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3 years ago

Please Help! (Image below)

Mathematics

1 answer:

Ivan3 years ago

7 0

Answer:

A is the correct answer

Step-by-step explanation:

I just did this problem

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FOR BRAINLIST ANSWER AND 10 POINTS ANSWER ALL PLZ !!!!!

igor_vitrenko [27]

<h2>Greetings!</h2>

<h3>If you can see, every 4 minutes the amount of water is increasing at a constant rate. </h3>

197 - 153 = 44

285 - 241 = 44

<h3>This means that at 8 minutes, because there is 153 gallons already in, we can divide 8 by 4 to see how many groups of 4 minutes there were. </h3>

8 ÷ 4 = 2

So then we can subtract two lots of 44 from 153:

153 - (44 x 2) = 65

<h3>So the starting amount was 65.</h3>

<h3>The answer to B is the second option, because it is true that the amount of water is increasing. This is by 44 gallons per 4 minutes.</h3>

The amount of gallons per minute can be found out by dividing the amount by 4:

44 ÷ 4 = 11

<h3>So it is 11 gallons per minute.</h3>
<h2>Hope this helps!</h2>

3 0

3 years ago

Someone please help!

Mila [183]

Answer:

RU is 8

Step-by-step explanation:

4 0

3 years ago

Help me please I'm stuck on this question!!!!

shutvik [7]

Answer:

1: 2.8

Step-by-step explanation:

15:42

Divide each side by 15

15/15 : 42/15

1: 2.8

6 0

2 years ago

A giant hummingbird can flap its wings at a rate of 13 flaps per second, how many times

antoniya [11.8K]

Answer:

780 flaps per second

Step-by-step explanation:

just multiply 13×60

8 0

2 years ago

In a previous exercise we formulated a model for learning in the form of the differential equation dp dt = k(m − p) where p(t) m

DaniilM [7]

I assume you mean

$\dfrac{dP}{dt} = k(M-P)$

ANSWER
An expression for P(t) is

$P = M - Me^{-kt}$

EXPLANATION
This is a separable differential equation. Treat M and k as constants. Then we can divide both sides by M - P to get the P term with the differential dP and multiply both sides by dt to separate dt from the P terms

$\begin{aligned} \dfrac{dP}{dt} &= k(M-P) \\ \dfrac{dP}{M-P} &= k\, dt
\end{aligned}$

Integrate both sides of the equation.

$\begin{aligned}
\int \dfrac{dP}{M-P} &= \int k\, dt \\
-\ln|M-P| &= kt + C \\
\ln|M-P| &= -kt - C\end{aligned}$

Note that for the left-hand side, u-substitution gives us

$u = M - P \implies du = -1dP \implies dP = -du$

hence why $\int \frac{dP}{M-P} \ne \ln|M - P|$

Now we use the definition of the logarithm to convert into exponential form.

The definition is

$\ln(a) = b \iff \log_e(a) = b \iff e^b = a$

so applying it here, we get

$\begin{aligned} \ln|M-P| &= -kt - C \\ |M - P| &= e^{-kt - C} \\ 
M - P &= \pm e^{-kt - C} 
 \end{aligned}$

Exponent properties can be used to address the constant C. We use $x^{a} \cdot x^{b} = x^{a+b}$ here:

$\begin{aligned}
 M - P &= \pm e^{-kt - C} \\
M - P &= \pm e^{- C - kt} \\ 
M - P &= \pm e^{- C + (- kt)} \\ 
M - P &= \pm e^{- C} \cdot e^{- kt} \\ 
M - P &= Ke^{- kt} && (\text{\footnotesize Let $K = \pm e^{-kt}$ }) \\ 
M &= Ke^{- kt} + P\\
P &= M - Ke^{- kt}
\end{aligned}$

If we assume that P(0) = 0, then set t = 0 and P = 0

$\begin{aligned} 
0 &= M - Ke^{- k\cdot 0} \\
0 &= M - K \cdot 1 \\
M &= K
 \end{aligned}
$

Substituting into our original equation, we get our final answer of

$P = M - Me^{-kt}$

6 0

3 years ago

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