we have to choose the correct conic section to fit the equation.x ^2+ y^ 2= 9. we know the equation of a circle is x^2 + y^2 = r^2
and if we compare both the equations we find that both are similar and we can even find the value of r by comparing both the equations.
So the correct conic section is circle
I think that it would produce a circle i'm not sure though. do you have a picture to go along with it?
F(x)= ∫[√x, 1] (2t - 1)/(t + 2) dt for x ≥ 0
<span>Let f(x) = ∫(2t - 1)/(t + 2) dt = ∫[2 + -5/(t + 2)] dt = 2t - 5Ln(t + 2) + constant </span>
<span>F(x) = f(1) - f(√x) = 2 – 5Ln(3) - (2√x - 4Ln(√x + 2) </span>
<span>F(1) = 2 - 5Ln(3) - (2√1 - 5Ln(√1 + 2) = 2 - 5Ln(3) - (2 - 5Ln(3)) = 0
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Let the numbers be x and y
According to the data : xy=-15 and x+y=-7
xy=-15
x+y=-7
y=-7-x
x(-x-7)=-15
y=-7-x
-x^2-7x=-15
y=-7-x
x^2+7x-15=0
y=-7-x
x=(-7±sqrt(7*7+4*15))/2
y=-7-x
x=(-7±sqrt(109<span>))/2
</span>
Goes to two systems.
1)
y=-7-x
x=(-7+sqrt(109<span>))/2
</span>
y=-7+3.5-sqrt(109<span>))/2
</span>x=(-7+sqrt(109<span>))/2
</span>
y=-3.5-sqrt(109<span>))/2
</span>x=(-7+sqrt(109<span>))/2
</span>
2)
y=-7-x
x=(-7-sqrt(109<span>))/2
</span>
y=-7+3.5+sqrt(109))/2
x=(-7-sqrt(109))/2
y=-3.5+sqrt(109))/2
x=(-7-sqrt(109<span>))/2</span>