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Nadusha1986 [10]
3 years ago
9

Find the TWO integers whos product is -12 and whose sum is 1m​

Mathematics
2 answers:
svetoff [14.1K]3 years ago
5 0

Answer:

\rm Numbers = 4 \ and \ -3.

Step-by-step explanation:

Given :-

The sum of two numbers is 1 .

The product of the nos . is 12 .

And we need to find out the numbers. So let us take ,

First number be x

Second number be 1-x .

According to first condition :-

\rm\implies 1st \ number * 2nd \ number= -12\\\\\rm\implies x(1-x)=-12\\\\\rm\implies x - x^2=-12\\\\\rm\implies x^2-x-12=0\\\\\rm\implies x^2-4x+3x-12=0\\\\\rm\implies x(x-4)+3(x-4)=0\\\\\rm\implies (x-4)(x+3)=0\\\\\rm\implies\boxed{\red{\rm x = 4 , -3 }}

Hence the numbers are 4 and -3.

irga5000 [103]3 years ago
3 0

Answer:

-3,4 are the TWO integers whos product is -12 and whose sum is 1.

Step-by-step explanation:

<em>-3</em><em>×</em><em>4</em><em>=</em><em>1</em><em>2</em>

<em>-3</em><em>+</em><em>4</em><em>=</em><em>1</em>

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A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the
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Answer:

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

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

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

Step-by-step explanation:

For this case we have the following differential equation:

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

And if we rewrite the expression we got:

\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:

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And solving for t we got:

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

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(820-p_0) e^6 = 820

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

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