Correct answer is C.
For example 2, 4, 8, 16, 32...
![2=\frac{4}{2} = \frac{8}{4} = \frac{16}{8} = \frac{32}{16} =...](https://tex.z-dn.net/?f=%202%3D%5Cfrac%7B4%7D%7B2%7D%20%3D%20%5Cfrac%7B8%7D%7B4%7D%20%3D%20%5Cfrac%7B16%7D%7B8%7D%20%3D%20%5Cfrac%7B32%7D%7B16%7D%20%3D...)
Answer:
That looks hard
Step-by-step explanation:
We have that
<span>3x-2y=8 -----> equation 1
2x+3y=Q----> equation 2
the solution is the point </span><span>(4,2)
in the equation 2 substitute the value of
x=4
y=2
so
</span>2x+3y=Q------> 2*4+3*2=Q-------> Q=8+6------> Q=14
<span>
the answer is
Q=14
</span>
It depends on what you mean by the delimiting carats "^"...
Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for
![\sqrt x](https://tex.z-dn.net/?f=%5Csqrt%20x)
.
In that case, you want to find the antiderivative,
![\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint%5Cfrac%7B%5Cmathrm%20dx%7D%7B%5Csqrt%7B9-8x-x%5E2%7D%7D)
Complete the square in the denominator:
![9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2](https://tex.z-dn.net/?f=9-8x-x%5E2%3D25-%2816%2B8x%2Bx%5E2%29%3D5%5E2-%28x%2B4%29%5E2)
Now substitute
![x+4=5\sin y](https://tex.z-dn.net/?f=x%2B4%3D5%5Csin%20y)
, so that
![\mathrm dx=5\cos y\,\mathrm dy](https://tex.z-dn.net/?f=%5Cmathrm%20dx%3D5%5Ccos%20y%5C%2C%5Cmathrm%20dy)
. Then
![\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint%5Cfrac%7B%5Cmathrm%20dx%7D%7B%5Csqrt%7B9-8x-x%5E2%7D%7D%3D%5Cint%5Cfrac%7B5%5Ccos%20y%7D%7B%5Csqrt%7B5%5E2-%285%5Csin%20y%29%5E2%7D%7D%5C%2C%5Cmathrm%20dy)
which simplifies to
![\displaystyle\int\frac{5\cos y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint%5Cfrac%7B5%5Ccos%20%0Ay%7D%7B5%5Csqrt%7B1-%5Csin%5E2y%7D%7D%5C%2C%5Cmathrm%20dy%3D%5Cint%5Cfrac%7B%5Ccos%20y%7D%7B%5Csqrt%7B%5Ccos%5E2y%7D%7D%5C%2C%5Cmathrm%20dy)
Now, recall that
![\sqrt{x^2}=|x|](https://tex.z-dn.net/?f=%5Csqrt%7Bx%5E2%7D%3D%7Cx%7C)
. But we want the substitution we made to be reversible, so that
![x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)](https://tex.z-dn.net/?f=x%2B4%3D5%5Csin%20y%5Ciff%20y%3D%5Csin%5E%7B-1%7D%5Cleft%28%5Cdfrac%7Bx%2B4%7D5%5Cright%29)
which implies that
![-\dfrac\pi2\le y\le\dfrac\pi2](https://tex.z-dn.net/?f=-%5Cdfrac%5Cpi2%5Cle%20y%5Cle%5Cdfrac%5Cpi2)
. (This is the range of the inverse sine function.)
Under these conditions, we have
![\cos y\ge0](https://tex.z-dn.net/?f=%5Ccos%20y%5Cge0)
, which lets us reduce
![\sqrt{\cos^2y}=|\cos y|=\cos y](https://tex.z-dn.net/?f=%5Csqrt%7B%5Ccos%5E2y%7D%3D%7C%5Ccos%20y%7C%3D%5Ccos%20y)
. Finally,
![\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint%5Cfrac%7B%5Ccos%20y%7D%7B%5Ccos%20y%7D%5C%2C%5Cmathrm%20dy%3D%5Cint%5Cmathrm%20dy%3Dy%2BC)
and back-substituting to get this in terms of
![x](https://tex.z-dn.net/?f=x)
yields
A=Annual amount=2000
i=annual interest=0.0205
n=number of years=3
Present value
=A((1+i)^n-1)/(i(1+i)^n)
=2000(1.0205^3-1)/(.0205(1.0205^3))
=5762.15