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enot [183]
3 years ago
7

Solve the equation. 2x+4+8x=24

Mathematics
2 answers:
evablogger [386]3 years ago
6 0

Answer:

\boxed{\boxed{\sf x=2}}

Step-by-step explanation:

\sf 2x+4+8x=24

Combine like terms:

  • \sf 2x+8x+4=24
  • \sf 10x+4=24

Subtract 4 from both sides:

  • \sf 10x+4-4=24-4
  • \sf 10x=20

Divide both sides by 10:

  • \sf \cfrac{10x}{10}=\cfrac{20}{10}
  • \sf x=2

____________________________

Rasek [7]3 years ago
3 0

2x + 4 + 8x = 24

2x + 8x = 24 - 4

10x = 20

x = 20 ÷ 10

x = 2

____

Hope it helps!

꧁✿ ᴿᴬᴵᴺᴮᴼᵂˢᴬᴸᵀ2222 ✬꧂ ❤

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Answer:

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Step-by-step explanation:

For this case we have the following differential equation:

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\frac{dp}{p-820}= \frac{1}{2} dt

If we integrate both sides we have:

ln|P-820|= \frac{1}{2}t +c

Using exponential on both sides we got:

P= 820 + P_o e^{1/2t}

Part a

For this case we know that p(0) = 770 so we have this:

770 = 820 + P_o e^0

P_o = -50

So then our model would be given by:

P(t) = -50e^{1/2t} +820

And if we want to find at which time the population would be extinct we have:

0=-50 e^{1/2 t} +820

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Using natural log on both sides we got:

ln(\frac{82}{5}) = \frac{1}{2}t

And solving for t we got:

t = 2 *ln(\frac{82}{5}) =5.595

Part b

For this case we know that p(0) = p0 so we have this:

p_0 = 820 + P_o e^0

P_o = p_0 -820

So then our model would be given by:

P(t) = (p_o -820)e^{1/2t} +820

And if we want to find at which time the population would be extinct we have:

0=(p_o -820)e^{1/2 t} +820

-\frac{820}{p_0 -820} = e^{1/2 t}

Using natural log on both sides we got:

ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t

And solving for t we got:

t = 2 *ln(-\frac{820}{p_0 -820})

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

12 = 2 *ln(\frac{820}{820-p_0})

6 = ln (\frac{820}{820-p_0})

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e^6 = \frac{820}{820-p_0}

(820-p_0) e^6 = 820

820-p_0 = \frac{820}{e^6}

p_0 = 820-\frac{820}{e^6}

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I would use this website. Hope it helps!
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