Urgent
fluorine-20 has a half-life of 11 seconds. identify the amount of fluorine-20 left from a 25 gram sample after 55 seconds.

a 0.78125 g
b 78125 mg
c 0.6352 g
d 7.8124 g

A. 0.78 g  

Step-by-step explanation:

The half-life of F-20 (11 s) is the time it takes for half of it to decay.  

After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.  

We can construct a table as follows:  

  No. of                  Fraction           Mass

half-lives     t/s   Remaining   Remaining/g

      0             0          1                      25

      1             11           ½                     12.5

      2           22          ¼                      6.2

      3           33          ⅛                       3.1

      4           44          ¹/₁₆                      1.6

      5           55          ¹/₃₂                   0.78

      6           66          ¹/₆₄                    0.39

We see that 0.78 g of F-20 remain after five half-lives (55 s).

y = 5x + 17

step-by-step explanation:

a perpendicular line will be the opposite reciprocal slope of the given equation.   the slope given is -1/5 so the opposite reciprocal is 5

find b:

7 = 5 (-2) + b

7 = -10 + b

17 = b

write equation

y = 5x + 17

b. would be the answer : )

step-by-step explanation:

step 1: add -16x to both sides.

16x−4y+−16x=64+−16x  

−4y=−16x+64

step 2: divide both sides by -4.

−4y /-4 = -16x + 64/-4

answer:   y=4x−16

hope i'm right