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3 years ago

How much of the original air is present after 240 breaths

Mathematics

1 answer:

zhenek [66]3 years ago

8 0

I believe 2mL is still present

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Middle school math.

Nonamiya [84]

The last one the reason is 2 is how much weeks so it’d be 2x - 126 cause that’s how much she wants to lose

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3 years ago

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6. A line that passes through the point (–1, 3) has a slope of 2. Find another point on the line. A. (0, 5) B. (–3, 2) C. (1, 4)

Bezzdna [24]

The answer is A because if we stare with (-1,3) and go up by 1x, since the slope is 2, then 3 + 2 is 5 and we have (0,5). Hope this helps!

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3 years ago

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Graph the piecewise function given below. i need help please

yawa3891 [41]

Answer:

this may help you.look it once

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3 years ago

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Unite rate. 3 cans of soup for $0.99

soldi70 [24.7K]

Answer:

$0.33

Step-by-step explanation:

Unit rate=1

if 3 equal to $0.99, then

1 will equal $0.33.

0.99/3=$0.33

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3 years ago

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Leno4ka [110]

Answer:

$\frac{s^2-25}{(s^2+25)^2}$

Step-by-step explanation:

Let's use the definition of the Laplace transform and the identity given: $\mathcal{L}[t \cos 5t]=(-1)F'(s)$ with $F(s)=\mathcal{L}[\cos 5t]$ .

Now, $F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt$ . Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that $F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt$ .

Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that

$F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s)$ .

Solving for F(s) on the last equation, $F(s)=\frac{s}{s^2+25}$ , then the Laplace transform we were searching is $-F'(s)=\frac{s^2-25}{(s^2+25)^2}$

3 0

3 years ago