Answer:
Your Answer Would be the 3rd Option.
Step-by-step explanation:
The Initial Value is : 4. All you have to do is look for when the x = 0
ROC: 1/2. Minutes =x, Inches= Y. To find ROC, take
Y2 - Y1
----------
X2 - Y1
Replacing those with numbers:
5 - 4
-------
2 - 0
Your ROC is 1/2 .
Equation:
y= 1/2x + 4
Know that The 4 is the initial value, so it's always in that spot. Your ROC is always Next to the x, which also means it's the slope.
recall that the radius is the distance from the center of a circle to any point on the circle.
we know the center is at -16,-14, and we know that -8,-8 is a point on the circle, so the distance between both must be the radius.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-16}~,~\stackrel{y_1}{-14})\qquad (\stackrel{x_2}{-8}~,~\stackrel{y_2}{-8})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{radius}{r}=\sqrt{[-8-(-16)]^2+[-8-(-14)]^2} \\\\\\ r=\sqrt{(-8+16)^2+(-8+14)^2}\implies r=\sqrt{8^2+6^2} \\\\\\ r=\sqrt{100}\implies \boxed{r=10} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%0A%5C%5C%5C%5C%0A%28%5Cstackrel%7Bx_1%7D%7B-16%7D~%2C~%5Cstackrel%7By_1%7D%7B-14%7D%29%5Cqquad%0A%28%5Cstackrel%7Bx_2%7D%7B-8%7D~%2C~%5Cstackrel%7By_2%7D%7B-8%7D%29%5Cqquad%20%5Cqquad%0Ad%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bradius%7D%7Br%7D%3D%5Csqrt%7B%5B-8-%28-16%29%5D%5E2%2B%5B-8-%28-14%29%5D%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ar%3D%5Csqrt%7B%28-8%2B16%29%5E2%2B%28-8%2B14%29%5E2%7D%5Cimplies%20r%3D%5Csqrt%7B8%5E2%2B6%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ar%3D%5Csqrt%7B100%7D%5Cimplies%20%5Cboxed%7Br%3D10%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D)
![\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-16}{ h},\stackrel{-14}{ k})\qquad \qquad radius=\stackrel{10}{ r} \\\\\\\ [x-(-16)]^2+[y-(-14)]^2=10^2\implies \blacktriangleright (x+16)^2+(y+14)^2=100 \blacktriangleleft](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%0A%28x-%20h%29%5E2%2B%28y-%20k%29%5E2%3D%20r%5E2%0A%5Cqquad%0Acenter~~%28%5Cstackrel%7B-16%7D%7B%20h%7D%2C%5Cstackrel%7B-14%7D%7B%20k%7D%29%5Cqquad%20%5Cqquad%0Aradius%3D%5Cstackrel%7B10%7D%7B%20r%7D%0A%5C%5C%5C%5C%5C%5C%5C%0A%5Bx-%28-16%29%5D%5E2%2B%5By-%28-14%29%5D%5E2%3D10%5E2%5Cimplies%20%5Cblacktriangleright%20%28x%2B16%29%5E2%2B%28y%2B14%29%5E2%3D100%20%5Cblacktriangleleft)
Answer:
-13x + 10.3y
Step-by-step explanation:
Hello,
1) Conditions:
x+2>=0==>x>-2
x>=0
So x>=0
1/5 ln(x+2)^5 +1/2 [ln x -ln(x^2+3x+2)^2]
=ln(x+2)+ ln√x - ln(x+1)(x+2)