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Sergio [31]
3 years ago
15

A test tube initially holds 25 bacteria. The population of bacteria in the test tube doubles each hour. Without making a graph,

determine how many hours it will take for the population of bacteria in the test tube to reach 800. Explain how you solved this problem.
Thank You >=
Mathematics
2 answers:
9966 [12]3 years ago
5 0
5 hours because 800/25=32 32= 2^5 25 X 1 = 25 25 X 2 = 50 AFTER FIRST HOUR 25 X 2^5 = 800 AFTER FIFTH HOUR
padilas [110]3 years ago
4 0
25*2^X=800
2^X=32
If you put a radical sign over each side, the index for 2^X would have to be 5 for X to disappear and the equation to be true. In other words, ({5}(√2^X)=({5}(√32)), which means that It will take 5 hours

Answer: 5 hours.
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A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 43 ft/s. Its height
Lina20 [59]

Answer:

Instantaneous Velocity at t = 1 is 20 feet per second

Step-by-step explanation:

We are given he following information in the question:

y(t)=43t-23t^{2}

B) Instantaneous Velocity at t = 1

y(1) = 43(1)-23(1)^2 = 20

A) Formula:

Average velocity =

\displaystyle\frac{\text{Displacement}}{\text{Time}}

1) 0.01

y(1 + 0.01)=y(1.01) = 43(1.01)-23(1.01)^{2} = 19.9677\\\\\text{Average Velocity} = \dfrac{y(1.01)-y(1)}{1.01-1} =\dfrac{19.9677-20}{1.01-1}= \dfrac{-0.0323}{0.01} = -3.230000 \text{feet per second}

2) 0.005 s

y(1 + 0.005)=y(1.005) = 43(1.005)-23(1.005)^{2} = 19.984425\\\\\text{Average Velocity} = \dfrac{y(1.005)-y(1)}{1.005-1} =\dfrac{19.984425-20}{1.005-1} = -3.1150000 \text{feet per second}

3) 0.002 s

y(1 + 0.002)=y(1.002) = 43(1.002)-23(1.002)^{2} = 19.993908\\\\\text{Average Velocity} = \dfrac{y(1.002)-y(1)}{1.002-1} =\dfrac{19.993908-20}{1.002-1} = -3.0460000 \text{feet per second}

4) 0.001 s

y(1 + 0.001)=y(1.001) = 43(1.001)-23(1.001)^{2} = 19.996977\\\\\text{Average Velocity} = \dfrac{y(1.001)-y(1)}{1.001-1} =\frac{19.996977-20}{1.001-1} = -3.0230000 \text{feet per second}

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