Answer:
<u>x-intercept</u>
The point at which the curve <u>crosses the x-axis</u>, so when y = 0.
From inspection of the graph, the curve appears to cross the x-axis when x = -4, so the x-intercept is (-4, 0)
<u>y-intercept</u>
The point at which the curve <u>crosses the y-axis</u>, so when x = 0.
From inspection of the graph, the curve appears to cross the y-axis when y = -1, so the y-intercept is (0, -1)
<u>Asymptote</u>
A line which the curve gets <u>infinitely close</u> to, but <u>never touches</u>.
From inspection of the graph, the curve appears to get infinitely close to but never touches the vertical line at x = -5, so the vertical asymptote is x = -5
(Please note: we cannot be sure that there is a horizontal asymptote at y = -2 without knowing the equation of the graph, or seeing a larger portion of the graph).
Answer:
64,890
Step-by-step explanation:
105×3 =315
103×2= 206
315×206= 64,890
Answer:
34.3948
Step-by-step explanation:
Answer:
Neither
Step-by-step explanation:
Answer:
x = 2
Step-by-step explanation:
Taking antilogs, you have ...
2³ × 8 = (4x)²
64 = 16x²
x = √(64/16) = √4
x = 2 . . . . . . . . (the negative square root is not a solution)
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You can also work more directly with the logs, if you like.
3·ln(2) +ln(2³) = 2ln(2²x) . . . . . . . . . . . write 4 and 8 as powers of 2
3·ln(2) +3·ln(2) = 2(2·ln(2) +ln(x)) . . . . use rules of logs to move exponents
6·ln(2) = 4·ln(2) +2·ln(x) . . . . . . . . . . . . simplify
2·ln(2) = 2·ln(x) . . . . . . . . . . . subtract 4ln(2)
ln(2) = ln(x) . . . . . . . . . . . . . . divide by 2
2 = x . . . . . . . . . . . . . . . . . . . take the antilogs
