Jerry measured and timed the descent of a diving bell spider in a large fish tank. The spider reached a position 412 412 inches below the surface of the water after descending at a constant rate for 112 112 seconds. Jerry recorded this as −412 −412 inches. What is the position of the spider relative to the surface after 1 second? Enter your answer in the box.

Respuesta :

Step 1

Convert mixed numbers to an improper fractions

[tex]4\frac{1}{2} =\frac{4*2+1}{2}= \frac{9}{2}[/tex]

[tex]1\frac{1}{2} =\frac{1*2+1}{2}= \frac{3}{2}[/tex]

Step 2

Find the unit rate of the descending

[tex]unit\ rate= \frac{(9/2)}{(3/2)}=3 \frac{inches}{sec}[/tex]

therefore

the answer is

the position of the spider relative to the surface after [tex]1[/tex] second is [tex]-3\ inches[/tex]


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