Answer:
Step-by-step explanation:
Given y = coskt
y' = -ksinkt
y'' = -k²coskt
Substitute this y'' into the expression 25y'' = −16y
25(-k²coskt) = -16(coskt)
25k²coskt = 16(coskt)
25k² = 16
k² = 16/25
k = ±√16/25
k = ±4/5
b) from the DE 25y'' = −16y
Rearrange
25y''+16y = 0
Expressing using auxiliary equation
25m² + 16 = 0
25m² = -16
m² = -16/25
m = ±4/5 I
m = 0+4/5 I
Since the auxiliary root is complex number
The solution to the DE will be expressed as;
y = Asinmt + Bsinmt
Since k = m
y = Asinkt+Bsinkt where A and B are constants
A. (3,5) B. (-4,-3) C. (-4,4) D. (5,-3) E. (6,7) F. (-8,-8) G. (-2,7) H. (-2,-4) I. (3,-9) J. (1,0)
Answer:
See below
Step-by-step explanation:
![\sqrt[3]{49} = 3.65930571002 \approx3.66 \\ \\ so \: \sqrt[3]{49} should \: lie \: at \: 3.6 \: which \: \\ fall \: between \: 3 \: and \: 4.](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B49%7D%20%20%3D%203.65930571002%20%5Capprox3.66%20%5C%5C%20%20%5C%5C%20so%20%5C%3A%20%20%5Csqrt%5B3%5D%7B49%7D%20should%20%5C%3A%20lie%20%5C%3A%20at%20%5C%3A%203.6%20%5C%3A%20which%20%5C%3A%20%20%5C%5C%20fall%20%5C%3A%20between%20%5C%3A%203%20%5C%3A%20and%20%5C%3A%204.)
Actually Lin did a mistake, she obtained square root and not cube root.
Range: (-infinity, 4]
the ] signifies that the four is included in the range, rather than ) which means reaching but never approaching. this is why we use ( ) for - and + infinity, cause it can never be reached.
minimum: (-infinity, infinity)
keep it in same format as you have for the maximum
increasing
the interval -infinity
