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miskamm [114]
3 years ago
14

The area of rectangle ABCD is 28 cm^2

Mathematics
1 answer:
Sonbull [250]3 years ago
8 0

The length of AB is (x + 4) cm and the values of a and be are a = 10, b = 4

A rectangle is a quadrilateral in which opposite sides are equal and parallel to each other. The area of a rectangle is:

Area = length * width

From the image:

Length of AB = x + 4

Length of BC = x + 6

The area of rectangle ABCD = Length of AB * Length of BC

28 = (x + 4)(x + 6)

x² + 10x + 24 = 28

x² + 10x = 4

Comparing with x² + ax = b gives:

a = 10, b = 4

Therefore the length of AB is (x + 4) cm and the values of a and be are a = 10, b = 4

Find out more at: brainly.com/question/15019502

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I assume you mean

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   \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} \\ 
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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}
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