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BARSIC [14]
3 years ago
10

What are the solutions to 2(x-6)2 = 50 ?

Mathematics
2 answers:
yuradex [85]3 years ago
4 0

Answer:

2(x-6)=50

x-6=50/2

x-6=25

x=31

Evgesh-ka [11]3 years ago
4 0
The answer to this question is X=31
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Given the graph of the function F(x) below, what happens to F(x) when x is a
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c

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From the graph as x gets larger and becomes positive f(x) approaches 0 so the answer is C.

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Please help<br> What does X=??
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Sara and Collin's ages add up to 85. Sara is ten years older than foustimes
malfutka [58]

Answer:

Collin is 37.5 and Sara is 47.5

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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&#10;\end{aligned}

Integrate both sides of the equation.

   \begin{aligned}&#10;\int \dfrac{dP}{M-P} &= \int k\, dt \\&#10;-\ln|M-P| &= kt + C \\&#10;\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} \\ &#10;M - P &= \pm e^{-kt - C} &#10; \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}&#10; M - P &= \pm e^{-kt - C} \\&#10;M - P &= \pm e^{- C - kt} \\ &#10;M - P &= \pm e^{- C + (- kt)} \\ &#10;M - P &= \pm e^{- C} \cdot e^{- kt} \\ &#10;M - P &= Ke^{- kt} && (\text{\footnotesize Let $K = \pm e^{-kt}$ }) \\ &#10;M &= Ke^{- kt} + P\\&#10;P &= M - Ke^{- kt}&#10;\end{aligned}

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

   \begin{aligned} &#10;0 &= M - Ke^{- k\cdot 0} \\&#10;0 &= M - K \cdot 1 \\&#10;M &= K&#10; \end{aligned}&#10;

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

   P = M - Me^{-kt}
6 0
3 years ago
Read 2 more answers
I WILL MAKE YOU THE BRAINLIEST
GaryK [48]

Answer: The first number line.




7 0
3 years ago
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