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3 years ago

Is 0.9 0.06 11/20 0.21 greater than 1/2 or smaller

Mathematics

2 answers:

cupoosta [38]3 years ago

3 0

1/2 = 0.5

0.9 is greater

0.06 is smaller

11/20 is greater

0.21 is smaller

olganol [36]3 years ago

3 0

1/2=.5
0.9 >.5
0.06 <.5
11/20=.55 >.5

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rebecca wants to crochet a scarf that is at least 4.4 feet long. So far, she's completed 2.5 feet. How much more does Rebecca in

vlabodo [156]

Answer:

4.4 < 2.5 +x

She needs to crochet at least 1.9 more ft

Step-by-step explanation:

She wants the scarf to be at least 4.4 ft long

4.4< scarf

The scarf is 2.5 ft long now

She will crochet x more ft

4.4 < 2.5 +x

Subtract 2.5 from each side

4.4 -2.5 < 2.5+x-2.5

1.9 <x

She needs to crochet at least 1.9 more ft

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3 years ago

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

Diana's parents keep a chart on the wall where they mark how much Diana grows each year. When Diana was 15, she grew 5/6 of an i

otez555 [7]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Someone please help me!!

Nata [24]

Answer:

its a for 6, b for 7, and a for 8

Step-by-step explanation:

heress how i did it ₙₓₙ⁺⁽°⁴₊₋²₆ÜààÆŒ

8 0

3 years ago

A communications company offers a variety of calling card options. Card A has a 30¢ connection fee and then costs 2¢ per minute.

arlik [135]

Answer:

The length of the call that would cost the same with both cards is 5 minutes.

Step-by-step explanation:

Hi there!

The cost with card A can be expressed as follows:

cost A = 30 + 2 · m

Where "m" is the length of the call in minutes.

In the same way, the cost of card B will be:

cost B = 10 + 6 · m

Where "m" is the length of the call in minutes.

We have to find the value of "m" for which the call would cost the same with both cards.

Then:

cost A = cost B

30 + 2 · m = 10 + 6 · m

Subtract 10 and 2 · m to both sides of the equation:

30 - 10 = 6 · m - 2 · m

20 = 4 · m

Divide by 4 both sides of the equation:

20/4 = m

5 = m

The length of the call that would cost the same with both cards is 5 minutes.

Have a nice day!

3 0

4 years ago