Answer:
You should evaluate whatever is in the parenthesis first.
Answer:
17% i think
Step-by-step explanation:
Answer:
I think it's 7
Step-by-step explanation:
-6 < 2T (-5+5) < 3+5
-6 < 2T < 8
+6 < 2T < 8 + 6
0 < 2T < 14
0 < ÷ 2 < ÷ 2
0 < T < 7
-0 < T < -0
T = 7
It's 6x-18
Hope this helps
Answer:
Step-by-step explanation:
![f(x)=4\sqrt{2x^3-1}=4\left(2x^3-1\right)^\frac{1}{2}\\\\f'(x)=4\cdot\dfrac{1}{2}(2x^3-1)^{-\frac{1}{2}}\cdot3\cdot2x^2=\dfrac{12x^2}{(2x^3-1)^\frac{1}{2}}=\dfrac{12x^2}{\sqrt{2x^3-1}}\\\\\text{used}\\\\\sqrt{a}=a^\frac{1}{2}\\\\\bigg[f\left(g(x)\right)\bigg]'=f'(g(x))\cdot g'(x)\\\\\bigg[nf(x)\bigg]'=nf'(x)\\\\(x^n)'=nx^{n-1}](https://tex.z-dn.net/?f=f%28x%29%3D4%5Csqrt%7B2x%5E3-1%7D%3D4%5Cleft%282x%5E3-1%5Cright%29%5E%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5Cf%27%28x%29%3D4%5Ccdot%5Cdfrac%7B1%7D%7B2%7D%282x%5E3-1%29%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%5Ccdot3%5Ccdot2x%5E2%3D%5Cdfrac%7B12x%5E2%7D%7B%282x%5E3-1%29%5E%5Cfrac%7B1%7D%7B2%7D%7D%3D%5Cdfrac%7B12x%5E2%7D%7B%5Csqrt%7B2x%5E3-1%7D%7D%5C%5C%5C%5C%5Ctext%7Bused%7D%5C%5C%5C%5C%5Csqrt%7Ba%7D%3Da%5E%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%5Cbigg%5Bf%5Cleft%28g%28x%29%5Cright%29%5Cbigg%5D%27%3Df%27%28g%28x%29%29%5Ccdot%20g%27%28x%29%5C%5C%5C%5C%5Cbigg%5Bnf%28x%29%5Cbigg%5D%27%3Dnf%27%28x%29%5C%5C%5C%5C%28x%5En%29%27%3Dnx%5E%7Bn-1%7D)
