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Anit [1.1K]
3 years ago
10

The midpoints of an irregular quadrilateral ABCD are connected to form another quadrilateral inside ABCD. Complete the explanati

on of why the quadrilateral is a parallelogram.​
Mathematics
1 answer:
Juliette [100K]3 years ago
5 0

Answer:

Suppose: M, N, P, Q are the midpoints of AB, BC, CD, AD respectively

=> MNPQ is the quadrilateral inside ABCD

connect B to D, ΔABD has : M is the midpoint of AB

                                               Q is the midpoint of AD

=> MQ is the midpoint polygon of ΔABD

=> MQ // BD and MQ = 1/2.BD (1)

ΔBCD has: N is the midpoint of BC

                   P is the midpoint of DC

=> NP is the midpoint polygon of ΔBCD

=> NP // BD and NP = 1/2.BD    (2)

from (1) and (2) => MQ // NP ( //BD)

                             MQ = NP  (=1/2.BD)

=> MNPQ is a parallelogram.​

=>  the quadrilateral inside ABCD is a parallelogram.​

Step-by-step explanation:

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Exponential function representing final amount with compound interest compounded continuously,

A=Pe^{rt}

Here, A = Final amount

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For P = $9600

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By substituting these values in the formula,

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t = \frac{0.693147}{0.06}

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Any amount will get doubled (with the same rate of interest and duration of investment) in the same time.

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Step-by-step explanation:

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Litter such as leaves falls to the forest floor, where the action of insects and bacteria initiates the decay process. Let A be
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Answer:

D = L/k

Step-by-step explanation:

Since A represents the amount of litter present in grams per square meter as a function of time in years, the net rate of litter present is

dA/dt = in flow - out flow

Since litter falls at a constant rate of L  grams per square meter per year, in flow = L

Since litter decays at a constant proportional rate of k per year, the total amount of litter decay per square meter per year is A × k = Ak = out flow

So,

dA/dt = in flow - out flow

dA/dt = L - Ak

Separating the variables, we have

dA/(L - Ak) = dt

Integrating, we have

∫-kdA/-k(L - Ak) = ∫dt

1/k∫-kdA/(L - Ak) = ∫dt

1/k㏑(L - Ak) = t + C

㏑(L - Ak) = kt + kC

㏑(L - Ak) = kt + C'      (C' = kC)

taking exponents of both sides, we have

L - Ak = e^{kt + C'} \\L - Ak = e^{kt}e^{C'}\\L - Ak = C"e^{kt}      (C" = e^{C'} )\\Ak = L - C"e^{kt}\\A = \frac{L}{k}  - \frac{C"}{k} e^{kt}

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A = \frac{L}{k}  - \frac{L}{k} e^{kt}

So, D = R - A =

D = \frac{L}{k} - \frac{L}{k}  - \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{kt}

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D = \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{k0}\\D = \frac{L}{k} e^{0}\\D = \frac{L}{k}

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