On some other planet, Thor stands on a bridge and throws his hammer at his brother Loki, who is at ground level. The hammer’s height h (in meters above the ground), t seconds after Thor threw it, is modeled by:

h(0) = 3600

t=60 when it hits the ground

max height at 4050

at 15 seconds

Step-by-step explanation:

h(t) = -2t^2 + 60t +3600

When it is thrown is t=0

h(0) = -2(0) +60(0) +3600 = 3600

When it hits the ground is h(t) =0

0 =  -2t^2 + 60t +3600

Factor out -2

0 = -2 (t^2 -30t -1800)

Divide by -2

0 = (t^2 -30t -1800)

Factor.  What 2 numbers multiply to -1800 and add to -30

0 = (t+30) (t-60)

Using the zero product property

t+30 = 0   t-60 =0

t = -30     t = 60

Since we do not have negative time

t=60

The maximum occurs halfway between the zeros

(-30+60)/2 = 30/2 = 15

Put this into the function to find the maximum value

h(15) = -2(15)^2 + 60(15) +3600

       = -2 (225) +900+3600

       =4050

Thrown at: 3600

Ground after: 60

Max height: 4050

After: 15

Step-by-step explanation:

h(t) = -2t² + 60t + 3600

h(0) = 3600

-2t² + 60t + 3600 = 0

t² - 30t - 1800 = 0

t² - 60t + 30t - 1800 = 0

t(t - 60) + 30(t - 60) = 0

(t - 60)(t + 30) = 0

t = 60, -30(not possible)

h(t) = -2t² + 60t + 3600

-2(t² - 30t) + 3600

-2(t² - 2(t)(15) + 15² - 15²) + 3600

-2[(t - 15)² - 225] + 3600

-2(t - 15)² + 450 + 3600

-2(t - 15)² + 4050

the formula of a slope:

from a table we have

substitute:

the slope of f(x) is -1.

the graph of funcfion g(x) is a horizontal line. the slope of a horizontal line is equal 0. therefore the slope of g(x) is 0.

-1 < 0

you can first factor out the n to get n(1+5). we know that 1+5 is 6, so we just have 6n.