Answer:
no thank you
Step-by-step explanation:
Answer:
![\large\boxed{1.\ f^{-1}(x)=\sqrt[12]{3^x}}\\\\\boxed{2.\ f^{-1}(x)=\sqrt[4]{3^x}}\\\\\ \boxed{3.\ f^{-1}(x)=\sqrt[3]{4^{7-x}}}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B1.%5C%20f%5E%7B-1%7D%28x%29%3D%5Csqrt%5B12%5D%7B3%5Ex%7D%7D%5C%5C%5C%5C%5Cboxed%7B2.%5C%20f%5E%7B-1%7D%28x%29%3D%5Csqrt%5B4%5D%7B3%5Ex%7D%7D%5C%5C%5C%5C%5C%20%5Cboxed%7B3.%5C%20f%5E%7B-1%7D%28x%29%3D%5Csqrt%5B3%5D%7B4%5E%7B7-x%7D%7D%7D)
Step-by-step explanation:
![2.\\y=\log_3x^4\\\\\text{Exchange x and y. Solve for y:}\\\\\log_3y^4=x\Rightarrow3^{\log_3y^4}=3^x\Rightarrow y^{4}=3^x\\\\y=\sqrt[4]{3^x}\\-------------------------](https://tex.z-dn.net/?f=2.%5C%5Cy%3D%5Clog_3x%5E4%5C%5C%5C%5C%5Ctext%7BExchange%20x%20and%20y.%20Solve%20for%20y%3A%7D%5C%5C%5C%5C%5Clog_3y%5E4%3Dx%5CRightarrow3%5E%7B%5Clog_3y%5E4%7D%3D3%5Ex%5CRightarrow%20y%5E%7B4%7D%3D3%5Ex%5C%5C%5C%5Cy%3D%5Csqrt%5B4%5D%7B3%5Ex%7D%5C%5C-------------------------)
![3.\\y=-\log_4x^3+7\\\\\text{Exchange x and y. Solve for y:}\\\\-\log_4y^3+7=x\qquad\text{subtract 7 from both sides}\\\\-\log_4 y^3=x-7\qquad\text{change the signs}\\\\\log_4y^3=7-x\Rightarrow4^{\log_4y^3}=4^{7-x}\\\\y^3=4^{7-x}\Rightarrow y=\sqrt[3]{4^{7-x}}](https://tex.z-dn.net/?f=3.%5C%5Cy%3D-%5Clog_4x%5E3%2B7%5C%5C%5C%5C%5Ctext%7BExchange%20x%20and%20y.%20Solve%20for%20y%3A%7D%5C%5C%5C%5C-%5Clog_4y%5E3%2B7%3Dx%5Cqquad%5Ctext%7Bsubtract%207%20from%20both%20sides%7D%5C%5C%5C%5C-%5Clog_4%20y%5E3%3Dx-7%5Cqquad%5Ctext%7Bchange%20the%20signs%7D%5C%5C%5C%5C%5Clog_4y%5E3%3D7-x%5CRightarrow4%5E%7B%5Clog_4y%5E3%7D%3D4%5E%7B7-x%7D%5C%5C%5C%5Cy%5E3%3D4%5E%7B7-x%7D%5CRightarrow%20y%3D%5Csqrt%5B3%5D%7B4%5E%7B7-x%7D%7D)
(10x4x5)+(5x5x5) because you could make 2 shapes out of it. one being a cude that is 5x5x5 and the other being a rectangle that is 4x10x5 thenyou would just add them together. :) hope that helps
Answer:
Quadrants I and IV
Step-by-step explanation:
When we graph y^2 = x - 4, we see that it is on the x axis, and goes up into Quadrant I, and down into Quadrant IV
