28.12.2021

Use a calculator to find the value of each expression rounded to two decimal places. Make sure you are in Radian Mode. cos-(0.75)

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09.07.2023, solved by verified expert
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If you meant to write Use a calculator to find the value of each expression, №18009785, 28.12.2021 19:21, then using a calculator in radian mode would show Use a calculator to find the value of each expression, №18009785, 28.12.2021 19:21

Rounding to the nearest hundredth gets us Use a calculator to find the value of each expression, №18009785, 28.12.2021 19:21

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Mathematics
Step-by-step answer
P Answered by Master

If you meant to write \cos^{-1}(0.75), then using a calculator in radian mode would show \cos^{-1}(0.75) \approx 0.722734

Rounding to the nearest hundredth gets us \cos^{-1}(0.75) \approx 0.72

Mathematics
Step-by-step answer
P Answered by PhD

It will take 26.34 minutes for the population to reach 5 times its initial value

Step-by-step explanation:

Exponential Growing

The population of bacteria grows at a rate expressed by the equation:

f(t)=256e^{0.06111t}

Where t is in minutes.

We need to know when the population will reach 5 times its initial value. The initial value can be determined by setting t=0:

f(0)=256e^{0.06111\cdot 0}=256\cdot 1=256

Now we find the time when the population is 5*256=1,280. The equation to solve is:

1,280=256e^{0.06111t}

Dividing by 256:

e^{0.06111t}=1,280/256=5

Taking natural logarithms:

ln\left(e^{0.06111t}\right)=ln5

Applying the logarithm properties:

0.06111t=ln5

Solving for t:

t=ln5/0.06111=26.34

It will take 26.34 minutes for the population to reach 5 times its initial value

Mathematics
Step-by-step answer
P Answered by PhD

It will take 26.34 minutes for the population to reach 5 times its initial value

Step-by-step explanation:

Exponential Growing

The population of bacteria grows at a rate expressed by the equation:

f(t)=256e^{0.06111t}

Where t is in minutes.

We need to know when the population will reach 5 times its initial value. The initial value can be determined by setting t=0:

f(0)=256e^{0.06111\cdot 0}=256\cdot 1=256

Now we find the time when the population is 5*256=1,280. The equation to solve is:

1,280=256e^{0.06111t}

Dividing by 256:

e^{0.06111t}=1,280/256=5

Taking natural logarithms:

ln\left(e^{0.06111t}\right)=ln5

Applying the logarithm properties:

0.06111t=ln5

Solving for t:

t=ln5/0.06111=26.34

It will take 26.34 minutes for the population to reach 5 times its initial value

Mathematics
Step-by-step answer
P Answered by PhD

The bacteria has an exponential growth rate and the population of the

bacteria increases rapidly with time.

(a) The relative rate of growth is \underline{\dfrac{ln(64)}{4}}.

(b) The initial size of the culture is 75 bacteria.

(c) The function that models the number of bacteria n(t) is; \underline{n(t) = 75 \cdot e^{1.04 \cdot t}}

(d) The number of bacteria after 4.5 hours is approximately 8,100 bacteria.

(e) The number of hours after which the bacteria will reach 75,000 is approximately 6.64 hours.

Reasons:

The count in the culture of bacteria after 2 hours = 600

The count after 6 hours = 38,400

(a) The relative rate of growth, k is given by the formula;

y = \mathbf{C \cdot e^{k \cdot t}}

Therefore, we get;

600 = C \cdot e^{k \times 2}

38,400= C \cdot e^{k \times 6}

Which gives;

\dfrac{38,400}{600} = \mathbf{\dfrac{C \cdot e^{k \times 6}}{C \cdot e^{k \times 2}}}

64= \dfrac{e^{k \times 6}}{e^{k \times 2}} = e^{k \times 6- k \times 2} = e^{4\cdot k}

\mathbf{e^{4\cdot k}} = 64

ln\left(e^{4\cdot k}) = ln(64)

4·k = ㏑(64)

The \ relative \ rate \ of \ growth, \, \underline{ k = \dfrac{ln(64)}{4}}

(b) The initial size of the culture, C, is given by the relation;

C =  \mathbf{\dfrac{y}{e^{k \cdot t}}}

Therefore, we get;

C = \dfrac{600}{e^{\dfrac{ln(64)}{4} \times 2}} = \mathbf{ 75}

The initial size of the culture, C = 75

(c) The function is y = C \cdot e^{k \cdot t}

Where:

y = n(t)

C = n₀

k = r

We get;

n(t) =  \mathbf{n_0 \cdot e^{r \cdot t}}

n₀ = C = 75

r =  k = \dfrac{ln(64)}{4} \approx 1.04

Which gives the function as follows; \underline{n(t) = 75 \cdot e^{1.04 \cdot t}}

(d) The number of bacteria after 4.5 hours is n(4.5) = 75 \cdot e^{1.04 \times 4.5} ≈ 8,100 bacteria

(e) At n(t) = 75,000, we have;

n(t) = 75,000 =  \mathbf{75 \cdot e^{1.04 \times t}}

t = \dfrac{ln\left(\dfrac{75,000}{75} \right)}{1.04} \approx  \mathbf{6.64}

The time at which the bacteria population will reach 75,000, t ≈ 6.64 hours.

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Mathematics
Step-by-step answer
P Answered by PhD

Cost of 7 gallons=$24.50

Cost of 1 gallon=24.50/7=3.5

Cost of 15 gallons=15*3.5=52.5

Cost of 15 gallons will be $52.5

Mathematics
Step-by-step answer
P Answered by PhD

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