24x^2 +25x - 47 53
----------------------- = -8x -3 - ---------------
ax-2 ax-2
add 53/ax-2 to each side
24x^2 +25x - 47+53
----------------------- = -8x -3
ax-2
24x^2 +25x +6
----------------------- = -8x -3
ax-2
multiply each side by ax-2
24x^2 +25x +6 = (ax-2) (-8x-3)
multiply out the right hand side
24x^2 +25x +6 = -8ax^2 +16x-3ax +6
24 = -8a 25 = 16 -3a
a = -3 9 = -3a
a = -3
Choice B
-8-4y=-5x
-4y=-5x+8
y=5/4x+2
slope= 5/4
y-intercept=(0,2)
Answer:
There are only 6 numbers that fit that description. (4,12,24,32,44,52)
Answer:
Her initial position was:
-29ft
Where we use the minus sign because this is below the ocean's surface.
Now we also know that she keeps descending at a rate of -29ft per minute, then if she keeps descending for t minutes, her position will be:
P(x) = -29ft - 29ft/min*t
Now, we also know that she does not want to descend more than 81ft below the ocean's surface, then we have the inequality:
P(x) ≥ -81ft
-29ft - 29ft/min*t ≥ -81ft
Now let's isolate t in one side:
- 29ft/min*t ≥ -81ft + 29ft = -52 ft
- 29ft/min*t ≥-52 ft
t ≤ -52ft/(- 29ft/min) = 1.79 min
Then the maximum amount of time that she can keep descending is 1.79 minutes.
