Answer:
the answer is D
Step-by-step explanation:
the color changes aswell in the sequence
A=Value of investment.
A=12000(1+0.015)^2
A= <span>£12362.70</span>
Answer:
AE = 15 cm; ED = 18 cm; AD = 15 cm (given)
Step-by-step explanation:
ΔBEC ~ ΔAED so ...
AD/BC = AE/BE = (BE+AB)/BE = 1 + AB/BE
Substituting given numbers (lengths in centimeters), we have ...
15/10 = 1 + 5/BE
1/2 = 5/BE
BE = 10
Similarly, ...
1/2 = 6/CE
CE = 12
Then the unknown sides are ...
AE = AB + BE = 5 + 10 = 15 . . . cm
ED = CE + CD = 12 + 6 = 18 . . . cm
Answer:
Step-by-step explanation:
(a)
Consider the following:
Use sine rule,
![\frac{b}{a}=\frac{\sinB}{\sin A} \\\\=\frac{\sin{\frac{\pi}{3}} }{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}](https://tex.z-dn.net/?f=%5Cfrac%7Bb%7D%7Ba%7D%3D%5Cfrac%7B%5CsinB%7D%7B%5Csin%20A%7D%0A%5C%5C%5C%5C%3D%5Cfrac%7B%5Csin%7B%5Cfrac%7B%5Cpi%7D%7B3%7D%7D%0A%7D%7B%5Csin%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%7D%7D%5C%5C%5C%5C%3D%5Cfrac%7B%5B%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%5D%7D%7B%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%7D%5C%5C%5C%5C%3D%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5Ctimes%20%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B1%7D%3D%5Csqrt%7B%5Cfrac%7B3%7D%7B2%7D%7D)
Again consider,
![\frac{b}{a}=\frac{\sin{B}}{\sin{A}} \\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]](https://tex.z-dn.net/?f=%5Cfrac%7Bb%7D%7Ba%7D%3D%5Cfrac%7B%5Csin%7BB%7D%7D%7B%5Csin%7BA%7D%7D%0A%5C%5C%5C%5C%5Csin%7BB%7D%3D%5Cfrac%7Bb%7D%7Ba%7D%5Ctimes%20%5Csin%7BA%7D%5C%5C%5C%5C%5Csin%7BB%7D%3D%5Csqrt%7B%5Cfrac%7B3%7D%7B2%7D%7D%5Csin%20%7BA%7D%5C%5C%5C%5CB%3D%5Csin%5E%7B-1%7D%5B%5Csqrt%7B%5Cfrac%7B3%7D%7B2%7D%7D%5Csin%7BA%7D%5D)
Thus, the angle B is function of A is, ![B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]](https://tex.z-dn.net/?f=B%3D%5Csin%5E%7B-1%7D%5B%5Csqrt%7B%5Cfrac%7B3%7D%7B2%7D%7D%5Csin%7BA%7D%5D)
Now find 
Differentiate implicitly the function 
with respect to A to get,
b)
When 
, the value of is,
c)
In general, the linear approximation at x= a is,
Here the function ![f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]](https://tex.z-dn.net/?f=f%28A%29%3DB%3D%5Csin%5E%7B-1%7D%5B%5Csqrt%7B%5Cfrac%7B3%7D%7B2%7D%7D%5Csin%7BA%7D%5D)
At 
![f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}](https://tex.z-dn.net/?f=f%28%5Cfrac%7B%5Cpi%7D%7B4%7D%29%3DB%3D%5Csin%5E%7B-1%7D%5B%5Csqrt%7B%5Cfrac%7B3%7D%7B2%7D%7D%5Csin%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%7D%5D%5C%5C%5C%5C%3D%5Csin%5E%7B-1%7D%5B%5Csqrt%7B%5Cfrac%7B3%7D%7B2%7D%7D.%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%5D%5C%5C%5C%5C%5C%3D%5Csin%5E%7B-1%7D%28%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%29%5C%5C%5C%5C%3D%5Cfrac%7B%5Cpi%7D%7B3%7D)
And,
from part b
Therefore, the linear approximation at is,
![f(x)=f'(A).(x-A)+f(A)\\\\=f'(\frac{\pi}{4}).(x-\frac{\pi}{4})+f(\frac{\pi}{4})\\\\=\sqrt{3}.[x-\frac{\pi}{4}]+\frac{\pi}{3}](https://tex.z-dn.net/?f=f%28x%29%3Df%27%28A%29.%28x-A%29%2Bf%28A%29%5C%5C%5C%5C%3Df%27%28%5Cfrac%7B%5Cpi%7D%7B4%7D%29.%28x-%5Cfrac%7B%5Cpi%7D%7B4%7D%29%2Bf%28%5Cfrac%7B%5Cpi%7D%7B4%7D%29%5C%5C%5C%5C%3D%5Csqrt%7B3%7D.%5Bx-%5Cfrac%7B%5Cpi%7D%7B4%7D%5D%2B%5Cfrac%7B%5Cpi%7D%7B3%7D)
d)
Use part (c), when 
, B is approximately,
![B=f(46°)=\sqrt{3}[46°-\frac{\pi}{4}]+\frac{\pi}{3}\\\\=\sqrt{3}(1°)+\frac{\pi}{3}\\\\=61.732°](https://tex.z-dn.net/?f=B%3Df%2846%C2%B0%29%3D%5Csqrt%7B3%7D%5B46%C2%B0-%5Cfrac%7B%5Cpi%7D%7B4%7D%5D%2B%5Cfrac%7B%5Cpi%7D%7B3%7D%5C%5C%5C%5C%3D%5Csqrt%7B3%7D%281%C2%B0%29%2B%5Cfrac%7B%5Cpi%7D%7B3%7D%5C%5C%5C%5C%3D61.732%C2%B0)
Answer:
25.12
Step-by-step explanation:
Formula: 2πr, r = d/2
Given: π = 3.14, d = 8
Sub: r = 8/2
Simplify: r = 4
Sub: 2(3.14)(4)
Simplify: 6.28(4)
Solve: 25.12
