D. f(x)=4 (5/2)^x
Are really good tool for graphing problems is Demos.com
7.89 is greater than 7.189, 6.03 is equal to 6.030, because that last zero doesn't really change anything it cancels itself out
Answer:
-4
Step-by-step explanation:
The common difference is found by subtracting the second term from a previous term
So...
3-7 would give the common difference = -4
-1-3 would give the common difference= -4
-5-1 would give the common difference= -4
We have a right triangle with a 15m hypotenuse and a 8m leg. If we use x for the missing leg then the Pythagorean Theorem states that:

Then we have to solve that equation for x:
![\begin{gathered} x^2=15^2-8^2=225-64 \\ x^2=161 \\ x=\sqrt[]{161} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%5E2%3D15%5E2-8%5E2%3D225-64%20%5C%5C%20x%5E2%3D161%20%5C%5C%20x%3D%5Csqrt%5B%5D%7B161%7D%20%5Cend%7Bgathered%7D)
So the answer is the square root of 161.
Answer:
![f(x)=4\sqrt[3]{16}^{2x}](https://tex.z-dn.net/?f=f%28x%29%3D4%5Csqrt%5B3%5D%7B16%7D%5E%7B2x%7D)
Step-by-step explanation:
We believe you're wanting to find a function with an equivalent base of ...
![4\sqrt[3]{4}\approx 6.3496](https://tex.z-dn.net/?f=4%5Csqrt%5B3%5D%7B4%7D%5Capprox%206.3496)
The functions you're looking at seem to be ...
![f(x)=2\sqrt[3]{16}^x\approx 2\cdot2.5198^x\\\\f(x)=2\sqrt[3]{64}^x=2\cdot 4^x\\\\f(x)=4\sqrt[3]{16}^{2x}\approx 4\cdot 6.3496^x\ \leftarrow\text{ this one}\\\\f(x)=4\sqrt[3]{64}^{2x}=4\cdot 16^x](https://tex.z-dn.net/?f=f%28x%29%3D2%5Csqrt%5B3%5D%7B16%7D%5Ex%5Capprox%202%5Ccdot2.5198%5Ex%5C%5C%5C%5Cf%28x%29%3D2%5Csqrt%5B3%5D%7B64%7D%5Ex%3D2%5Ccdot%204%5Ex%5C%5C%5C%5Cf%28x%29%3D4%5Csqrt%5B3%5D%7B16%7D%5E%7B2x%7D%5Capprox%204%5Ccdot%206.3496%5Ex%5C%20%5Cleftarrow%5Ctext%7B%20this%20one%7D%5C%5C%5C%5Cf%28x%29%3D4%5Csqrt%5B3%5D%7B64%7D%5E%7B2x%7D%3D4%5Ccdot%2016%5Ex)
The third choice seems to be the one you're looking for.