Answer:
x value of vertical asymptote and y value of horizontal asymptote
Step-by-step explanation:
The graph of 1/x approaches infinity as x approaches 0 (the vertical asymptote)
As x gets either bigger or smaller, 1/x approaches the x-axis (from above on the positive side, from below on the negative side) (the horizontal asymptote)
Consider 1/(x-5) + 2, at what value of x does the graph 'go nuts' ?
When the bottom of the fraction becomes 0, x - 5 becomes 0 when x = 5, so the vertical asymptote of g(x) is at x=5
What value of y does f(x) approach as x gets more positive or more negative - as x gets bigger (as an example), y approaches 0
What y value does g(x) approach as x gets bigger? Well, as x gets big, 1/(x-5) gets small, approaching 0. The smallest 0 + 2 can get is 2, so y=2 is the horizontal asymptote
Answer:
true because I know hdhdnndnd
Step-by-step explanation:
ehbendndbdbdbbch
Answer: <em>81</em>
Step-by-step explanation:
<em>3^4</em>
<em>3x3x3x3</em>
<em>3x3=9</em>
<em>9x3=27</em>
<em>27x3=</em><em>81</em>
You can count from 34-28 and the numbers you counted (the numbers in between 34-28) is your answer.
9514 1404 393
Answer:
m = (5x -6)/(43y -4)
Step-by-step explanation:
Subtract 4m+6 to separate m-terms from other terms.
5x +4m -(4m +6) = 43my +6 -(4m +6)
5x -6 = 43my -4m
5x -6 = m(43y -4) . . . . factor out m
(5x -6)/(43y -4) = m . . . . divide by the coefficient of m
