A trig identity is <span>asinucosu=<span>a/2</span>sin(2u)</span>So you can write your equation as<span>y=sin(x)cos(x)=<span>1/2</span>sin(2x)</span>Use the crain rule here<span><span>y′</span>=<span>d/<span>dx</span></span><span>1/2</span>sin(2x)=<span>1/2</span>cos(2x)<span>d/<span>dx</span></span>2x=cos(2x)</span>The curve will have horizontal tangents when y' = 0.<span><span>y′</span>=0=cos(2x)</span>On the interval [-pi, pi], solution to that is<span><span>x=±<span>π4</span>,±<span><span>3π</span>4</span></span></span>
Hello :
<span>a point on the y-axis is : P( 0, y)
</span>PA = PB or PA² = PB².... A(7, -7) B(1,1)
(7-0)²+(-7-y)² = (1-0)² + (1-y)²
49+49+14y+y² = 1+1-2y+y²
16y = - 96
y= - 6
It would be the first one (x,-y) because say for example you take the point of v1 (4,-1) and reflect across x axis you get (4,1) for v
B. All rectangles are parallelograms. <span />
The solution to the problem is as follows:
If you use Ln you can get rid of e.
ln(e^0.4x) = ln(0.4)
0.4x = ln(0.4)
x = (ln(0.4))/0.4
<span>x = -2.29
</span>
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