)/2

Putting in the values of d1 = 9 and d2 = 3, we get:

Area = (9*3)\2 = 27/2 = 13.5

The equation of this hyperbola in standard form is ((x+8)^2)/16 - ((y-4)^2)/144 = 1.

Note that z^2 + 4 = (z + 2i)(z - 2i), so we can write the given equation as |z + 2i||z - 2i| = |z||z + 2i|. If |z + 2i| = 0, then z = -2i, in which case |z + i| = |-i| = 1. Otherwise |z + 2i| /= 0, so we can divide both sides by |z + 2i|, to get |z - 2i| = |z|. This condition states that z is equidistant from the origin and 2i in the complex plane. Thus, z must lie on the perpendicular bisector of these complex numbers, which is the set of complex numbers where the imaginary part is 1. In other words z = x + i for some real number i. Then |z + i| = |x + 2i| = sqrt(x^2 + 4) >= 2. Therefore, the smallest possible value of |z + i| is 1, which occurs for z = -2i.