Using the identity cos^2(A)=1-sin^2(A)
transform
integral of sin^2(πx)cos<span>^5(πx)dx
=</span>integral of sin^2(πx)[1-sin^2(πx)]^2 cos(πx)dx
=integral of [sin^2(πx)-2sin^4(πx)+sin^6(πx)]cos(πx)dx
using the substitution u=sin(πx), du=πcos(πx)dx
=integral of [u^2-2u^4+u^6] (du/π)
=1/π[u^3/3-(2/5)u^5+u^7/7] + C
back substitute u=sin(πx)
=1/π[sin(πx)^3/3-(2/5)sin(πx)^5+sin(πx)^7/7]
or
=sin(πx)^3/3π-2sin(πx)^5/5π+sin(πx)^7/7π
Answer:
35
y = -4(-8) +3
y= 35
yea
that is the out put so 35
1 4 7 I think
I don’t know for sure though
Answer:
P ( x ) = -0.7 (x - 2)²(x + 3)
Step-by-step explanation:
<u>We are given</u> :
P ( x ) , has a root of multiplicity 2 at x = 2
and a root of multiplicity 1 at x = − 3
Then
P ( x ) = a (x - 2)²(x + 3) ; where ‘a’ is a real number.
P ( x ) = a (x - 2)²(x + 3)
= a (x² - 4x + 4)(x + 3)
= a [x³ - 4x² + 4x + 3x² - 12x + 12]
P (0) = -8.4
⇔ a [(0)³ - 4(0)² + 4(0) + 3(0)² - 12(0) + 12] = -8.4
⇔ 12 a = -8.4
⇔ a = (-8,4) ÷ 12
⇔ a = -0,7
<u>Conclusion</u> :
P ( x ) = -0.7 (x - 2)²(x + 3)
There will be 42 in each row with 2 left over