8. While solving the equation log,(8x) – log,(x2 - 1) = log, 3, Karla found possible solutions at x=3
1 answer:
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The answer has to be in y-intercept form. Which is y=mx+b and b can also be neg. y=mx-b. The m is your slope and b is your y-intercept. So A’s slope counting from 3 on the y axis to 1 on the x-axis which is down 2 over 1 so it will be -2/1. And your y intercept is 3. The equation is y=-2x+3. For B its the same thing but no graph. Remember that slope is RISE over RUN. So since the number is on the bottom of the line the rise is 4 and the run is 1. So in this case 10 is your y-intercept. Which means you equation is y=4x+10.
Answer:
2) D: x = [0, 24]
3) R: y = [0, 384]
4) see graph
<u>Step-by-step explanation:</u>
Eric's regular wage is $12 per hour for all hours less than 9 hours.
The minimum number of hours Eric can work each day is 0.
f(x) = 12x for 0 ≤ x < 9
Eric's overtime wage is $18 per hour for 9 hours and greater.
The maximum number of hours Eric can work each day is 24 (because there are only 24 hours in a day).
f(x) = 18(x - 8) + 12(8)
= 18x - 144 + 96
= 18x - 48 for 9 ≤ x ≤ 24
The daily wage where x represents the number of hours worked can be displayed in function format as follows:
2) Domain represents the x-values (number of hours Eric can work).
The minimum hours he can work in one day is 0 and the maximum he can work in one day is 24.
D: 0 ≤ x ≤ 24 → D: x = [0, 24]
3) Range represents the y-values (wage Eric will earn).
Eric's wage depends on the number of hours he works. Use the Domain (given above) to find the wage.
The minimum hours he can work in one day is 0.
f(x) = 12x
f(0) = 12(0)
= 0
The maximum hours he can work in one day is 24 <em>(although unlikely, it is theoretically possible).</em>
f(x) = 18x - 48
f(24) = 18(24) - 48
= 432 - 48
= 384
D: 0 ≤ y ≤ 384 → D: x = [0, 384]
4) see graph.
Notice that there is an open dot at x = 9 for f(x) = 12x
and a closed dot at x = 9 for f(x) = 18x - 48
