From your earlier questions, we found
![2\sin(4\pi t)+5\cos(4\pi t)=\sqrt{29}\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)](https://tex.z-dn.net/?f=2%5Csin%284%5Cpi%20t%29%2B5%5Ccos%284%5Cpi%20t%29%3D%5Csqrt%7B29%7D%5Csin%5Cleft%284%5Cpi%20t%2B%5Ctan%5E%7B-1%7D%5Cleft%28%5Cdfrac52%5Cright%29%5Cright%29)
so the wave has amplitude √29. The weight's maximum negative position from equilibrium is then -√29, so you are solving for <em>t</em> in the given interval for which
![\sqrt{29}\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)=-\dfrac{\sqrt{29}}2](https://tex.z-dn.net/?f=%5Csqrt%7B29%7D%5Csin%5Cleft%284%5Cpi%20t%2B%5Ctan%5E%7B-1%7D%5Cleft%28%5Cdfrac52%5Cright%29%5Cright%29%3D-%5Cdfrac%7B%5Csqrt%7B29%7D%7D2)
Divide both sides by √29:
![\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)=-\dfrac12](https://tex.z-dn.net/?f=%5Csin%5Cleft%284%5Cpi%20t%2B%5Ctan%5E%7B-1%7D%5Cleft%28%5Cdfrac52%5Cright%29%5Cright%29%3D-%5Cdfrac12)
Take the inverse sine of both sides, noting that we get two possible solution sets because we have
![\sin\left(\dfrac{7\pi}6\right)=\sin\left(\dfrac{11\pi}6\right)=-\dfrac12](https://tex.z-dn.net/?f=%5Csin%5Cleft%28%5Cdfrac%7B7%5Cpi%7D6%5Cright%29%3D%5Csin%5Cleft%28%5Cdfrac%7B11%5Cpi%7D6%5Cright%29%3D-%5Cdfrac12)
and the sine wave has period 2π, so
.
![\implies 4\pi t+\tan^{-1}\left(\dfrac52\right)=\dfrac{7\pi}6+2n\pi](https://tex.z-dn.net/?f=%5Cimplies%204%5Cpi%20t%2B%5Ctan%5E%7B-1%7D%5Cleft%28%5Cdfrac52%5Cright%29%3D%5Cdfrac%7B7%5Cpi%7D6%2B2n%5Cpi)
OR
![\implies 4\pi t+\tan^{-1}\left(\dfrac52\right)=\dfrac{11\pi}6+2n\pi](https://tex.z-dn.net/?f=%5Cimplies%204%5Cpi%20t%2B%5Ctan%5E%7B-1%7D%5Cleft%28%5Cdfrac52%5Cright%29%3D%5Cdfrac%7B11%5Cpi%7D6%2B2n%5Cpi)
where <em>n</em> is any integer.
Now solve for <em>t</em> :
![t=\dfrac{\frac{7\pi}6+2n\pi-\tan^{-1}\left(\frac52\right)}{4\pi}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B%5Cfrac%7B7%5Cpi%7D6%2B2n%5Cpi-%5Ctan%5E%7B-1%7D%5Cleft%28%5Cfrac52%5Cright%29%7D%7B4%5Cpi%7D)
OR
![t=\dfrac{\frac{11\pi}6+2n\pi-\tan^{-1}\left(\frac52\right)}{4\pi}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B%5Cfrac%7B11%5Cpi%7D6%2B2n%5Cpi-%5Ctan%5E%7B-1%7D%5Cleft%28%5Cfrac52%5Cright%29%7D%7B4%5Cpi%7D)
We get solutions between 0 and 0.5 when <em>n</em> = 0 of <em>t </em>≈ 0.196946 and <em>t</em> ≈ 0.363613.
Answer:
Option (A).
Step-by-step explanation:
Statements Reasons
1). Draw a line ZY parallel to PQ 1).Construction
2). m∠ZRP + m∠PRQ + m∠QRY = 180° 2). Angle addition postulate
3). ∠ZRP ≅ ∠RPQ 3). Alternate interior angles theorem
4). ∠QRY ≅ ∠PQR 4). Alternate angle theorem
5). m∠RPQ + m∠PRQ + m∠PQR =m∠ZRY 5). Substitution
6). m∠ZRY = 180° 6). Definition of supplementary angles
7). m∠RPQ+m∠PRQ+m∠PQR = 180° 7). Substitution
Here reason of statement (6) is incorrect.
m∠ZRY = 180°, because angle ZRY is a straight angle not a supplementary angle.
Therefore, Option (A) will be the answer.
The share of first, second and third will be 22875, 7625 , 61000 respectively
Let the share of first, second and third on be x, y, z respectively
We know the total share between three friends = 91500
x+y+z = 91500 ....(i)
Then,
First gets share three times as much as second
x = 3y ......(ii)
Third gets share twice as much as the other two combined
z = 2(x+y) ......(iii)
Solving equation (i),(ii) and (iii) we get,
3y+y+2(3y+y) = 91500
12y = 91500
y = 7625
Similarly, putting value of y in (ii) and (iii) equation we get,
x = 22,875
z = 61,000
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Answer:
Range is 19.54
Step-by-step explanation:
I. If R = H - L
when R is Range, H is highest value and L is lowest value
R = 25.6 - 6.06
R = 19.54
Answer:
A. Both functions are positive and decreasing on the interval.
Step-by-step explanation:
The table shows that f(x) decreases when x increases in the interval (0,3).
All the values of f(x) are positive in the interval (0,3).
For the exponential function that passes through the points (0, 27) and (3, 0), we also see that f(x) is decreasing when x increases: when x goes from 0 to 3, f(x) goes from 27 to 0.
Also all the values of f(x) are positive in the interval.
Then, both functions are positive and dereasing in the interval.