Apopulation of 1,000 bacteria is treated with a medicine and begins to die 5% each hour. how can the bacteria population be determined after a number of hours, ℎ?

a. (ℎ)=0.95(ℎ+1,000)

b. (ℎ)=1,000·ℎ^0.95

c. (ℎ)=0.95ℎ+1,000

d. (ℎ)=1,000·0.95^ℎ

Step-by-step explanation:

is an exponential equation where A is the beginning amount and B is the rate that the population grows or dies.

So we start with 1000 bacteria, they are giving us A.

The bacteria population is decreasing because they are dying 5% each hour.

So that is after the first hour we have 1000-.05(1000) or 1000(1)-1000(.05)=1000(1-.05)=1000(.95).

We will keep multiplying by .95 per hour. 0.95 is the repeated factor.

That is the function is .

If you let h=0 which means 0 hours has happened, you will see the bacteria is 1000 as desired. 1000(.95)^0=1000(1)=1000.

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step-by-step explanation:

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x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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