Answer:
½ ln 3
Step-by-step explanation:
∫ sec²x / tan x dx
If u = tan x, then du = sec²x dx.
∫ du / u
ln|u| + C
ln|tan x| + C
Evaluate between π/4 and π/3.
ln|tan(π/3)| + C − (ln|tan(π/4)| + C)
ln|√3| + C − ln|1| − C
ln(√3)
½ ln 3
This is a refreshing question!
We are given that
f(r)=ar+b, and
Sum f(r) =125 for r=1 to 5
Sum f(r) = 475 for r=1 to 10.
and we know, using Gauss's method, that
G(n)=sum (1,2,3.....n) = n(n+1)/2 or
G(n)=n(n+1)/2
Sum f(r) =125 for r=1 to 5
=>
sum=a(sum of 1 to 5) + 5b => G(5)a+5b=125 [G(5)=15]
15a+5b=125 ...................................................(1)
Similarly, Sum f(r) = 475 for r=1 to 10 => G(10)a+5b=475 [G(10)=55]
=>
55a+10b=475.................................................(2)
Solve system of equations (1) and (2)
(2)-2(1)
55-2(15)a=475-2(125) => 25a=225 =>
a=9
Substitute a=9 in 1 => 15(9)+5b=125 => 5b=-10
b=-2
Substitute a and b into f(r),
f(r)=9r-2
check: sum f(r), r=1,5 = (9-2)+(18-2)+(27-2)+(36-2)+(45-2)=135-10=125 [good]
We define the sum of f(r) for r=1 to n as
S(n)=sum f(r) for r=1 to n = 9(sum 1,2,3....n)-2n = 9n(n+1)/2-2n = 9G(n)-2n
S(n)=9n(n+1)/2-2n
checks:
S(5)=9(15)-2(5)=135-10=125 [good]
S(10)=9(55)-2(10)=495-20=475 [good]
Hence
(a)
S(n)=sum f(r) for r=1,n
= 9(sum i=1,n)+n(-2)
= 9(n(n+1)/2 -2n
=(9(n^2+n)/2) -2n
(b) sum f(r) for i=8,18
=sum f(r) for i=1,18 - sum f(r) for i=1,7
=S(18)-S(7)
=(9(18^2-18)/2-2(18))-(9(7^2-7)/2-2(7))
=1503-238
=1265
Answer:
Y= -4X + 8
Step-by-step explanation:
Use rise/run to find slope (y2- y1/ x2-x1). Plug the slope into point-slope form [y-y1= m(x-x1)]. this is your answer
180 I believe because all angles in a triangles will ALWAYS add up to 180
