Answer:
The period, in seconds to two decimal places is, 2.84 sec
Step-by-step explanation:
Using formula:
......[1]
where
T represents the Time period in second
l represents the length of the pendulum/
As per the statement:
Use the gravitational constant g = 9.8 m/s^2 , length of the simple pendulum(l) = 2 meters and ![\pi = 3.14](https://tex.z-dn.net/?f=%5Cpi%20%3D%203.14)
Substitute the given values we have;
![T = 2\cdot 3.14 \cdot \sqrt{\frac{2}{9.8}}](https://tex.z-dn.net/?f=T%20%3D%202%5Ccdot%203.14%20%5Ccdot%20%5Csqrt%7B%5Cfrac%7B2%7D%7B9.8%7D%7D)
⇒![T = 6.28 \cdot \sqrt{\frac{1}{4.9}}=6.28 \cdot \frac{1}{2.21359436212}= \frac{6.28}{2.21359436212} = 2.83701481512](https://tex.z-dn.net/?f=T%20%3D%206.28%20%5Ccdot%20%5Csqrt%7B%5Cfrac%7B1%7D%7B4.9%7D%7D%3D6.28%20%5Ccdot%20%5Cfrac%7B1%7D%7B2.21359436212%7D%3D%20%5Cfrac%7B6.28%7D%7B2.21359436212%7D%20%3D%202.83701481512)
Therefore, the period, in seconds to two decimal places is, 2.84 sec
Answer:
starting point Quadrant Q
Finishing point Quadrant S
B- the check point will be in Quadrant p, since the two values of x and y are positive. the two points are on a horizontal line 10.6 apart.
Step-by-step explanation:
A- because x value is negative and y- is positive it is Q
x value is positive and y negative it is S
Solution
Part A
The general form of the n-th term = a + (n-1)d
Here we are given the 8th term A8 = 28
, common difference (d) = 6
A8 = a + (n - 1) 6
28 = a + (8- 1)6
28 = a + 7*6
28 = a + 42
a = 28 - 42
a = -14
Part B
n-th term is 760. a = -14 and d = 6.
Plugging in the values in the general form, we get
760 = -14 + (n -1)6
760 = -14 + 6n - 6
760 = -20 + 6n
6n = 760 + 20
6n = 780
Dividing both sides by 6, we get
n = 780/6 = 130
The value of n is 130
Answer:
Step-by-step explanation:
4
3
6
5
2
1
Which statement is true about the potential solutions for this equation?
D. Neither solution is extraneous.
Answer:
1)option b
2)option a
Step-by-step explanation:
1) circumference = 62.8 in
=> 2πr= 62.8
=> r= 62.8/(2×3.14)= 10 in
diameter = 2×10 in = 20 in
2)diameter = 14m
radius = 7 m
Area of the circle= πr² =3.14(7)²=3.14×49
= 153.86
=153.9 m²