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3 years ago

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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3 years ago

What is the measure of

Morgarella [4.7K]

Answer:

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Step-by-step explanation:

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3 years ago

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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3 years ago

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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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3 years ago

Read 2 more answers

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

Evaluate the function​

Evaluate the function