Iodine-123 has a half-life of about 13 hours, emits gamma-radiation, and is readily absorbed by the thyroid.
because of these facts, it is regularly used in nuclear imaging.
a. write a formula that gives you the percent pp of iodine-123 left after t half-lives.
b. what is the decay rate per hour of iodine-123? approximate to the nearest millionth.
c. use your result to part (b). how many hours h would it take for you to have less than 1% of an initial dose of
iodine-123 in your system? round your answer to the nearest tenth of an hour.

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.

x = 17

step-by-step explanation:

1st step: isolate x.

24 - 3x = -27

-24         -24

-3x   =   -51

2nd step: divide.

-3x     =   -51

  3             3

  x       =     17

hope this

3. (5x + 1) - (-10x + 6)

simplify the answer choice. combine like terms. remember to distribute the negative to the terms inside the second parenthesis. also remember that two negatives = one positive, and one of each sign = negative.

- (-10x + 6) = + 10x - 6

simplify. combine like terms:

5x + 1 + 10x - 6

5x + 10x + 1 - 6

15x - 5

∴ it is your answer

~

given the equation is 8+9z=-6, the first step would be to subtract the 8 from both sides:

8+9z-8=-6-8

since 8 minus 8 = 0, this would cancel out the 8. and since -6 minus 8 equals -14, the equation would look like this

9z = -14