Ask question

Ask question

All categories

3 years ago

10

PLEASE HELP ASAP!!!!!!!

1 answer:

natita [175]3 years ago

8 0

Answer:

142

Step-by-step explanation:

You might be interested in

"Rebecca walks 100 feet in a straight line. She then turns 20 degrees to her left and walks another 100 feet, and then turns 20

iragen [17]

Answer:

1,800 ft

Step-by-step explanation:

Rebecca will walk in the pattern of a regular polygon with 'n' sides and internal angles of 160 degrees (180 at a straight line - 20 degrees from each turn).

The equation that describes the internal angle of a regular polygon is:

$A = \frac{(n-2)*180}{n}$

For A = 160 degrees:

$160n=(n-2)*180\\(180-160)n=360\\n=18\ sides$

If each side is 100 ft long, the total distance that Rebecca has walked is:

$d = 18*100\\d=1,800\ ft$

She walked 1,800 ft.

5 0

3 years ago

Find the area of the circle. Leave your answers in terms of pi.

Savatey [412]

Answer:

49pi m^2

Step-by-step explanation:

Area of circle=pi r^2

D=2r=14

r=7

Area of circle =7^2pi=49pim^2

3 0

3 years ago

Read 2 more answers

Am I the only one whose parents yell their name and when you tell back yes mom they don’t answer?

arsen [322]

No your not the only one

6 0

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

. If f(x) = 3x - x, what is the value of f(7 - 2)?

kifflom [539]

Answer:

10

Step-by-step explanation:

f(x) =3x-x=2x,

f(7-2)=f(5)= 2*5=10

4 0

3 years ago