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Dividing by a fraction is the same as multiplying by the reciprocal. so dividing by 1/2 is the same as multiplying by 2/1, or 2. so (3/4)/(1/2) = (3/4)*2 = 6/4 = 1 and 1/2
Answer:
13. 169π ft²
16. 34π mm
17. 12π in
18. 441π m²
Step-by-step explanation:
Circumference of a circle is calculated as follows:
C = 2π*r
Area of a circle is calculated as follows:
A = π*r²
where <em>r</em> is the radius of the circle
13. circumference of the circle: 26π ft
Then, its radius is:
26π = 2π*r
r = 26π/(2π) = 13 ft
A = π*13² = 169π ft²
16. area of the circle: 289π mm²
Then, its radius is:
289π = π*r²
r = √289
r = 17 mm
C = 2π*17 = 34π mm
17. area of the circle: 113.1 in²
Then, its radius is:
113.1 = π*r²
r = √(113.1/π)
r = 6 in
C = 2π*6 = 12π in
18. circumference of the circle: 131.95 m
Then, its radius is:
131.95 = 2π*r
r = 131.95/(2π)
r = 21 m
A = π*21² = 441π m²
I assume there's a plus sign missing above...
Convert to polar coordinates, using
![\begin{cases}x(r,\theta)=r\cos\theta\\y(r,\theta)=r\sin\theta\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dx%28r%2C%5Ctheta%29%3Dr%5Ccos%5Ctheta%5C%5Cy%28r%2C%5Ctheta%29%3Dr%5Csin%5Ctheta%5Cend%7Bcases%7D)
Then the Jacobian is
![\dfrac{\partial(x,y)}{\partial(r,\theta)}=\begin{vmatrix}x_r&y_r\\x_\theta&y_\theta\end{vmatrix}=\begin{vmatrix}\cos\theta&\sin\theta\\r\sin\theta&-r\cos\theta\end{vmatrix}=-r](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%28x%2Cy%29%7D%7B%5Cpartial%28r%2C%5Ctheta%29%7D%3D%5Cbegin%7Bvmatrix%7Dx_r%26y_r%5C%5Cx_%5Ctheta%26y_%5Ctheta%5Cend%7Bvmatrix%7D%3D%5Cbegin%7Bvmatrix%7D%5Ccos%5Ctheta%26%5Csin%5Ctheta%5C%5Cr%5Csin%5Ctheta%26-r%5Ccos%5Ctheta%5Cend%7Bvmatrix%7D%3D-r)
Then
![\mathrm dA=\mathrm dx\,\mathrm dy=|-r|\,\mathrm dr\,\mathrm d\theta=r\,\mathrm dr\,\mathrm d\theta](https://tex.z-dn.net/?f=%5Cmathrm%20dA%3D%5Cmathrm%20dx%5C%2C%5Cmathrm%20dy%3D%7C-r%7C%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta%3Dr%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta)
so the integral can be written as
![\displaystyle\iint_R\sin(x^2+y^2)\,\mathrm dA=\int_0^{\pi/2}\int_1^5r\sin(r^2)\,\mathrm dr\,\mathrm d\theta](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_R%5Csin%28x%5E2%2By%5E2%29%5C%2C%5Cmathrm%20dA%3D%5Cint_0%5E%7B%5Cpi%2F2%7D%5Cint_1%5E5r%5Csin%28r%5E2%29%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta)
Let
![s=r^2](https://tex.z-dn.net/?f=s%3Dr%5E2)
, so that
![\dfrac{\mathrm ds}2=r\,\mathrm dr](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20ds%7D2%3Dr%5C%2C%5Cmathrm%20dr)
.