now that we know what are the x-values, what are the y-values? well, we can just use the 2nd equation, since we know that y = x - 28, then
![\bf y = x - 28\implies \stackrel{\textit{when x = 16}}{y = 16 - 28}\implies y = -12 \\\\\\ y = x - 28\implies \stackrel{\textit{when x = 12}}{y = 12 - 28}\implies y = -16 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (\stackrel{x}{16}~~,~~\stackrel{y}{-12})\qquad,\qquad (\stackrel{x}{12}~~,~~\stackrel{y}{-16})~\hfill](https://tex.z-dn.net/?f=%5Cbf%20y%20%3D%20x%20-%2028%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bwhen%20x%20%3D%2016%7D%7D%7By%20%3D%2016%20-%2028%7D%5Cimplies%20y%20%3D%20-12%20%5C%5C%5C%5C%5C%5C%20y%20%3D%20x%20-%2028%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bwhen%20x%20%3D%2012%7D%7D%7By%20%3D%2012%20-%2028%7D%5Cimplies%20y%20%3D%20-16%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20~%5Chfill%20%28%5Cstackrel%7Bx%7D%7B16%7D~~%2C~~%5Cstackrel%7By%7D%7B-12%7D%29%5Cqquad%2C%5Cqquad%20%28%5Cstackrel%7Bx%7D%7B12%7D~~%2C~~%5Cstackrel%7By%7D%7B-16%7D%29~%5Chfill)
F(x)=1/(x-3)^3
The denominator can't be equal to zero (we can't divide by zero), then:
(x-3)^3 different 0
cubic root both sides:
cubic root [ (x-3)^3 ] different cubic root (0)
x-3 different 0
Adding 3 both sides:
x-3+3 different 0+3
x different 3
Then the function f is not defined at x=3
Answer: Option <span>b) f is not defined at x= 3</span>
The third one
x,y —>-x,-y
Because 2,-1–>-2,1
9514 1404 393
Explanation:
A)
Let t and f represent the numbers of two-point and four-point questions, respectively.
__
B) The number of points for each type of question is the product of the point-value per question and the number of questions of that type.
2t + 4f = 100 . . . . . the test is worth 100 points
t + f = 40 . . . . . . . . there are 40 questions on the test
_____
<em>Additional comment</em>
The test will have 30 two-point questions and 10 four-point questions.
