Area of a circle equation:
A=(3.14)r^2
40(3.14)=(3.14)*r^2
40=r^2
r= square root of 40
simplified radius is 2 times root 10
(See the photo I attached for the work)
Answer:
y = 6cos(π/2)t
Step-by-step explanation:
Given in the question that,
in a simple harmonic motion in which
at t=0 the amplitude is 6 cm, so the simple harmonic motion will be in cosine form. As in case of sine function, the value of function at t=0 is 0.
And,
the period is given as 4 sec
We know that model equation of simple harmonic motion is
<h3>y = a. cosb(t + c) + d</h3>
here,
a is amplitude
c is Horizontal shift or phase shift
d is Vertical shift
<h3>Step1</h3><h3>Find b</h3>
Period = 2 π / b
4 = 2π / b
b = 2π/4
b = 1π/2
<h3>Step2</h3><h3>Find shifts</h3>
Horizontal and vertical shifts are 0.
So, c = 0 and d = 0
<h3>Step3</h3><h3>Plug those values in the above in the equation</h3>
y = 6cos π/2 (t + 0) + 0
y = 6cos(π/2)t
The answer is a. elements
Answer:
Y = 4 and X = -6 , So the equation would be 4 + -6
Step-by-step explanation:
Look at were the line go across the numbers in Line Y and Line X this will help you understand more. And i also answered first :)
