Answer:
(1-cosA)/(1+cosA)
=(1-cosA)/(1+cosA) ×(1-cosA)/(1-cosA)
=(1-cosA)^2 /(1-cos^2A)
=(1-cosA)^2 / (sin^2A)
=(1/sinA - cosA/sinA )^2
=cosecA - cotA)^2
Hope it helps
Have a great Day ; )
A population of size P increasing at the rate of 2% may be modelled as follows
P = P0<span> e</span>0.02 t<span> , where t is the number of years after t = 0 and P</span>0<span> is the population at t = 0. </span>
Let t = 0 corresponds to January 1, 2000 and therefore t = 4 corresponds to December 31. But P is 2,000,000 when t = 4. Hence
2,000,000 = P0<span> e</span>0.02*4
<span>Solve the above for P0</span>
P0<span> = 2,000,000 / e</span>0.02*4<span> = 1 846 000 (rounded to the nearest thousand) </span>
<span>P0 is the population at t = 0 or on January 1, 2000.
A. is the correct answer</span>
Answer:
(u+4)(y-3)
Step-by-step explanation:
uy + 4y - 3u - 12=
uy + 4y - 3u - 3*4=
y(u+4)-3(u+4)=
(u+4)(y-3)
Answer:
6/8X+3
Step-by-step explanation: