First we need to convert the given equation to standard form, only then we can find the center and radius of the circle.
![x^{2} + y^{2} +18x+14y+105=0 \\ \\ x^{2} +18x+ y^{2}+14y=-105 \\ \\ x^{2} +2(x)(9)+ y^{2}+2(y)(7)=-105 \\ \\ x^{2} +2(x)(9)+ 9^{2} + [y^{2}+2(y)(7)+7^{2}] =-105+9^{2}+7^{2} \\ \\ (x+9)^{2}+ (y+7)^{2}=25 ](https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%2B%20y%5E%7B2%7D%20%2B18x%2B14y%2B105%3D0%20%5C%5C%20%20%5C%5C%20%0A%20x%5E%7B2%7D%20%2B18x%2B%20y%5E%7B2%7D%2B14y%3D-105%20%5C%5C%20%20%5C%5C%20%0A%20x%5E%7B2%7D%20%2B2%28x%29%289%29%2B%20y%5E%7B2%7D%2B2%28y%29%287%29%3D-105%20%5C%5C%20%20%5C%5C%20%0Ax%5E%7B2%7D%20%2B2%28x%29%289%29%2B%209%5E%7B2%7D%20%2B%20%5By%5E%7B2%7D%2B2%28y%29%287%29%2B7%5E%7B2%7D%5D%20%20%3D-105%2B9%5E%7B2%7D%2B7%5E%7B2%7D%20%20%5C%5C%20%20%5C%5C%20%0A%20%28x%2B9%29%5E%7B2%7D%2B%20%28y%2B7%29%5E%7B2%7D%3D25%20%20%0A%20%20)
The standard equation of circle is:
with center (a,b) and radius = r
Comparing our equation to above equation, we can write
Center of circle is (-9, -7) and radius of the given circle is 5
Answer:
Step-by-step explanation:
So the thing about adding of integers is that is both nos. are positive then you can simply add it and same with negative too.But with negative and positive we should subtract but that is for later.Luckily we have 2 positive integers so we do it like this:
+6 is the same as 6 and +2 is the same as 2 so we...
rule:add the numbers and add the sign.
So, it is asking for 150 is 23% of what
So, the answer is 652.17
I think
Hope this helps :D
Answer:
V = (About) 22.2, Graph = First graph/Graph in the attachment
Step-by-step explanation:
Remember that in all these cases, we have a specified method to use, the washer method, disk method, and the cylindrical shell method. Keep in mind that the washer and disk method are one in the same, but I feel that the disk method is better as it avoids splitting the integral into two, and rewriting the curves. Here we will go with the disk method.
![\mathrm{V\:=\:\pi \int _a^b\left(r\right)^2dy\:},\\\mathrm{V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy}](https://tex.z-dn.net/?f=%5Cmathrm%7BV%5C%3A%3D%5C%3A%5Cpi%20%5Cint%20_a%5Eb%5Cleft%28r%5Cright%29%5E2dy%5C%3A%7D%2C%5C%5C%5Cmathrm%7BV%5C%3A%3D%5C%3A%5Cint%20_1%5E3%5C%3A%5Cpi%20%5Cleft%5B%5Cleft%281%2B%5Cfrac%7B2%7D%7By%7D%5Cright%29%5E2-1%5Cright%5Ddy%7D)
The plus 1 in '1 + 2/x' is shifting this graph up from where it is rotating, but the negative 1 is subtracting the area between the y-axis and the shaded region, so that when it's flipped around, it becomes a washer.
![V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy,\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=\pi \cdot \int _1^3\left(1+\frac{2}{y}\right)^2-1dy\\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\= \pi \left(\int _1^3\left(1+\frac{2}{y}\right)^2dy-\int _1^31dy\right)\\\\](https://tex.z-dn.net/?f=V%5C%3A%3D%5C%3A%5Cint%20_1%5E3%5C%3A%5Cpi%20%5Cleft%5B%5Cleft%281%2B%5Cfrac%7B2%7D%7By%7D%5Cright%29%5E2-1%5Cright%5Ddy%2C%5C%5C%5C%5C%5Cmathrm%7BTake%5C%3Athe%5C%3Aconstant%5C%3Aout%7D%3A%5Cquad%20%5Cint%20a%5Ccdot%20f%5Cleft%28x%5Cright%29dx%3Da%5Ccdot%20%5Cint%20f%5Cleft%28x%5Cright%29dx%5C%5C%3D%5Cpi%20%5Ccdot%20%5Cint%20_1%5E3%5Cleft%281%2B%5Cfrac%7B2%7D%7By%7D%5Cright%29%5E2-1dy%5C%5C%5C%5C%5Cmathrm%7BApply%5C%3Athe%5C%3ASum%5C%3ARule%7D%3A%5Cquad%20%5Cint%20f%5Cleft%28x%5Cright%29%5Cpm%20g%5Cleft%28x%5Cright%29dx%3D%5Cint%20f%5Cleft%28x%5Cright%29dx%5Cpm%20%5Cint%20g%5Cleft%28x%5Cright%29dx%5C%5C%3D%20%5Cpi%20%5Cleft%28%5Cint%20_1%5E3%5Cleft%281%2B%5Cfrac%7B2%7D%7By%7D%5Cright%29%5E2dy-%5Cint%20_1%5E31dy%5Cright%29%5C%5C%5C%5C)
Our exact solution will be V = π(4In(3) + 8/3). In decimal form it will be about 22.2 however. Try both solution if you like, but it would be better to use 22.2. Your graph will just be a plot under the curve y = 2/x, the first graph.
