CubeyThePenguin

I will assume T_n means the nth term and S_n means the sum of the first nth terms.

a) T_1 = 3, S_1 = 3

T_1 = S_1

b) T_2 = 7, S_2 = 10

T_2 = S_2 - S_1

c) T_3 = 11, S_3 = 21

T_3 = S_3 - S_2

and so on.

\(h(t) = -16t^2 + 100t + 280\)

Maximum height is at -b/2a = -100/(-32) = 3.125 seconds. Plug that in:

\(h(3.125) = \boxed{136.25}\).

Part B

Use the quadratic formula to find the roots of the quadratic. We get t = -2.0966 or 8.3466. Time can't be negative, so the cannonball is in the air for 8 seconds.

By Vieta's Formulas, the sum of roots is \(-\frac{b}{a} = -\frac{17}{2}\) and the product of the roots is \(\frac{c}{a} = \frac{11}{2}\).

If the roots are a and b, then:

\(ab + (a + b) = \text{product of roots} + \text{sum of roots} = -\frac{17}{2} + \frac{11}{2} = \boxed{-3}\)

What am I supposed to do? The problem cuts off at $1.

1. Graphing: graph the system of equations and find the intersection points, which are the solutions.

2. Elimination. As an example:

x + 2y = 3            . (1)

3x + y = 4            . (2)

You can eliminate one variable by multiplying equation (2) by 2:

x + 2y = 3

6x + 2y = 8

-----------------

-5x = -5

By subtracting the equations, we have eliminated one variable, thus solving for the other.

3. Substitution. Let's use the same equations as last time. In equation (1), solve for x in terms of y:

x + 2y = 3            ----->        x = 3 - 2y

Plug this in the other equation:

3x + y = 4

3(3 - 2y) + y = 4

Thus, we can solve for y, and in turn, solve for x.

The mathematician solves \(pt\) problems every day.

18 = 1 * 18, 2 * 9, 3 * 6

(x - 1)(x - 18) = x^2 - 19x + 18

(x - 2)(x - 9) = x^2 - 11x + 18

(x - 3)(x - 6) = x^2 - 9x + 18

k = 19, 11, 9

average: 13

Triangle ABC is similar to triangle RST.

\(\frac{BC}{AC} = \frac{ST}{RT}\)

6/5 = ST/40

ST = 48 feet.