CubeyThePenguin

We know the vertex, so we can write the equation in vertex form: \(y = a(x+1)^2 + 7\)

The y-intercept is (0, 10), so plug that point in: \(10 = a + 7\)

So, a is equal to 3, and we now have the equation: \(y = 3(x+1)^2 + 7\)

I. The range of both distributions is 7 - 0 = 7 hours, so this is true.

II. This is also true, since the seniors graph is skewed.

III. The median is 1 and the mean is roughly 1.667, so this is false.

1. \(10001^2 -9999^2 = (10001 + 9999)(10001 - 9999) = (20000)(2) = \boxed{40000}\)

2. \(\frac{-5+2i}{1+7i} \cdot \frac{1-7i}{1-7i} = \frac{-5+2i+35i-14i^2}{50} = \boxed{\frac{9+37i}{50}}\)

ABCD is a parallelogram, so angle x = w and angle y = z.

w = 180 - 53 = 127 degrees

x = w = 127 degrees

y = z = 53 degrees

Make sure you understand why this is true.

Start by evaluating O(3) in the middle and working outwards. It's a tedious problem, but straightforward.

O(3) = 3^2 = 9

N(9) = 2/sqrt(9) = 2/3

O(2/3) = 4/9

...

and so on.

1. \((x+2)(x^2+3) = \boxed{x^3 + 2x^2 + 3x + 6}\)

2. \(x^2 - 9x - 36 = \boxed{(x-12)(x+3)}\)

3. \(3x^2 +14x - 5 = 0\)

(quadratic formula)

\(x = {-14 \pm \sqrt{196-4(3)(-5)} \over 2(3)} = {-14 \pm 16 \over 6}\)

\(\boxed{x=-5, -\frac{1}{3}}\)

This question has been posted before:

\(3(2-0.9h) + (-1.3h - 4) = (6-2.7h) -1.3h - 4 = \boxed{-4h+2}\)

Let the height be h. Then, we have \(h(h+8)(h-2) = 96.\) (volume is height x length x width)

Solving, we get h = 4 inches.

\(\text{slope} = m = \frac{3-1}{3+2} = \frac{2}{5}\)

Plug in (3, 3) into the equation y = mx + b.

\(3 = \frac{2}{5}(3) + b \quad \rightarrow \quad b = \frac{9}{5}\)

The equation is \(\boxed{y = \frac{2}{5}x + \frac{9}{5}.}\)