9 days ago

Describe how you regroup when you find the sum of 64 + 43

1 answer:

4 0

When regrouping, you're essentially simplifying the problem into smaller parts. For instance, you would calculate 60 + 40=? and then solve 4 + 3=? The final answer would be 107.

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The mass of a colony of bacteria, in grams, is modeled by the function P given by P(t)=2+5tan^−1.(t/2), where t is measured in d

AnnZ [9071]

Answer:

1.21 g/day

Step-by-step explanation:

We start with the fact that

The mass of the bacterial colony (in grams) is described by

$P(t)=2+5tan^{-1}(\frac{t}{2})$

Where t=Time(in days)

Next, we differentiate with respect to t

$P'(t)=5(\frac{1}{1+\frac{t^2}{4}}\times \frac{1}{2})$

Using the formula $\frac{d(tan^{-1}(x)}{dx}=\frac{1}{1+x^2}$

$P'(t)=\frac{5}{2}(\frac{4}{4+t^2})$

$P'(t)=\frac{10}{4+t^2}$

We know that P(t)=6

Now, substitute this value

$6=2+5tan^{-1}(\frac{t}{2})$

$5tan^{-1}(\frac{t}{2})=6-2=4$

$tan^}{-1}(\frac{t}{2})=\frac{4}{5}$

$\frac{t}{2}=tan(\frac{4}{5})$

$t=2tan(\frac{4}{5})$

Insert the given value of t

$P'(2tan\frac{4}{5})=\frac{10}{4+4tan^2(\frac{4}{5})}$

$P'(2tan\frac{4}{5})=\frac{10}{4}\times \frac{1}{1+tan^2(\frac{4}{5})}$

We understand that $1+tan^2\theta=sec^2\theta$

Applying the formula

$P'(2tan(\frac{4}{5})=\frac{5}{2}\times \frac{1}{sec^2(\frac{4}{5})}$

$P'(2tan\frac{4}{5})=\frac{5}{2}\times cos^2(\frac{4}{5})$

By employing $cos^2x=\frac{1}{sec^2x}$

$P'(2tan\frac{4}{5})=\frac{5}{2}\times (0.696)^2=1.21$ g/day

Consequently, the instantaneous rate at which the mass of the colony changes is=1.21g/day

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Response:

Detailed breakdown:

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Jon wants to put a circular decorative glass in a table. He cuts a hole in the table that is 20in in diameter. He uses a thin me

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