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djverab [1.8K]
2 years ago
15

A tank is filled with 20 litres of water. Half the water is removed and replaced with anti freece and thoroughly mixed. Half thi

s mixture is then removed and replaced with anti-freeze This process continues. find the first five terms in the sequence of amounts of water in the tank at each stage​
Mathematics
1 answer:
mr Goodwill [35]2 years ago
7 0

Answer:

The sequence of amountof water is geometric(first term

20, commonratio

0.5) but the se-quence of antifreezeis not (a geometricsequence cannot havezero as its first termsince that would implythat all terms are zero).

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floor 14

Step-by-step explanation:

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Answer:

2): (b) 80

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3 years ago
What are the values of m and n in the matrix addition below?
Aleksandr [31]

ANSWER

The correct answer is m=45,n=12.

<u>EXPLANATION</u>

We were given the matrix equation;

\left[\begin{array}{cc}n-1&6\\-19&m+3\end{array}\right] +\left[\begin{array}{cc}-1&0\\16&-8\end{array}\right] =\left[\begin{array}{cc}10&6\\-3&40\end{array}\right].


We must first simplify the Left Hand Side of the equation by adding corresponding entries.


\left[\begin{array}{cc}n-1+-1&6+0\\-19+16&m+3-8\end{array}\right]=\left[\begin{array}{cc}10&6\\-3&40\end{array}\right].


That is;


\left[\begin{array}{cc}n-2&6\\-3&m-5\end{array}\right]=\left[\begin{array}{cc}10&6\\-3&40\end{array}\right].

Since the two matrices are equal, their corresponding entries are also equal. we equate corresponding entries and solve for m and n.


This implies that;


n-2=10


We got this equation from row one-column one entry of both matrices.


n=12


Also, the row three-column three entries of both matrices will give us the equation;


m-5=40


m=45


Hence the correct answer is m=45,n=12.


The correct option is option 2






6 0
3 years ago
Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio
JulijaS [17]

Answer:

The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step explanation:

Given

Initial =1\ million

6\ years\ later = 500,000

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

t = Time

A = Amount left in the well

So:

\frac{dA}{dt} = kA

Where k represents the constant of proportionality

\frac{dA}{dt} = kA

Multiply both sides by dt/A

\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}

\frac{dA}{A}  = k\ dt

Integrate both sides

\int\ {\frac{dA}{A}  = \int\ {k\ dt}

ln\ A = kt + lnC

Make A, the subject

A = Ce^{kt}

t = 0\ when\ A =1\ million i.e. At initial

So, we have:

A = Ce^{kt}

1000000 = Ce^{k*0}

1000000 = Ce^{0}

1000000 = C*1

1000000 = C

C =1000000

Substitute C =1000000 in A = Ce^{kt}

A = 1000000e^{kt}

To solve for k;

6\ years\ later = 500,000

i.e.

t = 6\ A = 500000

So:

500000= 1000000e^{k*6}

Divide both sides by 1000000

0.5= e^{k*6}

Take natural logarithm (ln) of both sides

ln(0.5) = ln(e^{k*6})

ln(0.5) = k*6

Solve for k

k = \frac{ln(0.5)}{6}

k = \frac{-0.693}{6}

k = -0.1155

Recall that:

\frac{dA}{dt} = kA

Where

\frac{dA}{dt} = Rate

So, when

A = 600000

The rate is:

\frac{dA}{dt} = -0.1155 * 600000

\frac{dA}{dt} = -69300

<em>Hence, the amount of oil was decreasing at 69300 barrels, yearly</em>

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2 years ago
Given 6x+5y=12 find the equation of the line parallel to it passing through (1,10).
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Answer:

y = -6/5x+56/5

Step-by-step explanation:

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