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Kamila [148]
3 years ago
15

Never mind don't need help

Mathematics
1 answer:
madreJ [45]3 years ago
6 0

Answer:

thanks for the points

Step-by-step explanation:

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Please delete my question?
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Multiply: <br>(2sin B+cos B)by 3cosec B.secB​
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Factor each expression.<br> 8.) 8 + 16=
sveta [45]

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24

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4 0
3 years ago
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What is -212+(-102)?​
hram777 [196]

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-212+(-102) = -212-102

= -314

Hope this helps!

4 0
3 years ago
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A species of beetles grows 32% every year. Suppose 100 beetles are released into a field. How many beetles will there be in 10 y
siniylev [52]

Given that a species of beetles grows 32% every year.

So growth rate is given by

r=32%= 0.32


Given that 100 beetles are released into a field.

So that means initial number of beetles P=100


Now we have to find about how many beetles will there be in 10 years.

To find that we need to setup growth formula which is given by

A=P(1+r)^n where A is number of beetles at any year n.

Plug the given values into above formula we get:

A=100(1+0.32)^n

A=100(1.32)^n


now plug n=10 years

A=100(1.32)^{10}=100(16.0597696605)=1605.97696605

Hence answer is approx 1606 beetles will be there after 20 years.


Now we have to find about how many beetles will there be in 20 years.

To find that we plug n=20 years

A=100(1.32)^{20}=100(257.916201549)=25791.6201549

Hence answer is approx 25791 beetles will be there after 20 years.



Now we have to find time for 100000 beetles so plug A=100000

A=100(1.32)^n

100000=100(1.32)^n

1000=(1.32)^n

log(1000)=n*log(1.32)

\frac{\log\left(10000\right)}{\log\left(1.32\right)}=n

33.174666862=n

Hence answer is approx 33 years.

4 0
3 years ago
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