4x = 8, so x = 2
5y + 4 = 4, so 5y = 0, so y = 0
(2,0)
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
The answer is c
1/6 divided by 1/2
i had this question on my own
Answer:
substituting the given values we get,
3(1-1)+2*2
using bodmas we get
3(0)+2*2
0+2*2
0+4
4
therefore the answer is 4,thank you and mark the answer as brainliest if u find my answer helpful
Answer: i - j - k
Step-by-step explanation:
Taking the cross product between two vectors will give you a third vector that is orthogonal(perpendicular) to both vectors.
<1,1,0> x <1,0,1>
![det(\left[\begin{array}{ccc}i&j&k\\1&1&0\\1&0&1\end{array}\right] )](https://tex.z-dn.net/?f=det%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C1%261%260%5C%5C1%260%261%5Cend%7Barray%7D%5Cright%5D%20%29)
the determinate of the matrix: <1,-(1),-1>
or: i - j - k