Given data, cos(A - B) = cosAcosB + sinAsinB
<span>let, A=60' and B=30' ( here the ' sign bears degree) </span>
<span>L.H.S = cos(A - B) </span>
<span>=cos (60'-30) ( using value of A and B ) </span>
<span>=cos30' </span>
<span>= sqrt3/2 </span>
<span>R.H.S= cosAcosB + sinAsinB </span>
<span>=cos60' cos30' + sin60' sin30' </span>
<span>= 1/2* sqrt3/2+ sqrt3/2*1/2 </span>
<span>= sqrt3/4 + sqrt3/4 </span>
<span>=2 sqrt3/4 </span>
<span>= sqrt3/2 </span>
<span>so L.H.S =R.H.S or cos(A - B) = cosAcosB + sinAsinB</span>
Answer:

Step-by-step explanation:
given is the Differential equation in I order linear as

Take Laplace on both sides
![L(y') +4L(y) = 48L(t)\\sY(s)-y(0) +4Y(s) = 48 *\frac{1}{s^2} \\Y(s) [s+4]=\frac{48}{s^2}+9\\Y(s) = \frac{1}{s^2(s+4)}+\frac{9}{s+4}](https://tex.z-dn.net/?f=L%28y%27%29%20%2B4L%28y%29%20%3D%2048L%28t%29%5C%5CsY%28s%29-y%280%29%20%2B4Y%28s%29%20%3D%2048%20%2A%5Cfrac%7B1%7D%7Bs%5E2%7D%20%5C%5CY%28s%29%20%5Bs%2B4%5D%3D%5Cfrac%7B48%7D%7Bs%5E2%7D%2B9%5C%5CY%28s%29%20%3D%20%5Cfrac%7B1%7D%7Bs%5E2%28s%2B4%29%7D%2B%5Cfrac%7B9%7D%7Bs%2B4%7D)
Now if we take inverse we get y(t) the solution
Thus the algebraic equation would be
Answer:
7.56 km²
Step-by-step explanation:
Given data:
Width of the fjord, w = 6.3 km
Retreated terminus of the glacier between may 2001 and June 2005, d = 7.5 km
thus, the length lost , y = 7.5 - 6.3 = 1.2 km
now, the area is given as:
A = Length × width
on substituting the values, we get
A = 1.2 × 6.3
or
A = 7.56 km²
Hence, the surface area lost by the glacier in the fjord is 7.56 km²
Anything to the power of 0 is one.
Therefore we have:
(1)(y^-7z)
Which simplifies to:
y^-7z