A. all real numbers for both
Explanation:
Domain is the x-values. Since the graph goes forever horizontally, any x-value is on the graph.
Range is the same but with y-values. If the graph went on further than what the picture shows, then it also would show that every y value is on the graph
hope this helps
Answer:
false all the quadratic equations cannot be solved by factorising there is also graphical method to get the solution
Answer:
A
Step-by-step explanation:
Note that for (3, 2) → (5, 3)
The x- coordinate of the image is 2 more (+ 2) than the original and the y- coordinate of the image is 1 more (+ 1) than the original, that is
(x, y ) → (x + 2, y + 1 ) ← rule to obtain image, hence
(2, 4 ) → (2 + 2, 4 + 1 ) → (4, 5 ) → A
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.
Answer:
see work attached and shown
