We know that :



Using above ideas we can solve the Problem :
⇒ 
⇒ ![ln(x - 3) - ln(x + 3)^\frac{3}{8} = ln[\frac{(x - 3)}{(x + 3)^\frac{3}{8}}]](https://tex.z-dn.net/?f=ln%28x%20-%203%29%20-%20ln%28x%20%2B%203%29%5E%5Cfrac%7B3%7D%7B8%7D%20%3D%20ln%5B%5Cfrac%7B%28x%20-%203%29%7D%7B%28x%20%2B%203%29%5E%5Cfrac%7B3%7D%7B8%7D%7D%5D)
⇒ ![4ln[\frac{(x - 3)}{(x + 3)^\frac{3}{8}}] = ln[\frac{(x - 3)}{(x + 3)^\frac{3}{8}}]^4 = ln[\frac{(x - 3)^4}{(x + 3)^\frac{3}{2}}]](https://tex.z-dn.net/?f=4ln%5B%5Cfrac%7B%28x%20-%203%29%7D%7B%28x%20%2B%203%29%5E%5Cfrac%7B3%7D%7B8%7D%7D%5D%20%3D%20ln%5B%5Cfrac%7B%28x%20-%203%29%7D%7B%28x%20%2B%203%29%5E%5Cfrac%7B3%7D%7B8%7D%7D%5D%5E4%20%3D%20ln%5B%5Cfrac%7B%28x%20-%203%29%5E4%7D%7B%28x%20%2B%203%29%5E%5Cfrac%7B3%7D%7B2%7D%7D%5D)
⇒ ![\frac{1}{3}lnx + ln[\frac{(x - 3)^4}{(x + 3)^\frac{3}{2}}] = ln(x)^\frac{1}{3} + ln[\frac{(x - 3)^4}{(x + 3)^\frac{3}{2}}] = ln[\frac{\sqrt[3]{x}(x - 3)^4}{\sqrt{(x + 3)^{3}}}]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7Dlnx%20%2B%20ln%5B%5Cfrac%7B%28x%20-%203%29%5E4%7D%7B%28x%20%2B%203%29%5E%5Cfrac%7B3%7D%7B2%7D%7D%5D%20%3D%20ln%28x%29%5E%5Cfrac%7B1%7D%7B3%7D%20%2B%20ln%5B%5Cfrac%7B%28x%20-%203%29%5E4%7D%7B%28x%20%2B%203%29%5E%5Cfrac%7B3%7D%7B2%7D%7D%5D%20%3D%20ln%5B%5Cfrac%7B%5Csqrt%5B3%5D%7Bx%7D%28x%20-%203%29%5E4%7D%7B%5Csqrt%7B%28x%20%2B%203%29%5E%7B3%7D%7D%7D%5D)
Option 3 is the Answer
Answer:
Im not sure but i believe it is A. 5
Step-by-step explanation:
Hope this helps please mark brainlist if im right.
Answer:
4 congruent triangles
Step-by-step explanation:
since there is no one of the sum of two sides is less then the third side, so it's possible to form a triangle
Use straightedge and compass, it can draw 4 triangles at a common base
Volume of a cylinder is given by V = πr^2h
Since the content of the storage tank completely filled the 25 identical cylinders, this means that the volume of the 25 cylinders combined is equal to the volume of the storage tank.
25(πr^2h) = 235,500
πr^2h = 235,500/25 = 9,420
h = 9,420/(3.14 x 5^2) = 9,420/78.5 = 120.
Therefore, the height of the cylinderical containers is 120 in
1/2 is the probability the coin would land heads up both times.