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Ksju [112]
3 years ago
6

Evaluate the function​

Mathematics
1 answer:
Ira Lisetskai [31]3 years ago
3 0

Answer:

y = 20

Step-by-step explanation:

take the 2's multiply them =4 (2 x 2 = 4) so 5 x 4 = 20

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Find the quotient: 3/5 ÷ 1 /2 <br><br> Can someone pls explain??
ANEK [815]

Answer:

2/5

Step-by-step explanation:

expressing 1 1/2 in an improper fraction:

1 1/2 = 3/2

Hence

(3/5) ÷ (1 1/2)    

= (3/5) ÷ (3/2)   (convert divide to multiply by flipping divisor fraction)

= (3/5) x (2/3)

= (3 x 2) / (5 x 3)

= 6/15 (divide both numerator and denominator by 3)

= 2/5  

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What is the measure of
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Answer:

measure of what

Step-by-step explanation:

let me know

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Hannah has a small circular table. The radius of the tabletop is 23 cm.
ale4655 [162]
Area of circle = πr2 = π(23)^2
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Choose the equation of the horizontal line that passes through the point (−5, 9).
shtirl [24]
The correct answer is:  [B]:  " y = 9 " .
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6 0
3 years ago
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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
ohaa [14]

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

\frac{820}{50} = e^{1/2 t}

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})

Using exponentials we got:

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}

8 0
3 years ago
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