<span>i Ace</span>hello :
let : A(-6,6) B(6,-2)
the center is w((-6+6)/2 , (6-2)/2)....(midel <span>[ AB]
w(0 ,2)
the ridus is : r = AB/2
AB = </span>√(-6-6)²+(6+2)² = √(144+64) =<span>√208/2
</span>an equation in <span>standard form equation of the circle :
</span>(x+0)²+(y-2)² = (√208/2)²=208/4 = 52
Answer is 4 .............
First one single solution second no solution 3rd one single forth one infinity and last 2 are single and infinity
Answer:
1728 for A
7290000 for B
Step-by-step explanation:
Answer:
Step-by-step explanation:
- If f(x) is in th form of f(x)=g(x)-h(x) then f'(x)=g'(x) - h'(x)
- When f(x)=z(g(x)) then f'(x)= z'(g(x))g'(x) (called as chain rule)
<u>using these information</u>:
g(x)=ln2x then g'(x)=
h(x)=In(3x - 1) then h'(x)=![\frac{(3x-1)'}{3x-1} =\frac{3}{3x-1}f'(x)=g'(x) - h'(x) =[tex]\frac{1}{x} - \frac{3}{3x-1} =\frac{-1}{3x^2-x}](https://tex.z-dn.net/?f=%5Cfrac%7B%283x-1%29%27%7D%7B3x-1%7D%20%3D%5Cfrac%7B3%7D%7B3x-1%7D%3C%2Fp%3E%3Cp%3Ef%27%28x%29%3Dg%27%28x%29%20-%20h%27%28x%29%20%3D%5Btex%5D%5Cfrac%7B1%7D%7Bx%7D%20-%20%5Cfrac%7B3%7D%7B3x-1%7D%20%3D%5Cfrac%7B-1%7D%7B3x%5E2-x%7D)
