No it’s D because it is rise over run and you rise 3 and run 1. 3/1 is simplified to 3.
Answer:
Please check the explanation
Step-by-step explanation:
Given the function
Given that the output = -3
i.e. y = -3
now substituting the value y=-3 and solve for x to determine the input 'x'
switch sides
Add 1 to both sides
![\mathrm{For\:}g^3\left(x\right)=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}](https://tex.z-dn.net/?f=%5Cmathrm%7BFor%5C%3A%7Dg%5E3%5Cleft%28x%5Cright%29%3Df%5Cleft%28a%5Cright%29%5Cmathrm%7B%5C%3Athe%5C%3Asolutions%5C%3Aare%5C%3A%7Dg%5Cleft%28x%5Cright%29%3D%5Csqrt%5B3%5D%7Bf%5Cleft%28a%5Cright%29%7D%2C%5C%3A%5Csqrt%5B3%5D%7Bf%5Cleft%28a%5Cright%29%7D%5Cfrac%7B-1-%5Csqrt%7B3%7Di%7D%7B2%7D%2C%5C%3A%5Csqrt%5B3%5D%7Bf%5Cleft%28a%5Cright%29%7D%5Cfrac%7B-1%2B%5Csqrt%7B3%7Di%7D%7B2%7D)
Thus, the input values are:
![x=-\sqrt[3]{2}+5,\:x=\frac{\sqrt[3]{2}\left(1+5\cdot \:2^{\frac{2}{3}}\right)}{2}-i\frac{\sqrt[3]{2}\sqrt{3}}{2},\:x=\frac{\sqrt[3]{2}\left(1+5\cdot \:2^{\frac{2}{3}}\right)}{2}+i\frac{\sqrt[3]{2}\sqrt{3}}{2}](https://tex.z-dn.net/?f=x%3D-%5Csqrt%5B3%5D%7B2%7D%2B5%2C%5C%3Ax%3D%5Cfrac%7B%5Csqrt%5B3%5D%7B2%7D%5Cleft%281%2B5%5Ccdot%20%5C%3A2%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%5Cright%29%7D%7B2%7D-i%5Cfrac%7B%5Csqrt%5B3%5D%7B2%7D%5Csqrt%7B3%7D%7D%7B2%7D%2C%5C%3Ax%3D%5Cfrac%7B%5Csqrt%5B3%5D%7B2%7D%5Cleft%281%2B5%5Ccdot%20%5C%3A2%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%5Cright%29%7D%7B2%7D%2Bi%5Cfrac%7B%5Csqrt%5B3%5D%7B2%7D%5Csqrt%7B3%7D%7D%7B2%7D)
And the real input is:
![x=-\sqrt[3]{2}+5](https://tex.z-dn.net/?f=x%3D-%5Csqrt%5B3%5D%7B2%7D%2B5)
Answer:
2 or 8is the answer..............
Answer:
I hope that you are doing well, during this weird time; and that this message finds you in a good place.
To get you started on this problem, you will want to think about how to go about graphing this linear function. You want to keep in mind that you want to plot your dependent variable on the y-axis, as it changes with respect to your independent variable (plotted on the x-axis). In this problem, the cost of the water bill depends on the amount of water used, in HCF. So the graph should be plotted with cost on the y-axis, and HCF on the x-axis.
It follows that you are given the coordinates of (16, $48.37) and (22, y2); where y2 represents the cost when using 22 HCF. You are also given the slope of your line (m, in y = mx + b format). You may recall that you can set-up a slope as the change in y over the change in x, or the "rise over the run" [m = (y2-y1) / (x2-x1)]. We can set this problem up, using this equation to solve for y2 (the cost at 22 HCF) algebraically.
So we would set up the problem as 1.65 = (y2-48.37)/(22-16). We then solve by combining like-terms and using inverse operations as follows:
1.65 = (y2 - 48.37) / 6
1.65(6) = y2 - 48.37
9.9 = y2 - 48.37
y2 = $58.27
Step-by-step explanation:
