To evaluate 17 int (sin^2 (x) cos^3(x))
From Trig identity. Cos^2(x) + sin^2(x) =1. Cos^2(x) = 1 - sin^2 (x)
Cos^3(x) = cosx * (1 - sin^2 (x)) = cosx - cosxsin^2x
So we have 17 int (sin^2x(cosx - cosxsin^2x))
int (sin^2x(cosx)dx - int (sin^4xcosx)dx. ----------(1)
Let u = sinx then du = cosxdx
Substituting into (1) we have
int (u^2du) - int (u^4du)
u^3/3 - u^5/5
Substitute value for u we have
(sinx)^3/3 - (sinx)^5/5
Hence we have 17 [ sin^3x/3 - sin^5x/5]
Answer:
Total Cost (T)= .80a+1.25c
T= a+c
Step-by-step explanation:
This equation has 2 variables. It also has 2 equations
Answer:
Step-by-step explanation:
* Look to the attached file
If you complete the square you get
and as any number squared is positive and 4 is positive, the result must be positive
Answer:
<h2><em>
B. (b+3c)+(b+3c) </em></h2><h2><em>C. </em><em>
2(b)+2(3c)</em></h2>
Step-by-step explanation:
Given this expression 2(b+3c), its equivalent expression is derived by simply opening up the bracket as shown below;
Open the parenthesis by multiplying the constant outside the bracket with all the variables in parenthesis.
= 2(b+3c)
= 2(b)+ 2(3c)
= 2b +2*3*c
= 2b +6c
It can also be written as sum of b+3c in 2 places i.e (b+3c)+(b+3c) because multiplying the function b+3c by 2 means we are to add the function by itself in two places.
<em>Hence the equivalent expression are (b+3c)+(b+3c) and 2(b)+2(3c) or 2b+6c</em>
