Answer:
5
Step-by-step explanation:
5+7
Answer:
3
Step-by-step explanation:
lim(t→∞) [t ln(1 + 3/t) ]
If we evaluate the limit, we get:
∞ ln(1 + 3/∞)
∞ ln(1 + 0)
∞ 0
This is undetermined. To apply L'Hopital's rule, we need to rewrite this so the limit evaluates to ∞/∞ or 0/0.
lim(t→∞) [t ln(1 + 3/t) ]
lim(t→∞) [ln(1 + 3/t) / (1/t)]
This evaluates to 0/0. We can simplify a little with u substitution:
lim(u→0) [ln(1 + 3u) / u]
Applying L'Hopital's rule:
lim(u→0) [1/(1 + 3u) × 3 / 1]
lim(u→0) [3 / (1 + 3u)]
3 / (1 + 0)
3
The picture can be divided into 2 same triangles and one rectangular.
The triangle areas is {1/2 *base * height}*2 (because 2 same triangles)
and you need to sum rectangular area (width * base).
The result will be area of 2 triangle +area of rectangular.
=2(1/2*base of triangle* height)+ (base of rectangular * height)
= (base of triangle* height)+(base of rectangular * height)
=(base of triangle +base of rectangular)*height
b1 is top length
b2 is bottom length, and so
Base of triangle =(b1-b2/2)
Base of rectangular =b2
Height =h
Thus A ={(b1-b2/2)+b2 }*h
A= [(b1+b2)/2]*h
Answer:
n=209 answer for number 1.
Answer:
B
Step-by-step explanation:
it between B and c not really sure though
