A pendulum is swinging next to a wall. The distance D(t)D(t)D, left parenthesis, t, right parenthesis (in \text{cm}cmstart text, c, m, end text) between the bob of the pendulum and the wall as a function of time ttt (in seconds) can be modeled by a sinusoidal expression of the form a\cdot\sin(b\cdot t)+da⋅sin(b⋅t)+da, dot, sine, left parenthesis, b, dot, t, right parenthesis, plus, d. At t=0t=0t, equals, 0, when the pendulum is exactly in the middle of its swing, the bob is 5\text{ cm}5 cm5, start text, space, c, m, end text away from the wall. The bob reaches the closest point to the wall, which is 3\text{ cm}3 cm3, start text, space, c, m, end text from the wall, 111 second later. Find D(t)D(t)D, left parenthesis, t, right parenthesis. \textit{t}tstart text, t, end text should be in radians.

answer

106 emergencies

step-by-step explanation:

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

about 0.75 hours

step-by-step explanation: