The first step before graphing these is to ensure they are both in y=mx+b (slope-intercept) format.
Only the first one (y-2x=0) is not.
To convert this to slope-intercept form, just add 2x to both sides so y is isolated!
This leaves us with y=2x+0, which is in slope-intercept form.
When graphing these, you must identify the slope and the y-intercept.
In y=mx+b form, the m is slope and b is y-intercept.
In y=2x+0, the slope is 2 and the y-intercept is 0.
This means there is a slope of positive 2 and the y-intercept is (0, 0).
The y-intercept is the y-value where ever x=0.
In y=6x-10, the slope is 6 and the y-intercept is -10.
This means there is a slope of positive 6 and the y-intercept is (0, -10).
Hope this helps! :)
Answer:
C. D: -2≤x≤3 ; R: -1≤y≤4
Step-by-step explanation:
The domain is all the values on the x-axis (horizontally) and the range is all the values on the y-axis (vertically).
The x-axis values (domain) go from -2 to +3, which is -2≤x≤3.
The y-axis values (range) go from -1 to +4, which is -1≤y≤4.
Answer:
Hello,
Step-by-step explanation:
![A)\\g(x)=\dfrac{x-5}{-3} =\dfrac{-x}{3} +\dfrac{5}{3} \\\\(gog)(x)=g(g(x))=g(\dfrac{-x}{3} +\dfrac{5}{3})\\\\=\dfrac{\dfrac{-x}{3} +\dfrac{5}{3} }{3}+\dfrac{5}{3} \\\\\\=\dfrac{-x}{9} +\dfrac{5}{9} +\dfrac{5}{3}\\\\=-\dfrac{x}{9}+\dfrac{20}{9} \\\\\\(gog)(2)=-\dfrac{2}{9}+\dfrac{20}{9} =\dfrac{18}{9}=2 \\\\](https://tex.z-dn.net/?f=A%29%5C%5Cg%28x%29%3D%5Cdfrac%7Bx-5%7D%7B-3%7D%20%3D%5Cdfrac%7B-x%7D%7B3%7D%20%2B%5Cdfrac%7B5%7D%7B3%7D%20%5C%5C%5C%5C%28gog%29%28x%29%3Dg%28g%28x%29%29%3Dg%28%5Cdfrac%7B-x%7D%7B3%7D%20%2B%5Cdfrac%7B5%7D%7B3%7D%29%5C%5C%5C%5C%3D%5Cdfrac%7B%5Cdfrac%7B-x%7D%7B3%7D%20%2B%5Cdfrac%7B5%7D%7B3%7D%20%7D%7B3%7D%2B%5Cdfrac%7B5%7D%7B3%7D%20%20%5C%5C%5C%5C%5C%5C%3D%5Cdfrac%7B-x%7D%7B9%7D%20%2B%5Cdfrac%7B5%7D%7B9%7D%20%2B%5Cdfrac%7B5%7D%7B3%7D%5C%5C%5C%5C%3D-%5Cdfrac%7Bx%7D%7B9%7D%2B%5Cdfrac%7B20%7D%7B9%7D%20%5C%5C%5C%5C%5C%5C%28gog%29%282%29%3D-%5Cdfrac%7B2%7D%7B9%7D%2B%5Cdfrac%7B20%7D%7B9%7D%20%3D%5Cdfrac%7B18%7D%7B9%7D%3D2%20%5C%5C%5C%5C)
![B)\\f(x)=y=-3x-5\\exchanging\ y\ and\ x\\x=-3y-5\\3y=-x-5\\\\y=\dfrac{-x}{3} -\dfrac{5}{3} \\\\f^{-1}(x)=\dfrac{-x}{3} -\dfrac{5}{3} \\\\](https://tex.z-dn.net/?f=B%29%5C%5Cf%28x%29%3Dy%3D-3x-5%5C%5Cexchanging%5C%20y%5C%20and%5C%20x%5C%5Cx%3D-3y-5%5C%5C3y%3D-x-5%5C%5C%5C%5Cy%3D%5Cdfrac%7B-x%7D%7B3%7D%20-%5Cdfrac%7B5%7D%7B3%7D%20%5C%5C%5C%5Cf%5E%7B-1%7D%28x%29%3D%5Cdfrac%7B-x%7D%7B3%7D%20-%5Cdfrac%7B5%7D%7B3%7D%20%5C%5C%5C%5C)
![C)\\\\(fog)(x)\ must\ be\ equal\ to\ x\\\\\\(fog)(x)=g(f(x))=g(-3x-5)\\\\=\dfrac{-(-3x-5)}{3} +\dfrac{5}{3} \\\\=x+\dfrac{5}{3} +\dfrac{5}{3} \\\\\\=x+\dfrac{10}{3}\ and\ not\ x\ !!!\\](https://tex.z-dn.net/?f=C%29%5C%5C%5C%5C%28fog%29%28x%29%5C%20must%5C%20be%5C%20equal%5C%20to%5C%20x%5C%5C%5C%5C%5C%5C%28fog%29%28x%29%3Dg%28f%28x%29%29%3Dg%28-3x-5%29%5C%5C%5C%5C%3D%5Cdfrac%7B-%28-3x-5%29%7D%7B3%7D%20%2B%5Cdfrac%7B5%7D%7B3%7D%20%5C%5C%5C%5C%3Dx%2B%5Cdfrac%7B5%7D%7B3%7D%20%2B%5Cdfrac%7B5%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%3Dx%2B%5Cdfrac%7B10%7D%7B3%7D%5C%20and%5C%20not%5C%20x%5C%20%21%21%21%5C%5C)
f(x) and g(x) are not inverse functions.
Answer:
They did not run at the same speed. Calculation of their speed shows that Jada was running at a higher speed.
Step-by-step explanation:
Let's first calculate the speed of each person using the units provided so that we do not complicate the question.
speed = distance covered/time taken
Andre's speed = 2 km/15 min = 0.133 km/min
Jada's speed = 3km/20 min = 0.15 km/min
Clearly we can see that Jada was running at a higher speed than Andre. The question says they were running at constant speed not equal speed. That just means each person maintained their speed, they did not get faster or slower.