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erik [133]
2 years ago
5

Help help help blah blah blah

Mathematics
2 answers:
Zanzabum2 years ago
6 0

Answer:

It is c blah blah blah

Step-by-step explanation:

pls give brainliest blah blah blah

uysha [10]2 years ago
6 0

Answer:

?

Step-by-step explanation:

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Jerome, Kevin, and Seth shared a submarine sandwich.
Elanso [62]
Seth at 1 - 1/2 - 1/3 = 1/6 of the sandwich.

Therefore the ratio is:

1/2 : 1/3 : 1/6

Multiply everything by 6:

3: 2: 1 D
6 0
3 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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I can’t figure this out someone please
marishachu [46]

Answer:

a^6/(b^12c^3d)

Step-by-step explanation:

7 0
3 years ago
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José writes this problem: 675 divided by 1600 type ALL expressions which show José's problem.
irina1246 [14]
(A.675 divided by 1600), and (B.675/1600) are the answers
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3 years ago
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A mobile phone carrier charges an extra fee if a customer uses more than 500 minutes each month. Tom has already used 230 minute
tigry1 [53]
500<span> ≤ 230+x


x will be the amount of minutes Tom has left.</span>
3 0
3 years ago
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