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RUDIKE [14]
2 years ago
11

2x+2y=6 3x-y=5 equivalent equation

Mathematics
1 answer:
galina1969 [7]2 years ago
4 0

Answer:

x = 2, y = 1

Step-by-step explanation:

<u>Given:</u>

\begin{bmatrix}2x+2y=6\\ 3x-y=5\end{bmatrix}

<u>Solve:</u>

\mathrm{Isolate\;x\;for\;2x+2y=6:x=3-y}

\mathrm{Substitute\:}

\begin{bmatrix}3\left(3-y\right)-y=5\end{bmatrix}

\mathrm{Simplfy}

\begin{bmatrix}9-4y=5\end{bmatrix}

\mathrm{Isolate\;y\;for\;9-4y=5:y=1}

\mathrm{For\:}x=3-y\;\mathrm{Substitute\:}y=1

x=3-1

\mathrm{Simplify}

x=2

\mathrm{The\:Answers\:to\:the\:system\:of\:equations\:are:}

x=2,\:y=1

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

<u>Check Answer:</u>

Since x =2 , y =1

2(2)+2(1)

4 + 2

= 6

<u><em>True</em></u>

3(2) - (1)

6 -1

5

<u><em>True</em></u>

Hence, x = 2, y = 1

<u><em>~Lenvy~</em></u>

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A florist delivers flowers to anywhere in town. D is the distance from the delivery address to the florist shop in miles. The co
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Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th
trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

    t = 20: 999 fishes

    t = 30: 1168 fishes

(c)

P(\infty) = 1200

Step-by-step explanation:

Given

P(t) =\frac{d}{1+ke^-{ct}}

d = 1200\\k = 11\\c=0.2

Solving (a): Fishes at t = 0

This gives:

P(0) =\frac{1200}{1+11*e^-{0.2*0}}

P(0) =\frac{1200}{1+11*e^-{0}}

P(0) =\frac{1200}{1+11*1}

P(0) =\frac{1200}{1+11}

P(0) =\frac{1200}{12}

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Solving (a): Fishes at t = 10, 20, 30

t = 10

P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483

t = 20

P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999

t = 30

P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168

Solving (c): \lim_{t \to \infty} P(t)

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of e^{-ct} decreases

This implies that:

{t \to \infty} = {e^{-ct} \to 0}

So:

The value of P(t) for large values is:

P(\infty) = \frac{1200}{1 + 11 * 0}

P(\infty) = \frac{1200}{1 + 0}

P(\infty) = \frac{1200}{1}

P(\infty) = 1200

5 0
3 years ago
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