Answer:400
Step-by-step explanation:it didn’t say it gives away and souvenirs so the 6$ can be out the equation and 4000 divided by 10 is 400
$12.72 Because 6% of 12 is 12 multiplied by 0.06 = 0.72. Then 12 + 0.72 = $12.72
Hope this helps!!!
![y=\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt](https://tex.z-dn.net/?f=y%3D%5Cdisplaystyle%5Cint_1%5Ex%5Csqrt%7Bt%5E3-1%7D%5C%2C%5Cmathrm%20dt)
By the fundamental theorem of calculus,
![\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm d}{\mathrm dx}\displaystyle\int_1^x\sqrt{t^3-1}\,\mathrm dt=\sqrt{x^3-1}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cdisplaystyle%5Cint_1%5Ex%5Csqrt%7Bt%5E3-1%7D%5C%2C%5Cmathrm%20dt%3D%5Csqrt%7Bx%5E3-1%7D)
Now the arc length over an arbitrary interval
![(a,b)](https://tex.z-dn.net/?f=%28a%2Cb%29)
is
![\displaystyle\int_a^b\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx=\int_a^b\sqrt{1+x^3-1}\,\mathrm dx=\int_a^bx^{3/2}\,\mathrm dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_a%5Eb%5Csqrt%7B1%2B%5Cleft%28%5Cfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%5Cright%29%5E2%7D%5C%2C%5Cmathrm%20dx%3D%5Cint_a%5Eb%5Csqrt%7B1%2Bx%5E3-1%7D%5C%2C%5Cmathrm%20dx%3D%5Cint_a%5Ebx%5E%7B3%2F2%7D%5C%2C%5Cmathrm%20dx)
But before we compute the integral, first we need to make sure the integrand exists over it.
![x^{3/2}](https://tex.z-dn.net/?f=x%5E%7B3%2F2%7D)
is undefined if
![x](https://tex.z-dn.net/?f=x%3C0)
, so we assume
![a\ge0](https://tex.z-dn.net/?f=a%5Cge0)
and for convenience that
![a](https://tex.z-dn.net/?f=a%3Cb)
. Then
Answer:
(A)
Step-by-step explanation:
M=-2
therefore
x¹=3, y¹=-12, x²=6 y²=k
M=(y²-y¹)/(x²-x¹)
-2=(k+12)/(6-3)
-2×3=k+12
-6=k+12
k=-18