Where? Are the expressions?
The HA Theorem is a special case of the AAS postulate.
Answer:
B
Step-by-step explanation:
The second graph is the correct choice because the y-intercept is two, the graph is negative, and has a slope of m=3/1.
ANSWER:
Let t = logtan[x/2]
⇒dt = 1/ tan[x/2] * sec² x/2 × ½ dx
⇒dt = 1/2 cos² x/2 × cot x/2dx
⇒dt = 1/2 * 1/ cos² x/2 × cosx/2 / sin x/2 dx
⇒dt = 1/2 cosx/2 / sin x/2 dx
⇒dt = 1/sinxdx
⇒dt = cosecxdx
Putting it in the integration we get,
∫cosecx / log tan(x/2)dx
= ∫dt/t
= log∣t∣+c
= log∣logtan x/2∣+c where t = logtan x/2
Answer:
Area of plot =Area of Square +Area of triangle
=300²+1/2×104×180
=90000+9360
=99360m²
