A scuba diver dove from the surface of the ocean to an elevation of −49 9/10 feet at a rate of −9.5 feet per minute. After spending 10 minutes at that elevation, the diver ascended to an elevation of −28 9/10 feet. The total time for the dive so far was 17 1/8 minutes. What was the rate of change in the diver's elevation during the ascent? Round your answer to the nearest hundredth.

Rate of change  =  4,06 f/min

Step-by-step explanation:

The ascendent length  28 9/10   =  289/ 10

The time for that ascendent maneuver is total time 17 1/8 minutes  - spended time in the elevation 10 minutes, then

17 1/8  =  137/8

137/8  - 10   =  ( 137 - 80 ) /8   =  57/8 minutes

The rate of change in the diver´s elevation is

289 / 10 / 57/8

Rate of change  =  289*8/ 57*10

Rate of change  =  2312 / 570 ft/min

Rate of change  =  4,06 f/min

hello from mrbilldoesmath!

answer:

(7 x + 6 y) (49 x^2 - 42 x y + 36 y^2)

discussion:

the trick is to recognize that 343 = 7^3 and 216 = 6^3 and, in general,

a^3 + b^3 =   (a + b) (a^2 - a b + b^2).

in our case

343x^3+216y^3   =   (7x)^3 + (6y)^3

so take a = 7x and b = 6y in the above equation to get the answer.

you,

mrb

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