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leva [86]
4 years ago
13

How many 1.1 centimeter lengths can be made from a string that is 5.093 centimeters long? Enter a numerical answer only.

Mathematics
1 answer:
Alexandra [31]4 years ago
5 0
The number of 1.1 cm in 5.093 cm can be calculated through finding the ratio of:
(longer length)/(shorter length)
longer length=5.093 cm
shorter length=1.1 cm
thus the ratio will be:
5.093/1.1
=4.63
This means that 1.1 cm length will go into 5.093 cm length by a factor of 4.63

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B. Solve for the missing side.<br> 5<br> 43<br> Y
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Answer:

My answer is 215

Step-by-step explanation:

I multiply the 43 by 5 and get 215 for my answer

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3 years ago
for employees of Papa Tony's pizza or cleaning up at the end of a busy night. There is a list of 43 complete a task that need to
nydimaria [60]
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4 years ago
Write the number as a question of integers to show that it is rational.<br> Please reply quickly!
snow_lady [41]

Answer:

\frac{-37}{99}

Step-by-step explanation:

0.555555... becomes \frac{5}{9}

Using this logic, we can do the following:

-0.37373737... = \frac{-37}{99}

Have a nice day!

     I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.

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3 0
3 years ago
Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th
trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

    t = 20: 999 fishes

    t = 30: 1168 fishes

(c)

P(\infty) = 1200

Step-by-step explanation:

Given

P(t) =\frac{d}{1+ke^-{ct}}

d = 1200\\k = 11\\c=0.2

Solving (a): Fishes at t = 0

This gives:

P(0) =\frac{1200}{1+11*e^-{0.2*0}}

P(0) =\frac{1200}{1+11*e^-{0}}

P(0) =\frac{1200}{1+11*1}

P(0) =\frac{1200}{1+11}

P(0) =\frac{1200}{12}

P(0) = 100

Solving (a): Fishes at t = 10, 20, 30

t = 10

P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483

t = 20

P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999

t = 30

P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168

Solving (c): \lim_{t \to \infty} P(t)

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of e^{-ct} decreases

This implies that:

{t \to \infty} = {e^{-ct} \to 0}

So:

The value of P(t) for large values is:

P(\infty) = \frac{1200}{1 + 11 * 0}

P(\infty) = \frac{1200}{1 + 0}

P(\infty) = \frac{1200}{1}

P(\infty) = 1200

5 0
3 years ago
What are si units of measure for length mass volume denstiy time and temperature?
natali 33 [55]
Hi !

International System Units : 

Lenght : meter (m)
Mass : kilogram (kg)
Volume : cubic meter (m³)
Density : kilogram per cubic meter (kg/m³)
Time : seconds (s)
Temperature : kelvin (K)

4 0
4 years ago
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