Answer:
Step-by-step explanation:
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Answer: Table H would be the correct answer;
The rule of a function is that for each x-value given there can't be more than 1 y-value
<u>In Table F:</u>
x = -13, then y = -2
x = -13, then y = 0
x = -13, then y = 5
x = -13, then y = 7
For the x-value -13, there are 4 different y-values, so <em>it's not a function.</em>
<u>In Table G:</u>
x = -6, then y = 3
x = -1, then y = -1
x = -1, then y = 5
x = 10, then y = -9
For the x-value -1, there are 2 different y-values, hence <em>this isn't a function.</em>
<u>In Table H:</u>
x = 1, then y = 4
x = 3, then y = 4
x = 7, then y = 4
x = 12, then y = 4
For each x-value, there is only 1 y-value, so <em>this is a function.</em>
<u>In table J:</u>
x = -9, then y = -7
x = -2, then y = -5
x = 0, then y = 0
x = 0, then y = 6
For the x-value 0, there are 2 different y-value therefore <em>this isn't a function</em>
Hope this helps!
Check the picture below.
well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.
bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.
![\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}](https://tex.z-dn.net/?f=%5Cbf%20A%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B4%7D%29%5Cqquad%20B%28%5Cstackrel%7Bx_2%7D%7B5%7D~%2C~%5Cstackrel%7By_2%7D%7B1%7D%29%20~%5Chfill%20%5Cstackrel%7Bslope%7D%7Bm%7D%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%20%7B%5Cstackrel%7By_2%7D%7B1%7D-%5Cstackrel%7By1%7D%7B4%7D%7D%7D%7B%5Cunderset%7Brun%7D%20%7B%5Cunderset%7Bx_2%7D%7B5%7D-%5Cunderset%7Bx_1%7D%7B1%7D%7D%7D%5Cimplies%20%5Ccfrac%7B-3%7D%7B4%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bslope%20of%20AB%7D%7D%7B-%5Ccfrac%7B3%7D%7B4%7D%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7B%5Cunderline%7Bnegative%20reciprocal%7D%20and%20slope%20of%20the%20diameter%7D%7D%7B%5Ccfrac%7B4%7D%7B3%7D%7D)
so, it passes through the midpoint of AB,
so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)
The graph g(x) is the graph of f(x) translated (5,2,3) units (down,up,left,right) , and g(x) =(f(x-3),f(x)-5,f(x)+3,f(x-2),f(x)+
marusya05 [52]
Answer:
The graph g(x) is the graph of f(x) translated <u>2</u> units <u>right</u>, and g(x) = <u>f(x-2)</u>
Step-by-step explanation:
g(x) passes through points (0, -5) and (1, -2), then the slope of g(x) is the same as the slope of f(x), which is 3.
f(x) passes through (0, 1) and g(x) passes through (2, 1). Therefore, the graph g(x) is the graph of f(x) translated 2 units right.
f(x - c) translates f(x) c units to the right, therefore g(x) = f(x-2)
In order to check this result, we make:
f(x) = 3x + 1
f(x-2) = 3(x-2) + 1
f(x-2) = 3x - 6 + 1
f(x-2) = 3x - 5 = g(x)
