The formula
in solving the integral of the infinity of 3 is ∫3<span>∞?</span>(1<span>)÷((</span>x−2<span><span>)<span><span>(3/</span><span>2)</span></span></span>)</span><span>dx</span>
Substitute the numbers given
then solve
limn→inf∫3n(1/((n−2)(3/2))dn
limn→inf[−2/(n−2−−−−−√)−(−2/3−2−−−−√)
=0+2=2
Solve for the integral of 2 when 2 is approximate to 0.
Transpose 2 from the other side to make it -2
∫∞3(x−2)−3/2dx=(x−2)−1/2−1/2+C
(x−2)−1/2=1x−2−−−−√
0−(3−2)−1/2−1/2=2
<span> </span>
I think it is the first one -2
Step-by-step explanation:
tanA=8/15 and tan B=40/9
tan(A+B)=(tanA+tanB)/1-tanAtanB
(8/15+40/9)/1-(8/15×40/9)
=-672/185
Answer:
cos=24/25
tan=7/24
Step-by-step explanation:
sin=opposite/hypotenuse
therefore 7 is the opposite and 25 is the hypotenuse.
using pythagorean theorem we can find the adjacent. i.e hypotenuse squared= sum of adjacent squared and opposite squared.
h^2= a^2+ O^2
25 squared - 7 squared= 576
which is equal to 24 squared.
therefore adjacent=24
cos=adjacent/hypotenuse
cos=24/25
tan=opposite/adjacent
tan=7/24.
