Answer:
![\displaystyle 9\:metres\:wide \\ 11\:metres\:long](https://tex.z-dn.net/?f=%5Cdisplaystyle%209%5C%3Ametres%5C%3Awide%20%5C%5C%2011%5C%3Ametres%5C%3Along)
Step-by-step explanation:
{−2w + 2l = 4
{40 = 2w + 2l
__________
![\displaystyle \frac{44}{4} = \frac{4l}{4}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B44%7D%7B4%7D%20%3D%20%5Cfrac%7B4l%7D%7B4%7D)
[Plug this back into both equations above to get the width of 9 metres]; ![\displaystyle 9 = w](https://tex.z-dn.net/?f=%5Cdisplaystyle%209%20%3D%20w)
I am joyous to assist you anytime.
Answer:
(c) 2.034 s; (d) 8.944 cm
Step-by-step explanation:
Velocity and acceleration
s = 8cos(t) + 4sin(t)
v = -8sin(t) + 4cos(t)
a = -8cos(t) + 4sin(t)
(c) Time to first equilibrium position
The equilibrium position is where the mass hangs before it is pulled downward, that is, at s = 0.
Set s = 0 and solve for t.
![\begin{array}{rcl}0 & = & 8\cos t + 4\sin t\\-8\cos t & = & 4\sin t\\-8 & = & 4\dfrac{\sin t}{\cos t}\\\\-2 & = & \tan t\\t & = & \arctan(-2)\\& = & -1.107 \pm n\pi\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D0%20%26%20%3D%20%26%208%5Ccos%20t%20%2B%204%5Csin%20t%5C%5C-8%5Ccos%20t%20%26%20%3D%20%26%204%5Csin%20t%5C%5C-8%20%26%20%3D%20%26%204%5Cdfrac%7B%5Csin%20t%7D%7B%5Ccos%20t%7D%5C%5C%5C%5C-2%20%26%20%3D%20%26%20%5Ctan%20t%5C%5Ct%20%26%20%3D%20%26%20%5Carctan%28-2%29%5C%5C%26%20%3D%20%26%20-1.107%20%5Cpm%20n%5Cpi%5C%5C%5Cend%7Barray%7D)
If n = 1,
t = -1.107 + π = 2.034 s
(d) Distance from equilibrium position
The mass will reach its maximum distance when v = 0, that is, when it is at the peak or trough of its oscillation.
Set v = 0 and solve for t.
![\end{array}\begin{array}{rcl}0 & = &- 8\sin t + 4\cos t\\8\sin t & = & 4\cos t\\\dfrac{\sin t}{\cos t} & = & \dfrac{1}{2}\\\\\tan t & = & 0.5\\t & = & \arctan(0.5)\\& = & 0.4636 \pm n\pi\\\end{array}](https://tex.z-dn.net/?f=%5Cend%7Barray%7D%5Cbegin%7Barray%7D%7Brcl%7D0%20%26%20%3D%20%26-%208%5Csin%20t%20%2B%204%5Ccos%20t%5C%5C8%5Csin%20t%20%26%20%3D%20%26%204%5Ccos%20t%5C%5C%5Cdfrac%7B%5Csin%20t%7D%7B%5Ccos%20t%7D%20%26%20%3D%20%26%20%5Cdfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%5Ctan%20t%20%26%20%3D%20%26%200.5%5C%5Ct%20%26%20%3D%20%26%20%5Carctan%280.5%29%5C%5C%26%20%3D%20%26%200.4636%20%5Cpm%20n%5Cpi%5C%5C%5Cend%7Barray%7D)
If n = 0,
t = 0.4636
Then
s = 8cos(0.4636) + 4sin(0.4636) = 8×0.8944 + 4×0.4472 = 7.156 + 1.789 = 8.944 cm
The figure below shows the graphs of s and v vs t. They indicate that the mass first reaches its equilibrium position at 2.034 s, and the amplitude of its vibration is 8.944 cm.
The one that should be performed according to the order of PEMDAS is 14 times 8.
P= PARENTHESIS
E= EXPONENTS
M= MULTIPLICATION
D= DIVISION
A= ADDITION
S= SUBTRACTION
Hope this helps! (:
Arithmetic sequence has a common difference between the terms.
for 1,4,7,10,...
4-1 = 3
7-4 = 3
10-7 = 3
so we do see a common difference between the terms.
Hence, True