First if a>0 sqrta=a^(1/2), and (a^p)^q=a^pxq
sqrt(a^4/9)= (a^4/9)^1/2=a^4/9x1/2=<span>a^(2/9)
so the answer is </span><span>a^(2/9)</span>
<span>∫(sinx cosx)^2dx = ∫(1/2sin 2x)^2 dx = 1/4∫sin^2 2x dx = 1/4∫1/2(1 - cos 4x)dx = 1/8∫(1 - cos 4x) dx = 1/8[x - sin 4x / 4] + c = 1/32(4x - sin 4x) + c
</span>
Answer:
V=-15
Step-by-step explanation:
What you do is you take the negative 3 and multiply it by the 5 at the top which makes -15/v
|AC| = 2|DE|
⇒
4x - 2 = 2(x + 6) = 2x + 12
2x = 14
x = 7
⇒
|AC| = 26
Step-by-step explanation:
first u need to pick a random coordinate to substitute the X . example 0
y = 2(0) + 1
y =1 , X= 0
X= 1
y= 2(1) + 1
y= 3, X= 1
then you continue this same process and again until u have at least two points to draw a straight line. it kinda depended if u want to fill the entire graph.
